NCERT Solutions For Class 10 Maths: Quadratic Equations

EXERCISE 4.1

Question 1 (i): Check whether the following are quadratic equations :
(x + 1)2 = 2(x – 3)

Solution-

The given equation is $(x + 1)^2 = 2(x – 3)$.
Expanding the terms, we get $x^2 + 2x + 1 = 2x - 6$.
Rewriting the equation by bringing all terms to the LHS:
$x^2 + 2x - 2x + 1 + 6 = 0$.
$x^2 + 7 = 0$.
This equation is of the standard form $ax^2 + bx + c = 0$, where $a=1$, $b=0$, and $c=7$. Since $a \neq 0$, the degree of the polynomial is 2.
Therefore, it is a quadratic equation.

Question 1 (ii): x2 – 2x = (–2) (3 – x)

Solution-

The given equation is $x^2 – 2x = (–2) (3 – x)$.
Expanding the RHS: $x^2 – 2x = -6 + 2x$.
Rewriting in the standard form:
$x^2 – 2x - 2x + 6 = 0$.
$x^2 – 4x + 6 = 0$.
Since this is of the form $ax^2 + bx + c = 0$ with $a=1 \neq 0$,
Therefore, it is a quadratic equation.

Question 1 (iii): (x – 2)(x + 1) = (x – 1)(x + 3)

Solution-

The given equation is $(x – 2)(x + 1) = (x – 1)(x + 3)$.
Expanding LHS: $x^2 + x - 2x - 2 = x^2 - x - 2$.
Expanding RHS: $x^2 + 3x - x - 3 = x^2 + 2x - 3$.
Setting LHS = RHS: $x^2 - x - 2 = x^2 + 2x - 3$.
Simplifying the terms by canceling $x^2$ on both sides:
$-x - 2 = 2x - 3$.
$3x - 1 = 0$.
Since the degree of this equation is 1, it is not of the form $ax^2 + bx + c = 0$ where $a \neq 0$.
Therefore, it is not a quadratic equation.

Question 1 (iv): (x – 3)(2x +1) = x(x + 5)

Solution-

The given equation is $(x – 3)(2x +1) = x(x + 5)$.
Expanding LHS: $2x^2 + x - 6x - 3 = 2x^2 - 5x - 3$.
Expanding RHS: $x^2 + 5x$.
Setting LHS = RHS: $2x^2 - 5x - 3 = x^2 + 5x$.
Rewriting in the standard form:
$2x^2 - x^2 - 5x - 5x - 3 = 0$.
$x^2 - 10x - 3 = 0$.
Since this equation is of the standard form $ax^2 + bx + c = 0$ with degree 2,
Therefore, it is a quadratic equation.

Question 1 (v): (2x – 1)(x – 3) = (x + 5)(x – 1)

Solution-

The given equation is $(2x – 1)(x – 3) = (x + 5)(x – 1)$.
Expanding LHS: $2x^2 - 6x - x + 3 = 2x^2 - 7x + 3$.
Expanding RHS: $x^2 - x + 5x - 5 = x^2 + 4x - 5$.
Setting LHS = RHS: $2x^2 - 7x + 3 = x^2 + 4x - 5$.
Rewriting in the standard form:
$2x^2 - x^2 - 7x - 4x + 3 + 5 = 0$.
$x^2 - 11x + 8 = 0$.
Since the polynomial $p(x)$ has degree 2,
Therefore, it is a quadratic equation.

Question 1 (vi): x2 + 3x + 1 = (x – 2)2

Solution-

The given equation is $x^2 + 3x + 1 = (x – 2)^2$.
Expanding RHS: $(x - 2)^2 = x^2 - 4x + 4$.
Setting LHS = RHS: $x^2 + 3x + 1 = x^2 - 4x + 4$.
Simplifying by canceling $x^2$:
$3x + 1 = -4x + 4$.
$7x - 3 = 0$.
Since the degree of this equation is 1, it is not a polynomial of degree 2.
Therefore, it is not a quadratic equation.

Question 1 (vii): (x + 2)3 = 2x (x2 – 1)

Solution-

The given equation is $(x + 2)^3 = 2x (x^2 – 1)$.
Expanding LHS: $(x + 2)^3 = x^3 + 6x^2 + 12x + 8$.
Expanding RHS: $2x (x^2 – 1) = 2x^3 - 2x$.
Setting LHS = RHS: $x^3 + 6x^2 + 12x + 8 = 2x^3 - 2x$.
Rewriting in the standard form:
$2x^3 - x^3 - 6x^2 - 12x - 2x - 8 = 0$.
$x^3 - 6x^2 - 14x - 8 = 0$.
Since the highest degree of the polynomial is 3, it is a cubic equation.
Therefore, it is not a quadratic equation.

