NCERT Solutions for Class 10 Maths Chapter 4 - Quadratic Equations

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Exercise 4.1

1. Check whether the following are quadratic equations :
(i) (x + 1)² = 2(x – 3)
(ii) x² – 2x = (– 2)(3 – x)
(iii) (x – 2)(x + 1) = (x – 1)(x + 3)
(iv) (x – 3)(2x + 1) = x(x + 5)
(v) (2x – 1)(x – 3) = (x + 5)(x – 1)
(vi) x² + 3x + 1 = (x – 2)²
(vii) (x + 2)³ = 2x(x² –1)
(viii) x³ – 4x² – x + 1 = (x – 2)³

Sol. (i) Given equation is
(x + 1)² = 2(x – 3)
⇒ x² + 2x + 1 = 2x – 6
[... (a + b)² = a² + 2ab + b²]
⇒ x² + 1 + 6 = 0
⇒ x² + 7 = 0
which is of the form ax²+ bx + c = 0, where a ≠ 0 and b = 0.
Thus, (x + 1)² = 2(x – 3) is a quadratic equation.

(ii) Given equation is
x² – 2x = (– 2)(3 – x)
⇒ x² – 2x = – 6 + 2x
⇒ x² – 2x – 2x + 6 = 0
⇒ x² – 4x + 6 = 0
which
is of the form ax² + bx + c = 0, where
a ≠ 0.
Thus, x² – 2x = (– 2)(3 – x) is a quadratic equation.

(iii) Given equation is
(x – 2)(x + 1) = (x – 1)(x + 3)
⇒ x² – 2x + x – 2 = x² – x + 3x – 3
⇒ x² – 2x + x – 2 – x² + x – 3x + 3 = 0
⇒ – 3x + 1 = 0
which is not of the form ax² + bx + c = 0, as
a = 0 and b and c are any real numbers.
It is not a quadratic equation.

(iv) Given equation is
(x – 3)(2x + 1) = x(x + 5)
⇒ 2x² – 6x + x – 3 = x² + 5x
⇒ 2x² – 6x + x – 3 – x² – 5x = 0
⇒ x² – 10x – 3 = 0
which
is of the form ax² + bx + c = 0, where
a ≠ 0
Thus, x² – 10x – 3 = 0 is a quadratic equation.

(v) Given equation is
(2x – 1)(x – 3) = (x + 5)(x – 1)
⇒ 2x² – x – 6x + 3 = x² + 5x – x – 5
⇒ 2x² – x – 6x + 3 – x² – 5x + x + 5 = 0
⇒ x² – 11x + 8 = 0
which
is of the form ax² + bx + c = 0, where
a ≠ 0
Thus, x² – 11x + 8 = 0 is a quadratic equation.

(vi) Given equation is
x₂ + 3x + 1 = (x – 2)²
⇒ x² + 3x + 1 = x² + 4 – 4x
[... (a – b)² = a² – 2ab + b²]
⇒ x² + 3x + 1 – x² – 4 + 4x = 0
⇒ 7x – 3 = 0
which is not of the form ax² + bx + c = 0, as
a = 0 and b, c are any real numbers.
It is not a quadratic equation.

(vii) Given equation is
(x + 2)³ = 2x(x² – 1)
⇒ x³ + 8 + 3x²(2) + 3x(2)² = 2x³ – 2x
[... (a + b)³ = a³ + b³ + 3ab² + 3a²b]
⇒ x³ + 8 + 6x² + 12x = 2x³ – 2x
⇒ x³ + 8 + 6x² + 12x – 2x³ + 2x = 0
⇒ – x³ + 6x² + 14x + 8 = 0
which is not of the form ax² + bx + c = 0, as it includes the term with coefficient x³.
Thus,
(x + 2)³ = 2(x² – 1) is not a quadratic equation.

(viii) Given equation is
x³ – 4x² – x + 1 = (x – 2)³
⇒ x³ – 4x² – x + 1 = x³ – 3(x²) 2 + 3x(2)² – (2)³
[... (a – b)³ = a³ – b³ + 3ab² – 3a²b]
⇒ x³ – 4x² – x + 1 = x³ – 6x² + 12x – 8
⇒ x³ – 4x² – x + 1 – x³ + 6x² – 12x + 8 = 0
⇒ 2x² – 13x + 9 = 0
which
is of the form ax² + bx + c = 0, where
a ≠ 0.
Thus, x³ – 4x² – x + 1 = (x – 2)³ is a quadratic equation.

2. Represent the following situations in the form of quadratic equations
(i) The area of a rectangular plot is 528 m². The length of the plot (in metre) is one more than twice its breadth. We need to find the length and breadth of the plot.
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
(iii) Rohan's mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan's present age.
(iv) A train travels a distance of 480 km at an uniform speed. If the speed had been 8 kmh^–1 less, than it would have taken 3 hours more to cover the same distance.We need to find the speed of the train.