Question 1 (viii): x3 – 4x2 – x + 1 = (x – 2)3

Solution-

The given equation is $x^3 – 4x^2 – x + 1 = (x – 2)^3$.
Expanding RHS: $(x – 2)^3 = x^3 - 6x^2 + 12x - 8$.
Setting LHS = RHS: $x^3 – 4x^2 – x + 1 = x^3 - 6x^2 + 12x - 8$.
Simplifying by canceling $x^3$ terms:
$- 4x^2 – x + 1 = - 6x^2 + 12x - 8$.
Rewriting in the standard form:
$- 4x^2 + 6x^2 - x - 12x + 1 + 8 = 0$.
$2x^2 - 13x + 9 = 0$.
Since the polynomial $p(x)$ has degree 2, it is a quadratic equation.

Question 2 (i): Represent the following situations in the form of quadratic equations :
The area of a rectangular plot is 528 m2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.

Solution-

Let the breadth of the plot be $x$ metres.
Then the length of the plot is $(2x + 1)$ metres.
Area of the plot is 528 m$^2$.
Area = Length $\times$ Breadth.
$x(2x + 1) = 528$.
$2x^2 + x = 528$.
Representing the situation mathematically in the standard form $ax^2 + bx + c = 0$:
$\mathbf{2x^2 + x – 528 = 0}$.

Question 2 (ii): The product of two consecutive positive integers is 306. We need to find the integers.

Solution-

Let the first positive integer be $x$.
Since the integers are consecutive, the second integer is $(x + 1)$.
The product of the two consecutive positive integers is 306.
$x(x + 1) = 306$.
$x^2 + x = 306$.
Representing the situation mathematically:
$\mathbf{x^2 + x – 306 = 0}$.

Question 2 (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.

Solution-

Let Rohan’s present age be $x$ years.
Rohan’s mother’s present age is $(x + 26)$ years.

In 3 years, Rohan’s age will be $(x + 3)$ years.
In 3 years, his mother’s age will be $(x + 26 + 3) = (x + 29)$ years.
The product of their ages 3 years from now is 360.
$(x + 3)(x + 29) = 360$.
Expanding the product: $x^2 + 29x + 3x + 87 = 360$.
$x^2 + 32x + 87 - 360 = 0$.
Representing the situation mathematically:
$\mathbf{x^2 + 32x – 273 = 0}$.

Question 2 (iv): A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.

Solution-

Let the uniform speed of the train be $x$ km/h.
Distance $= 480$ km.
Time taken at uniform speed ($T_1$) $= \frac{480}{x}$ hours.

If the speed had been 8 km/h less, new speed $= (x - 8)$ km/h.
New time taken ($T_2$) $= \frac{480}{x - 8}$ hours.

We are given that $T_2 = T_1 + 3$.
$\frac{480}{x - 8} = \frac{480}{x} + 3$.
Multiplying by $x(x - 8)$:
$480x = 480(x - 8) + 3x(x - 8)$.
$480x = 480x - 3840 + 3x^2 - 24x$.
$0 = 3x^2 - 24x - 3840$.
Dividing the entire equation by 3:
Representing the situation mathematically:
$\mathbf{x^2 - 8x – 1280 = 0}$.

EXERCISE 4.2

Question 1 (i): Find the roots of the following quadratic equations by factorisation:
x2 – 3x – 10 = 0

Solution-

The quadratic equation is $x^2 – 3x – 10 = 0$.
We split the middle term $-3x$ into $-5x$ and $+2x$ such that their product is $-10x^2$.
$x^2 + 2x - 5x – 10 = 0$.
Factorising by grouping:
$x(x + 2) - 5(x + 2) = 0$.
$(x - 5)(x + 2) = 0$.
Equating each linear factor to zero to find the roots:
$x - 5 = 0 \implies x = 5$.
$x + 2 = 0 \implies x = -2$.
The roots of the equation are 5 and –2.

Question 1 (ii): 2x2 + x – 6 = 0

Solution-

The quadratic equation is $2x^2 + x – 6 = 0$.
We split the middle term $x$ into $+4x$ and $-3x$ such that their product is $(2x^2)(-6) = -12x^2$.
$2x^2 + 4x - 3x – 6 = 0$.
Factorising by grouping:
$2x(x + 2) - 3(x + 2) = 0$.
$(2x - 3)(x + 2) = 0$.
Equating each linear factor to zero:
$2x - 3 = 0 \implies x = \frac{3}{2}$.
$x + 2 = 0 \implies x = -2$.
The roots of the equation are $\mathbf{\frac{3}{2}}$ and –2.

Question 1 (iii): $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$

Solution-

The quadratic equation is $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$.
We split the middle term $7x$ such that the product is $(\sqrt{2}x^2)(5\sqrt{2}) = 10x^2$. We use $2x$ and $5x$.
$\sqrt{2}x^2 + 2x + 5x + 5\sqrt{2} = 0$.
Factorising by grouping, noting $2 = \sqrt{2} \times \sqrt{2}$:
$\sqrt{2}x(x + \sqrt{2}) + 5(x + \sqrt{2}) = 0$.
$(\sqrt{2}x + 5)(x + \sqrt{2}) = 0$.
Equating each factor to zero:
$\sqrt{2}x + 5 = 0 \implies x = -\frac{5}{\sqrt{2}}$.
$x + \sqrt{2} = 0 \implies x = -\sqrt{2}$.
The roots of the equation are $\mathbf{-\frac{5}{\sqrt{2}}}$ and $\mathbf{-\sqrt{2}}$.

Question 1 (iv): $2x^2 – x + \frac{1}{8} = 0$

Solution-

The quadratic equation is $2x^2 – x + \frac{1}{8} = 0$.
Multiply the entire equation by 8 to simplify: $16x^2 – 8x + 1 = 0$.
We split the middle term $-8x$ such that the product is $(16x^2)(1) = 16x^2$. We use $-4x$ and $-4x$.
$16x^2 – 4x - 4x + 1 = 0$.
$4x(4x - 1) - 1(4x - 1) = 0$.
$(4x - 1)(4x - 1) = 0$.
Equating the repeated linear factor to zero:
$4x - 1 = 0 \implies x = \frac{1}{4}$.
The equation has two equal real roots.
The roots of the equation are $\mathbf{\frac{1}{4}}$ and $\mathbf{\frac{1}{4}}$.

Question 1 (v): $100 x^2 – 20x + 1 = 0$

Solution-

The quadratic equation is $100 x^2 – 20x + 1 = 0$.
We split the middle term $-20x$ such that the product is $(100x^2)(1) = 100x^2$. We use $-10x$ and $-10x$.
$100 x^2 – 10x - 10x + 1 = 0$.
$10x(10x - 1) - 1(10x - 1) = 0$.
$(10x - 1)(10x - 1) = 0$.
Equating the repeated linear factor to zero:
$10x - 1 = 0 \implies x = \frac{1}{10}$.
The roots of the equation are $\mathbf{\frac{1}{10}}$ and $\mathbf{\frac{1}{10}}$.

Question 2: Solve the problems given in Example 1.

Solution-

(i) The situation concerns John ($x$ marbles) and Jivanti ($45-x$ marbles).
The quadratic equation representing the situation is $x^2 – 45x + 324 = 0$.
We find the roots by factorisation. We split $-45x$ into $-36x$ and $-9x$.
$x^2 - 9x - 36x + 324 = 0$.
$x(x - 9) - 36(x - 9) = 0$.
$(x - 36)(x - 9) = 0$.
The roots are $x = 36$ or $x = 9$.
The number of marbles John had to start with is 9 or 36.

(ii) The situation concerns the number of toys produced ($x$).
The quadratic equation representing the situation is $x^2 – 55x + 750 = 0$.
We find the roots by factorisation. We split $-55x$ into $-30x$ and $-25x$.
$x^2 - 30x - 25x + 750 = 0$.
$x(x - 30) - 25(x - 30) = 0$.
$(x - 30)(x - 25) = 0$.
The roots are $x = 30$ or $x = 25$.
The number of toys produced that day is 25 or 30.