Sol. (i) Let the breadth of the plot = x m
Then, the length of the plot = (2x + 1) m
(By condition)
Area of the rectangular plot = 528 m²
(Given)
Area of the rectangular plot
= Length × Breadth
(2x + 1)x = 528
⇒ 2x² + x = 528
2x² + (33 – 32)x – 528 = 0
2x² – 32x + 33x – 528 = 0
Solve the quadratic equation, we get
2x(x – 16) + 33(x – 16) = 0
(x – 16)(2x + 33) = 0
x = 16 or -33/2
Since, the breadth of the rectangular plot can't be negative' the breadth will be 16 m.
Now, the length of the plot = (2x + 1) m
Here, x = 16 m
= 2 × 16 + 1
= 33 m
Thus,
the breadth of the plot is 16 m and length of the plot is 33.

(ii) Let the two consecutive positive integers be x and x + 1.
Then, according to the question, Product of two consecutive integers = 306
⇒ x(x + 1) = 306
⇒ x² + x = 306
⇒ x² + x – 306 = 0
Now, solve the quadratic equation, we get
⇒ x² + x – 306 = 0
⇒ x² + 18x – 17x – 306 = 0
⇒ x(x + 18) – 17(x + 18) = 0
⇒ (x – 17)(x + 18) = 0
x = 17 and – 18
Here, we take positive integer, so the positive integer is 17.

(iii) Let the age of Rohan = x years
Then, his mother's age = (x + 26) year
After three years,
Rohan's age = (x + 3)years
Rohan's mother's age = [(x + 26) + 3] years
= (x + 29)yr
According to the questions
(x + 3)(x + 29) = 360
⇒ x² + 29x + 3x + 87 = 360
⇒ x² + 32x – 273 = 0
On comparing with ax² + bx + c = 0, we get
a = 4, b =√4, c = 3
Factorization of the above quadratic equation using quadratic formula.
$$x=\frac{-b±\sqrt{b^2-4ac}}{2a}\\\text{Put a = 1, b = 32 and c = – 273,}\\x=\frac{-32±\sqrt{32^2-4×1×(-273)}}{2×1}\\x=\frac{-32±\sqrt{2116}}{2}\\=\frac{-32±46}{2}\\x=\frac{14}{2}=7\\\text{Taking negative part}\\x=\frac{-32-46}{2}\\=\frac{-78}{2}\\= – 39\\\text{Since age can't be negative.}\\\text{Hence, Rohan's age is 7 years.}$$

(iv) Let the speed of the train = x km/h
Distance travelled by the train = 480 km
Therefore, time taken for travelling 480 km
$$=\frac{480}{x}h (speed=\frac{Distance}{Time})\\\text{If the speed had been 8 km/h less i.e. (x – 8) km/h, then time taken for travelling}\\480 km=\frac{480}{x-8}hr\\\text{According to the question,}\\\frac{480}{x-8}-\frac{480}{x}=3\\⇒\frac{480x-480(x-8)}{x(x-8)}=3\\⇒ 480x – 480x + 3840 = 3x(x – 8)\\⇒ 3840 = 3x(x – 8)\\⇒ 1280 = x(x – 8)\\⇒ 1280 = x^2 – 8x\\⇒ x^2 – 8x – 1280 = 0\\\text{Thus, above equation represents the required quadratic equation. Now, solve the quadratic equation by quadratic formula.}\\x=\frac{-b±\sqrt{b^2-4ac}}{2×1}\\x=\frac{8±\sqrt{8^2-4×1×(-1280)}}{2a}\\x=\frac{8±\sqrt{64+5120}}{2}\\x=\frac{8±\sqrt{5184}}{2}\\=\frac{8±\sqrt{72}}{2}\\\text{By solving, we get}\\\text{x = 40 and – 32}\\\text{Speed cannot be negative.}\\\text{Thus, the speed of the train is 40 km/h.}$$

Exercise 4.2

1. Find the roots of the following quadratic equations by factorisation

$$(i) x^3 – 3x – 10 = 0\\(ii) 2x^2 + x – 6 = 0\\(iii)\sqrt{2}x^2+7x+5\sqrt{2}=0\\(iv) 2x^2-x+\frac{1}{8}=0\\(v) 100x2 – 20x + 1 = 0$$

Sol. (i) Given equation is
x² – 3x – 10 = 0
on splitting the middle term,
⇒ x² – (5x – 2x) – 10 = 0
⇒ x² – 5x + 2x – 10 = 0
⇒ x(x – 5) + 2(x – 5) = 0
⇒ (x + 5)(x + 2) = 0
Now, x – 5 = 0 ⇒ x = 5
and x + 2 = 0 ⇒ x = – 2
Hence, the roots of the equation x² – 3x – 10 = 0 are – 2 and 5.

(ii) Given equation is
2x² + x – 6 = 0
on splitting the middle term, we get
⇒ 2x² + 4x – 3x – 6 = 0
⇒ 2x(x + 2) – 3(x + 2) = 0
⇒ (x + 2)(2x – 3) = 0
Now,
x + 2 = 0 ⇒ x = – 2
and 2x – 3 = 0 ⇒ x = 32
Hence, the roots of the equation 2x² + x – 6 = 0 are – 2 and 3/2

(iii) Given equation is

$$\sqrt{2}x^2+7x+5\sqrt{2}=0\\\text{On splitting the middle term,}\\\sqrt{2}x^2+(5x+2x)+5\sqrt{2}=0\\⇒\sqrt{2}x^2+5x+2x+5\sqrt{2}=0\\⇒x(\sqrt{2}x+5)+\sqrt{2}(\sqrt{2}x+5)=0\\⇒(\sqrt{2}x+5)(x+\sqrt{2})\\\text{Now,}\sqrt{2}x+5=0⇒x=-\frac{5}{\sqrt{2}}\\\text{and}\space x+\sqrt{2}=0⇒x = −\sqrt{2}\\\text{Hence, the roots of the equation}\\\sqrt{2}x^2+7x+5\sqrt{2}=0\space\text{are}-\frac{5}{\sqrt{2}}\space\text{and}-\sqrt{2}$$

(iv) Given equation is
$$2x^2-x+\frac{1}{8}=0\\\text{On multiplying both sides by 8, we get}\\⇒ 16x^2 – 8x + 1 = 0\\\text{On splitting the middle term,}\\⇒ 16x^2 – (4x + 4x) + 1 = 0\\⇒ 16x^2 – 4x – 4x + 1 = 0\\⇒ 4x(4x – 1) – 1(4x – 1) = 0\\⇒ (4x – 1)(4x – 1) = 0\\Now, 4x – 1 = 0 ⇒ x =\frac{1}{4}\\and, 4x – 1 = 0 ⇒ x =\frac{1}{4}\\\text{Hence, the roots of the equation}\\2x^2-x+\frac{1}{8}=0\space\text{are}\frac{1}{4}\text\space{and}\frac{1}{4}$$

(v) Given equation is
100x² – 20x + 1 = 0
On splitting the middle term, we get
⇒ 100x² – (10x + 10x) + 1 = 0
⇒ 100x² – 10x – 10x + 1 = 0
⇒ 10x(10x – 1) – 1(10x – 1) = 0
⇒ (10x – 1)(10x – 1) = 0
Now, 10x – 1 = 0
$$⇒x=\frac{1}{10}\\\text{and} 10x – 1 = 0 ⇒ x =\frac{1}{10}\\\text{Hence, the roots of the equation}\\100x^2 – 20x + 1 = 0 \text{are}\frac{1}{10}\text{and}\frac{1}{10}$$

2. Solve the following quadratic equations
(i) x2 – 45x + 324 = 0
(ii) x2 – 55x + 750 = 0

Sol. (i) Given equation is
x² – 45x + 324 = 0
On splitting the middle term,
x² – (36x + 9x) + 324 = 0
⇒ x² – 36x – 9x + 324 = 0
⇒ x(x – 36) – 9(x – 36) = 0
⇒ (x – 36)(x – 9) = 0
Now, x – 36 = 0 ⇒ x = 36
and x – 9 = 0 ⇒ x = 9
Hence,
the roots of the equation x² – 45x + 324 = 0 are 9 and 36.

(ii) Given equation is
x² – 55x + 750 = 0
On splitting the middle term, we get
⇒ x² – (30x + 25x) + 750 = 0
⇒ x² – 30x – 25x + 750 = 0
⇒ x(x – 30) – 25(x – 30) = 0
⇒ (x – 30)(x – 25) = 0
Now, x – 30 = 0 ⇒ x = 30
and x – 25 = 0 ⇒ x = 25
Hence,
the roots of the equation x² – 55x + 750 = 0 are 25 and 30 .

3. Find two numbers whose sum is 27 and product is 182.

Sol. Let the numbers be x and y.
According to the question,
Sum of the numbers = x + y = 27 ...(i)
Product of the numbers = xy = 182 ...(ii)
$$\text{From equation (ii), put} y = \frac{182}{x}\text{in equation (i),}\\\text{we get}x+\frac{182}{x}=27$$
⇒ x² + 182 = 27x
⇒ x² – 27x + 182 = 0
On splitting the middle term, we get
⇒ x² – (14x + 13x) + 182 = 0
⇒ x² – 14x – 13x + 182 = 0
⇒ x(x – 14) – 13(x – 14) = 0
⇒ (x – 14)(x – 13) = 0
Now,
x – 14 = 0 ⇒ x = 14
and
x – 13 = 0 ⇒ x = 13
x = 13, 14
Putting x = 13 in equation (i), we get y = 14
Putting x = 14 in equation (i), we get y = 13
Therefore, in both cases, the numbers are 13 and 14.

4. Find two consecutive positive integers, sum of whose squares is 365.

Sol. Let the two consecutive positive integers be x and x + 1.
According to the question,
Sum of the squares of two consecutive positive integers = 365
x² + (x + 1)² = 365
⇒ x² + x² + 2x + 1 = 365
[... (a + b)² = a² + 2ab + b²]
⇒ 2x² + 2x – 364 = 0
⇒ x² + x – 182 = 0
On splitting the middle term, we get
⇒ x² + (14x – 13x) – 182 = 0
⇒ x² + 14x –13x – 182 = 0
⇒ x(x + 14) – 13(x + 14) = 0
⇒ (x + 14)(x – 13) = 0
Now,
x + 14 = 0 ⇒ x = – 14
and
x – 13 = 0 ⇒ x = 13
x = – 14, 13
Since, x is a positive integer which cannot be negative, therefore x = 13.
Hence, the two consecutive positive integer are 13 and 14.