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

Solution-

Let the first number be $x$. The second number is $(27 - x)$.
Product: $x(27 - x) = 182$.
$27x - x^2 = 182$.
Rewriting in the standard form: $x^2 - 27x + 182 = 0$.
We find the roots by factorisation. We split $-27x$ into $-14x$ and $-13x$.
$x^2 - 13x - 14x + 182 = 0$.
$x(x - 13) - 14(x - 13) = 0$.
$(x - 14)(x - 13) = 0$.
The roots are $x = 14$ or $x = 13$.
The two numbers are 13 and 14.

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

Solution-

Let the two consecutive positive integers be $x$ and $(x + 1)$.
Sum of squares: $x^2 + (x + 1)^2 = 365$.
$x^2 + x^2 + 2x + 1 = 365$.
$2x^2 + 2x - 364 = 0$.
Dividing by 2: $x^2 + x - 182 = 0$.
We find the roots by factorisation. We split $x$ into $+14x$ and $-13x$.
$x^2 + 14x - 13x - 182 = 0$.
$x(x + 14) - 13(x + 14) = 0$.
$(x - 13)(x + 14) = 0$.
The roots are $x = 13$ or $x = -14$.
Since the integers must be positive, $x = 13$.
The integers are $x=13$ and $x+1=14$.
The two consecutive positive integers are 13 and 14.

Question 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.

Solution-

Let the base of the right triangle be $x$ cm.
Altitude $= (x - 7)$ cm.
Hypotenuse $= 13$ cm.
By the Pythagoras theorem: $(\text{Base})^2 + (\text{Altitude})^2 = (\text{Hypotenuse})^2$.
$x^2 + (x - 7)^2 = 13^2$.
$x^2 + x^2 - 14x + 49 = 169$.
$2x^2 - 14x - 120 = 0$.
Dividing by 2: $x^2 - 7x - 60 = 0$.
We find the roots by factorisation. We split $-7x$ into $-12x$ and $+5x$.
$x^2 - 12x + 5x - 60 = 0$.
$x(x - 12) + 5(x - 12) = 0$.
$(x + 5)(x - 12) = 0$.
The roots are $x = -5$ or $x = 12$.
Since distance cannot be negative, Base $x = 12$ cm.
Altitude $= x - 7 = 12 - 7 = 5$ cm.
The other two sides are 12 cm and 5 cm.

Question 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 rupees) 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.

Solution-

Let the number of articles produced be $x$.
Cost of production of each article $= (2x + 3)$ rupees.
Total cost $= 90$.
$x(2x + 3) = 90$.
$2x^2 + 3x - 90 = 0$.
We find the roots by factorisation. We split $3x$ into $+15x$ and $-12x$.
$2x^2 + 15x - 12x - 90 = 0$.
$x(2x + 15) - 6(2x + 15) = 0$.
$(x - 6)(2x + 15) = 0$.
The roots are $x = 6$ or $x = -\frac{15}{2}$.
Since the number of articles cannot be negative, $x = 6$.
Number of articles produced = 6.
Cost of each article $= 2x + 3 = 2(6) + 3 = 15$ rupees.
The number of articles produced is 6, and the cost of each article is ₹ 15.

EXERCISE 4.3

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

Solution-

The equation is $2x^2 – 3x + 5 = 0$.
$a=2, b=-3, c=5$.
We find the discriminant, $D = b^2 – 4ac$.
$D = (-3)^2 – 4(2)(5)$.
$D = 9 – 40 = -31$.
Since $D < 0$ ($b^2 – 4ac < 0$), the quadratic equation has no real roots.

Question 1 (ii): 3x2 – 4$\sqrt{3}$x + 4 = 0

Solution-

The equation is $3x^2 – 4\sqrt{3}x + 4 = 0$.
$a=3, b=-4\sqrt{3}, c=4$.
We find the discriminant, $D = b^2 – 4ac$.
$D = (-4\sqrt{3})^2 – 4(3)(4)$.
$D = 48 – 48 = 0$.
Since $D = 0$ ($b^2 – 4ac = 0$), the quadratic equation has two equal real roots.
The roots are given by $x = \frac{-b}{2a}$.
$x = \frac{-(-4\sqrt{3})}{2(3)} = \frac{4\sqrt{3}}{6} = \frac{2\sqrt{3}}{3}$.
The real roots are $\mathbf{\frac{2\sqrt{3}}{3}}$ and $\mathbf{\frac{2\sqrt{3}}{3}}$.