5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Sol. Given that, hypotenuse of right angled triangle = 13 cm
Let the base of the right angled triangle = x
According to the question,
Then, altitude of the right angled triangle
= (x – 7) cm
By applying pythagoras theorem,
Hypotenuse² = Base² + Altitude²
⇒ (13)² = x² + (x – 7)²
⇒ 169 = x² + x² – 14x + 49
[... (a – b)² = a² – 2ab + b²]
⇒ 2x² – 14x – 120 = 0
⇒ x² – 7x – 60 = 0
On splitting the middle term,
⇒ x² – (12x – 5x) – 60 = 0
⇒ x² – 12x + 5x – 60 = 0
⇒ x(x – 12) + 5(x – 12) = 0
⇒ (x – 12)(x + 5) = 0
Now,
x – 12 = 0 ⇒ x = 12
and
x + 5 = 0 ⇒ x = – 5
Since, base of the triangle cannot be negativ
e, hence x ≠ – 5.
Hence, base of the
triangle = 12 cm and altitude of the triangle = 12 – 7 = 5 cm.

6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in `) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was ` 90, find the number of articles produced and the cost of each article.

Sol. Let the number of pottery articles produced on a particular day = x
Cost of production of each article = 2x + 3
So, the total cost of production
= Number of pottery articles produced
× Cost of production of each article
= x(2x + 3)
According to the question,
x(2x + 3) = 90
⇒ 2x² + 3x = 90
⇒ 2x² + 3x – 90 = 0
On splitting the middle term,
⇒ 2x² + (15x – 12x) – 90 = 0
⇒ 2x² + 15x – 12x – 90 = 0
⇒ x(2x + 15) – 6(2x + 15) = 0
⇒ (2x + 15)(x – 6) = 0
Now,
2x + 15 = 0 ⇒ x =-15/2
and
x – 6 = 0 ⇒ x = 6
x = −-15/2 and 6
But x cannot be negative i.e., number of pottery articles should be positive.
x = 6
Hence, the number of articles produced is 6 and the cost of each article is
(2 × 6 + 3) = ₹ 15

Exercise 4.3

1. Find the roots of the following quadratic equations, if they exist, by the method of completing the square :
(i) 2x² – 7x + 3 = 0
(ii) 2x² + x – 4 = 0
(iii) 4x²+4√3x+3=0
(iv) 2x² + x + 4 = 0

Sol. (i) Given equation is
2x² – 7x + 3 = 0
On dividing both sides by 2, we get

$$x^2=-\frac{7}{2}x+\frac{3}{2}=0\\\text{On adding}(\frac{1}{2}\text{coefficient of x})^2\\\text{i.e,}(\frac{1}{2}×\frac{7}{2})^2=\frac{49}{16}\space\text{on both sides,we get}\\x^2-\frac{7}{2}x+\frac{49}{16}=-\frac{3}{2}+\frac{49}{16}\\⇒(x-\frac{7}{4})^2=\frac{25}{16}\space[... a^2 + 2ab + b^2 = (a – b)^2]\\⇒(x-\frac{7}{4})^2=(\frac{5}{4})^2\\x-\frac{7}{4}=±\frac{5}{4}\\x=\frac{7}{4}±\frac{5} {4}=\frac{7±5}{4}\\=\frac{12}{4},\frac{2}{4}=3,\frac{1}{2}\\\text{Hence, the roots of equation}2x2 – 7x + 3 = 0 \text{are}\frac{1}{2}\text{and}3.$$

(ii) Given equation is
$$2x^2 + x – 4 = 0\\\text{On dividing both sides by 2, we get}\\⇒ 2x^2 + 4 = 4\\⇒x^2+\frac{1}{2}x=2\\\text{Adding}(\frac{1}{2}\text{coefficient of x)}^2\\i.e,(\frac{1}{2}×\frac{7}{2})^2=\frac{49}{16}\text{on both sides, we get}\\x^2+\frac{1}{2}x+\frac{1}{16}=2+\frac{1}{16}\\⇒x^2+\frac{1}{2}x+\frac{1}{16}=\frac{33}{16}\\⇒(x+\frac{1}{4})^2=(\frac{\sqrt{33}}{4})^2\space[... a^2 + 2ab + b^2 = (a + b)^2]\\⇒x+\frac{1}{4}=±\frac{\sqrt{33}}{4}\\⇒x =-\frac{1}{4}±\frac{\sqrt{33}}{4}=\frac{-1±\sqrt{33}}{4}\\⇒ x =\frac{\sqrt{33-1}}{4},-(\frac{\sqrt{33+1}}{4})\\\text{Hence, the roots of equation} 2x^2 + x – 4 = 0\text{are}-(\frac{\sqrt{33+1}}{4})\text{and}\frac{\sqrt{33-1}}{4}$$

(iii) Given equation is
$$4x^2+4\sqrt{3x}+3=0\\\text{On dividing both sides by 4, we get}\\⇒x^2+\sqrt{3x}+\frac{3}{4}=0\\⇒x^2+\sqrt{3x}=-\frac{3}{4}\\\text{Adding}(\frac{1}{2}\text{coefficient of x)}^2\\\text{i.e.,}(\frac{1}{2}\sqrt{3})^2\\=\frac{3}{4}\text{on both sides, we get}\\x^2+\sqrt{3x}+\frac{3}{4}=-\frac{3}{4}+\frac{3}{4}\\⇒(x+\frac{\sqrt{3}}{2})^2=0\space[... a^2 + 2ab + b^2 = (a + b)^2]\\x+\frac{\sqrt{3}} {2}=0\text{and}\space x+\frac{\sqrt{3}}{2}=0\\⇒x=-\frac{\sqrt{3}} {2}=0\text{and}\space x=- \frac{\sqrt{3}}{2}=0\\\text{Hence, the roots of equation}\space4x^2+4\sqrt{3}x+3\\= 0 \space are −\frac{\sqrt{3}}{2}\text{and}-\frac{\sqrt{3}}{2}$$

(iv) Given equation is
$$2x^2 + x + 4 = 0\\\text{On dividing both sides by 2, we get}\\⇒x^2+\frac{1}{2}x+2=0\\⇒x^2+\frac{1}{2}x=-2\\\text{Adding}(\frac{1}{2}\text{coefficient of x})^2\\\text{i.e.,}(\frac{1}{2}×\frac{1}{2})^2=\frac{1}{16}\text{on both sides, we get}\\x^2+\frac{1}{2}x+\frac{1}{16}=-2+\frac{1}{16}\\(x+\frac{1}{4})^2=-(\frac{\sqrt{31}}{4})^2\\\text{which is not possible as the square of a real number cannot be negative. Therefore, the real roots of the equation}\space2x^2 + x + 4 = 0\space \text{do not exist.}$$

2. Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.

Sol. (i) Given equation is
2x² – 7x + 3 = 0
On comparing with ax2 + bx + c = 0, we get
a = 2, b = – 7 and c = 3
By using quadratic formula, we get
$$x=\frac{-b±\sqrt{b^2-4ac}}{2a}\\x=\frac{-(-7)±\sqrt{(-7)^2-4×2×3}}{2×2}\\x=\frac{7±\sqrt{49-24}}{4}\\x=\frac{7±\sqrt{25}}{4}\\=\frac{7+5}{4}\\=\frac{7+5}{4}\text{and}\frac{7-5}{4}\\\frac{12}{4}\text{and}\frac{2}{4}\\\text{Roots are x = 3 and}\frac{1}{2}$$

(ii) Given equation is
2x² + x – 4 = 0
On comparing with ax² + bx + c = 0, we get
a = 2, b = 1 and c = – 4
By using quadratic formula, we get
$$x=\frac{-b±\sqrt{b^2-4ac}}{2a}\\x=\frac{-(-1)±\sqrt{(1)^2-4×2×(-4)}}{2×2}\\x=\frac{-(1)±\sqrt{1+32}}{4}\\x=\frac{-1±\sqrt{33}}{4}\\\text{Roots are}=\frac{\sqrt{33}-1}{4}\text{and}\frac{\sqrt{33}-1}{4}$$

(iii) Given equation is
$$4x^2+\sqrt{3x}+3=0\\\text{On comparing with} \space ax^2 + bx + c = 0,\text{ we get}\\a = 4, b =4\sqrt{3},c = 3\\\text{By using quadratic formula, we get}\\=\frac{-b±\sqrt{b^2-4ac}}{2a}\\x=\frac{4\sqrt{3}±\sqrt{(4\sqrt{3})^2-4×2×3}}{2×4}\\=\frac{4\sqrt{3}±\sqrt{48-48}}{8}\\=\frac{4\sqrt{3}}{8}\\-\frac{\sqrt{3}}{2}\\\text{Roots are}=\frac{\sqrt{-3}}{2}\text{or}\frac{\sqrt{-3}}{4}$$

(iv) Given equation is
$$2x^2 + x + 4 = 0\\\text{On comparing with}ax^2 + bx + c = 0,\text{we get a = 2, b = 1 and c = 4}\\\text{By using quadratic formula, we get}\\\frac{-b±\sqrt{b^2-4ac}}{2×a}\\x=\frac{-1\sqrt{1^2-4×2×4}}{2×2}\\=\frac{-1±\sqrt{1-32}}{4}\\=\frac{-1±\sqrt{-31}}{4}\\\text{So, the given equation has no real roots, i.e., it has imaginary roots.}\\\text{Therefore, there is no real solution for the given equations.}$$

3. Find the roots of the following equations :
$$\textbf{(i)}x^2-\frac{1}{2}x=3,x ≠ 0\\\textbf{(ii)}\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30},x ≠ – 4, 7$$

Sol. (i) Given equation is
$$x-\frac{1}{x}=3,x≠0\\⇒\frac{x^2-1}{x}=3\\⇒ x2 – 3x – 1 = 0\\\text{On comparing with}\space ax^2 + bx + c = 0, \text{we get}\\a = 1, b = – 3 \text{and} c = – 1\\\text{Discriminant} D = (– 3)2 – 4 × (1) × (– 1)\\= 9 + 4 = 13 > 0\\\text{So, the given equation has real and different roots.}\\x=\frac{-b±\sqrt{D}}{2a}\\=\frac{-(-3)±\sqrt{13}}{2(1)}\\=\frac{3±\sqrt{13}}{2}\\=\frac{3+\sqrt{13}}{2},\frac{3-\sqrt{13}}{2}\\\text{Hence, the roots of equation}x-\frac{1}{x}= 3 \text{are}\\\frac{3+\sqrt{13}}{2}\text{and}\frac{3+\sqrt{13}}{2}$$