Question 1 (iii): 2x2 – 6x + 3 = 0

Solution-

The equation is $2x^2 – 6x + 3 = 0$.
$a=2, b=-6, c=3$.
We find the discriminant, $D = b^2 – 4ac$.
$D = (-6)^2 – 4(2)(3)$.
$D = 36 – 24 = 12$.
Since $D > 0$ ($b^2 – 4ac > 0$), the quadratic equation has two distinct real roots.
The roots are given by the quadratic formula $x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$.
$x = \frac{-(-6) \pm \sqrt{12}}{2(2)} = \frac{6 \pm 2\sqrt{3}}{4}$.
$x = \frac{3 \pm \sqrt{3}}{2}$.
The real roots are $\mathbf{\frac{3 + \sqrt{3}}{2}}$ and $\mathbf{\frac{3 - \sqrt{3}}{2}}$.

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

Solution-

The equation is $2x^2 + kx + 3 = 0$.
For the equation to have two equal roots (i.e., coincident roots), the discriminant must be zero ($D = b^2 – 4ac = 0$).
Here, $a=2, b=k, c=3$.
$k^2 – 4(2)(3) = 0$.
$k^2 – 24 = 0$.
$k^2 = 24$.
$k = \pm \sqrt{24} = \pm 2\sqrt{6}$.
The values of $k$ are $\mathbf{2\sqrt{6}}$ and $\mathbf{-2\sqrt{6}}$.

Question 2 (ii): kx (x – 2) + 6 = 0

Solution-

The equation is $kx (x – 2) + 6 = 0$.
First, write the equation in the standard form: $kx^2 - 2kx + 6 = 0$.
Here, $a=k, b=-2k, c=6$.
For the equation to have two equal roots, $D = b^2 – 4ac = 0$.
$(-2k)^2 – 4(k)(6) = 0$.
$4k^2 – 24k = 0$.
$4k(k - 6) = 0$.
This gives $k = 0$ or $k = 6$.
Since the given equation must be a quadratic equation, $a \neq 0$, so $k \neq 0$.
The value of $k$ is $\mathbf{6}$.

Question 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.

Solution-

Let the breadth be $x$ metres. Length $= 2x$ metres.
Area $= 2x \cdot x = 2x^2$.
Given Area $= 800$ m$^2$.
$2x^2 = 800$.
$x^2 - 400 = 0$.
To determine possibility, we check the discriminant, $D = b^2 – 4ac$. Here $a=1, b=0, c=-400$.
$D = 0^2 - 4(1)(-400) = 1600$.
Since $D > 0$, real roots exist. Yes, it is possible.
Solving $x^2 = 400$, we get $x = \pm 20$.
Since breadth cannot be negative, Breadth $x = 20$ m.
Length $= 2x = 40$ m.
The length is 40 m and the breadth is 20 m.

Question 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.

Solution-

Let the age of the first friend be $x$ years. The second friend's age is $(20 - x)$ years.
Four years ago, their ages were $(x - 4)$ and $(20 - x - 4) = (16 - x)$ years.
Product of ages four years ago: $(x - 4)(16 - x) = 48$.
$16x - x^2 - 64 + 4x = 48$.
$-x^2 + 20x - 112 = 0$.
Rewriting in standard form: $x^2 - 20x + 112 = 0$.
To check if the situation is possible, we find the discriminant $D = b^2 – 4ac$.
Here $a=1, b=-20, c=112$.
$D = (-20)^2 – 4(1)(112) = 400 – 448 = -48$.
Since $D < 0$ ($b^2 – 4ac < 0$), the equation has no real roots.
Therefore, the situation is not possible.

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

Solution-

Let the length be $l$ and breadth be $b$.
Perimeter $2(l + b) = 80$ m. $l + b = 40$. So, $b = 40 - l$.
Area $l \cdot b = 400$ m$^2$.
$l(40 - l) = 400$.
$40l - l^2 = 400$.
Rewriting in the standard form: $l^2 - 40l + 400 = 0$.
To check if the situation is possible, we find the discriminant $D = b^2 – 4ac$.
Here $a=1, b=-40, c=400$.
$D = (-40)^2 – 4(1)(400) = 1600 – 1600 = 0$.
Since $D = 0$, the equation has two equal real roots. Yes, it is possible.
The equal roots are $l = \frac{-b}{2a}$.
$l = \frac{-(-40)}{2(1)} = 20$ m.
Length $l = 20$ m.
Breadth $b = 40 - l = 40 - 20 = 20$ m.
The length is 20 m and the breadth is 20 m.