(ii) Given equation is
$$x=\frac{-b±\sqrt{D}}{2a}\\=\frac{-(-3)±\sqrt{1}}{2×1}\\=\frac{3±\sqrt{1}}{2}= \frac{3±1}{2}=\frac{4}{2},\frac{2}{2}= 2, 1\\\text{Hence, the roots of equation}\\\frac{1}{x+1}-\frac{1}{x-7}=\frac{11}{30}\text{are 2 and 1.}$$

4. The sum of the reciprocals of Rehman's ages, (in years) 3 years ago and 5 years from now is 1/3 Find his present age.

Sol. Let the present age of Rehman be x years
Rehman's age, 3 years ago = (x – 3)years
Rehman's age, after 5 years = (x + 5)years
According to the question,
$$\frac{1}{x-3}+\frac{1}{x+5}=\frac{1}{3}\\⇒\frac{x+5+x-3}{(x-3)(x+5)}=\frac{1}{3}\\⇒\frac{2x+2}{x^2+2x-15}=\frac{1}{3}\\6x + 6 = x^2 + 2x – 15\\⇒ x^2 + 2x – 6x – 15 – 6 = 0\\⇒ x^2 – 4x – 21 = 0\\\text{On splitting the middle term,}\\⇒ x^2 – (7x – 3x) – 21 = 0\\⇒ x^2 – 7x + 3x – 21 = 0\\⇒ x(x – 7) + 3(x – 7) = 0\\⇒ (x – 7) (x + 3) = 0\\x – 7 = 0 ⇒ x = 7 or x + 3 = 0 ⇒ x = – 3\\\text{which is not possible because age cannot be negative. So, present age of Rehman = 7 years.}$$

5. In a class test, the sum of Shefali's marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.

Sol. Let Shefali's marks in Mathematics and English will be x and (30 – x) respectively.
According to the questions,
(Marks in Mathematics + 2)
× (Marks in English – 3) = 210
(x + 2) × [(30 – x) – 3] = 210
⇒ (x + 2)(27 – x) = 210
⇒ 27x – x² + 54 – 2x = 210
⇒ x² – 25x + 156 = 0
On splitting the middle term, we get
⇒ x² – (13x + 12x) + 156 = 0
⇒ x² – 13x – 12x + 156 = 0
⇒ x(x – 13) – 12(x – 13) = 0
⇒ (x – 13)(x – 12) = 0
⇒ x = 12 and 13
When x = 12, marks in Mathematics will be 12 and marks in English will be 18
When x = 13, marks in Mathematics will be 13 and marks in English will be = 17

6. The diagonal of a rectangular field is 60 meters more than the shorter side. If the longer side is 30 meters more than the shorter side, find the sides of the field.

Sol. Let ABCD be the rectangle field. Let the shorter side BC of the rectangle be x m.

According to the question,
Diagonal of the rectangle
AC = (x + 60) m
Other Side of the rectangle
AB = (x + 30) m
By Pythagoras theorem, in DABC,
AC² = AB² + BC²
(... In rectangle every adjacent side makes an angle 90° to each other)
⇒ (x + 60)² = (x + 30)² + x²
⇒ x² + 120x + 3600 = x² + 60x + 900 + x²
[... (a + b)² = a² + 2ab + b²]
⇒ (2x² – x²) + (60x – 120x) + (900 – 3600) = 0
⇒ x² – 60x – 2700 = 0
On splitting the middle term, we get
⇒ x² – (90x – 30x) – 2700 = 0
⇒ x² – 90x + 30x – 2700 = 0
⇒ x(x – 90) + 30(x – 90) = 0
⇒ (x – 90)(x + 30) = 0
Either x – 90 = 0 ⇒ x = 90 or x + 30 = 0 ⇒ x = – 30 which is not possible because side cannot be negative.
x = 90
So, the shorter side of the rectangle field = 90 m
and the longer side of the rectangle field = 120 m.

7. The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.

Sol. Let the required numbers be x and y, where x > y.
Difference of squares of two numbers = 180
x² – y² = 180 ...(i)
Square of smaller number
= 8 × larger number
⇒ y² = 8x ...(ii)
From equations (i) and (ii), we have
x² – 8x = 180
⇒ x² – 8x – 180 = 0
On splitting the middle term, we get
⇒ x² – (18x – 10x) – 180 = 0
⇒ x² – 18x + 10x – 180 = 0
(By factorisation method)
⇒ x(x – 18) + 10(x – 18) = 0
⇒ (x – 18)(x + 10) = 0
x – 18 = 0 or x + 10 = 0
x = 18 or x = – 10
Now,
x = 18
⇒ y² = 8 × 18 = 144
[From equation (ii)]
⇒ y = ± 12
⇒ y = 12 or – 12
Again, x = – 10 ⇒ y² = [8 × (– 10)] = – 80 which is not possible i.e., imaginary value.
Hence, the numbers are (18 and 12) or (18 and – 12).

8. A train travels 360 km at a uniform speed. If the speed had been 5 kmh–1 more, if would have taken 1 hour less for the same journey. Find the speed of the train.

Sol. $$\text{Let the uniform speed of the train be x}kmh^{–1}\\\text{So, time taken to cover 360 km }=\frac{360}{x}h\\\text{and time taken to cover 360 km when the speed is increased as 5 km/h}=\frac{360}{x+5}h\\(Speed =\frac{Distance}{Time})\\\text{According to the questions,}\\\frac{360}{x}-\frac{360}{x+5}=1\\⇒\\\frac{360(x+5)-360x}{x(x+5)}=1\\⇒ 360(x + 5) – 360x = x(x + 5)\\⇒ 360x + 1800 – 360x = x^2 + 5x\\⇒ x^2 + 5x – 1800 = 0\\\text{On splitting the middle term, we get}\\⇒ x^2 + (45x – 40x) – 1800 = 0\\⇒ x^2 + 45x – 40x – 1800 = 0\\⇒ x(x + 45) – 40(x + 45) = 0\\⇒ x + 45 = 0 ⇒ x = – 45\\or x – 40 = 0 ⇒ x = 40\\\text{But speed of train cannot be negative.}\\x = 40$$

9. Two water taps together can fill a tank in 9(3/8)hhours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Sol. Let the time taken by the tap of smaller diameter to fill the tank be x hr.
Time taken by the tap of larger diameter = (x – 10) hr.
Part of tank filled by smaller diameter in 1 hour=1/x
P art of tank filled by larger diameter in 1 hour =1/x-10
It is given that the tank can be fillied in 9(9/8)=75/8
hours by both the taps together, the taps fill 8/75
Part of the tank in 1 hour. Then,
$$\frac{1}{x}+\frac{1}{x-10}=\frac{8}{75}\\⇒\frac{x-10+x}{x(x-10)}=\frac{8}{75}\\⇒\frac{2x-10}{x^2-10x}=\frac{8}{75}\\⇒ 75(2x – 10) = 8(x^2 – 10x)\\⇒ 150x – 750 = 8x^2 – 80x\\⇒ 8x^2 – 230x + 750 = 0\\⇒ 8x^2 – 200x – 30x + 750 = 0\\⇒ 8x(x – 25) – 30(x – 25) = 0\\⇒ (x – 25)(8x – 30) = 0\\⇒ x = 25, 30/8$$
Time taken by the tap of smaller diameter cannot be 30/8 = 3.75 hours. As in this case, the time taken by the tap of larger diameter will be negative, which is logically not possible.
Therefore, time taken individually by the tap of smaller diameter and the tap of larger diameter will be 25 and 25 – 10 = 15 hours respectively.

10. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 kmh–1 more than that of the passenger train, find the average speed of the two trains.

Sol. Let the average speed of the passenger train
be x kmh^–1
Then, the average speed of the express train
= (x + 11) kmh^–1
Time taken by passenger train to cover 132 km
$$=\frac{132}{x}h(speed=\frac{distance}{Time})$$
Time taken by express train to cover 132 km
$$=\frac{132}{x+11}h\\\text{According to question,}\\\frac{132}{x}-\frac{132}{x+11}=1\\\text{On dividing both sides by 132, we get}\\⇒\frac{1}{x}-\frac{1}{x+11}=\frac{1}{132}\\⇒\frac{(x+11)-x}{x(x+11)}=\frac{1}{132}\\⇒ x(x + 11) = 132 × 11\\⇒ x^2 + 11x – 1452 = 0\\⇒ x^2 + (44 – 33)x – 1452 = 0\\⇒ x^2 + 44x – 33x – 1452 = 0\\⇒ x(x + 44) – 33(x + 44) = 0\\⇒ (x + 44)(x – 33) = 0\\x + 44 = 0\\⇒ x = – 44\\or x – 33 = 0\\⇒ x = 33$$
Since, x ≠ – 44 because speed cannot be negative.
Hence, the average speed of the passenger train be 33 km/h and the average speed of the express train = (33 + 11) km/h = 44 km/h.

11. Sum of the areas of two squares is 468 m². If the difference of their perimeters is 24 m, find the sides of the two squares.

Sol. Suppose the side of the two squares be x metre and y metre.
Then, area of the first square = x² m²
[... Area of the square = (Side)²]
Area of the second square = y² m²
Perimeter of the first square = 4x
Perimeter of the second square = 4y
[... Perimeter of the square = 4 × (Side)]
According
to the question, sum of areas of two squares = 468 m²
x² + y² = 468 ...(i)
and difference of their perimeters = 24 m
⇒ 4x – 4y = 24 or x – y = 6 ...(ii)
From equation (ii),
y = x – 6
Put the v
alues of y in Equation (i), we get
x² + (x – 6)² = 468
⇒ x² + x² – 12x + 36 – 468 = 0
[... (a – b)² = a² – 2ab + b²]
⇒ 2x² – 12x – 432 = 0
⇒ x² – 6x – 216 = 0
⇒ x² – (18x – 12x) – 216 = 0
⇒ x² – 18x + 12x – 216 = 0
⇒ x(x – 18) + 12(x – 18) = 0
⇒ (x – 18) (x + 12) = 0
Either
x – 18 = 0 or x + 12 = 0
⇒ x = 18 or x = – 12
Since, side of square cannot be negative.
x = 18
Side of the first square = x m = 18 m
and side of the second square = y m = (18 – 6) m = 12 m

Exercise 4.4

1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them :
(i) 2x² – 3x + 5 = 0
(ii) 3x²-4√3x+4=0
(iii) 2x² – 6x + 3 = 0

Sol. (i) Given equation is
2x² – 3x + 5 = 0
On comparing with ax² + bx + c = 0, we get
a = 2, b = – 3 and c = 5
Discriminant, D = b² – 4ac
= (–3)² – 4(2)(5)
= 9 – 40 = – 31 < 0
Hence, the equation 2x² – 3x + 5 = 0 has no real roots i.e., imaginary roots.

(ii) Given equation is
3x²-4√3x+4=0 ...(i)
On comparing with ax² + bc + c = 0, we get
a = 3, b = −4√3 and c = 4
Discriminant, D = b² – 4ac
(−4√3)²-4(3)(4)
= 48 – 48 = 0
Hence, the equation 3x²-4√3x+4=0
has two equal and real roots.
Equation (i) can be written as
(√3x)²-2(√3x)(2)+(2)²= 0
⇒(√3x-2)²=0
[... a² – 2ab + b² = (a – b)²]
(√3x-2)(√3x-2)=0
⇒ x =2/√3,2/√3,
Hence, the equal and real roots are2/√3 and 2/√3,

(iii) Given equation is
2x² – 6x + 3 = 0
On comparing with ax² + bx + c = 0, we get
a = 2, b = – 6 and c = 3
Discriminant, D = b² – 4ac
= (– 6)² – 4(2)(3)
= 36 – 24 = 12 > 0
Hence,
the equation 2x² – 6x + 3 = 0 has two distinct real roots.
$$x=\frac{-b±\sqrt{D}}{2a}=\frac{-(-6)±\sqrt{12}}{4}\\=\frac{(6)±2\sqrt{3}}{4}=\frac{(3)±\sqrt{3}}{2}\\\text{Hence, the real roots are}\frac{(3)+\sqrt{3}}{2}\text{and}\frac{(3)-\sqrt{3}}{2}$$

2. Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(i) 2x² + kx + 3 = 0
(ii) kx(x – 2) + 6 = 0

Sol. (i) The given equation is
2x² + kx + 3 = 0
This equation is of the form ax² + bx + c = 0, where a = 2, b = k and c = 3.
Discriminant,
D = b² – 4ac = (k)² – 4 × 2 × 3
= k² – 24
For equal roots, D = 0
⇒ k² – 24 = 0
⇒ k² = 24
⇒k=±√24=±2√6
k = ±2√6

(ii) The given equation is
kx(x – 2) + 6 = 0 ...(i)
Equation (i) can be written as
kx² – 2kx + 6 = 0 ...(ii)
On comparing equation (ii) with standard equation ax² + bx + c = 0, we get
a = k, b = – 2k and c = 6
Discriminant, D = b² – 4ac
= (– 2k)² – 4 × k × 6
= 4k² – 24k = 4k(k – 6)
For equal roots,
D = 0 ⇒ 4k(k – 6) = 0
⇒ k = 0 or k = 6

3. Is it possible to design a rectangular mango grove whose length is twice its breadth and the area is 800 m²? If so, find its length and breadth.

Sol. Let breadth of a rectangular mango grove = x metre
Length of a rectangular mango grove = 2x metre
According to the question,
Area of rectangular mango grove = 800 m²
⇒ Length × Breadth = 2x(x) = 800
⇒ 2x² = 800
⇒ x² = 400
⇒ x = ± 20
Since, negative sign is not possible because breadth can never be negative.
Therefore,
breadth of mango grove x = 20 m and length of mango grove 2x = 2 × 20 = 40 m.

4. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.

Sol. Let age of one of the two friends is x years
Then, age of other friend
= (20 – x) years
4 years ago of one of two friends = (x – 4)years
4 years ago of the other friend = (20 – x – 4)years
= (16 – x)years
According to the question,
(x – 4)(16 – x) = 48
⇒ 16x –x² – 64 + 4x = 48
⇒ x² – 20x + 112 = 0
On comparing the above equation with ax² + bx + c = 0, we get
a = 1, b = – 20 and c = 112
Discriminant, D = b² – 4ac
= (– 20)² – 4 × 1 × 112
= 400 – 448 = – 48 < 0
which implies that the real roots are not possible, because this condition represent imaginary roots. So, this equation does not exist.

5. Is it possible to design a rectangular park of perimeter 80 m and area 400 m2? If so, find its length and breadth.

Sol. Let the breadth of the park = x metre
Then, according to the question,
Perimeter of a rectangular park = 80 m
⇒ 2(Length + Breadth) = 80 m
⇒ Length + Breadth = 40 m
⇒ Length = (40 – x) m
Area of a rectangular park
= Length × Breadth = (40 – x)x m
According to the question,
Area of the rectangular park is 400 m².
(40 – x)x = 400
⇒ x² – 40x + 400 = 0
⇒ x² – 2x × 20 + (20)² = 0
⇒ (x – 20)² = 0
[... a² – 2ab + b² = (a – b)²]
⇒ x = 20
Thus, breadth of the park = 20 m
and length of the park = (40 – 20) m = 20 m
So, it is possible to design the rectangular park having equal length and breadth i.e., 20 m.