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NCERT Solutions

Quadratic Equations

Madhya Pradesh Board · Class 10 · Mathematics

NCERT Solutions for Quadratic Equations — Madhya Pradesh Board Class 10 Mathematics.

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35 Questions Solved · 3 Sections

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EXERCISE 4.1

1Check whether the following are quadratic equations :Show solution
(i) (x+1)2=2(x−3)(x+1)^2=2(x-3) gives x2+2x+1=2x−6x^2+2x+1=2x-6, so x2+7=0x^2+7=0. It is not a quadratic equation in the standard real-coefficient form with real roots? But as an equation it is of degree 2, so it is quadratic.

(ii) x2−2x=−2(3−x)x^2-2x=-2(3-x) gives x2−2x=−6+2xx^2-2x=-6+2x, so x2−4x+6=0x^2-4x+6=0. Quadratic.

(iii) (x−2)(x+1)=(x−1)(x+3)(x-2)(x+1)=(x-1)(x+3) gives x2−x−2=x2+2x−3x^2-x-2=x^2+2x-3, so −3x+1=0-3x+1=0. Not quadratic.

(iv) (x−3)(2x+1)=x(x+5)(x-3)(2x+1)=x(x+5) gives 2x2−5x−3=x2+5x2x^2-5x-3=x^2+5x, so x2−10x−3=0x^2-10x-3=0. Quadratic.

(v) (2x−1)(x−3)=(x+5)(x−1)(2x-1)(x-3)=(x+5)(x-1) gives 2x2−7x+3=x2+4x−52x^2-7x+3=x^2+4x-5, so x2−11x+8=0x^2-11x+8=0. Quadratic.

(vi) x2+3x+1=(x−2)2=x2−4x+4x^2+3x+1=(x-2)^2=x^2-4x+4, so 7x−3=07x-3=0. Not quadratic.

(vii) (x+2)3=2x(x2−1)(x+2)^3=2x(x^2-1) gives x3+6x2+12x+8=2x3−2xx^3+6x^2+12x+8=2x^3-2x, so x3−6x2−14x−8=0x^3-6x^2-14x-8=0. Not quadratic.

(viii) x3−4x2−x+1=(x−2)3=x3−6x2+12x−8x^3-4x^2-x+1=(x-2)^3=x^3-6x^2+12x-8, so 2x2−13x+9=02x^2-13x+9=0. Quadratic.

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1(i)(x+1)2=2(x−3)(x + 1)^2 = 2(x - 3)Show solution
(x+1)2=2(x−3)(x+1)^2=2(x-3) becomes x2+2x+1=2x−6x^2+2x+1=2x-6, so x2+7=0x^2+7=0. This is a quadratic equation because it is of the form ax2+bx+c=0ax^2+bx+c=0 with a≠0a\neq 0.

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1(ii)x2−2x=(−2)(3−x)x^2 - 2x = (-2)(3 - x)Show solution
x2−2x=−2(3−x)x^2-2x=-2(3-x) gives x2−2x=−6+2xx^2-2x=-6+2x, so x2−4x+6=0x^2-4x+6=0. This is a quadratic equation, so the answer is Yes.

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1(iii)(x - 2)(x + 1) = (x - 1)(x + 3)Show solution
(x−2)(x+1)=(x−1)(x+3)(x-2)(x+1)=(x-1)(x+3) gives x2−x−2=x2+2x−3x^2-x-2=x^2+2x-3, hence −3x+1=0-3x+1=0. The x2x^2 terms cancel, so it is not a quadratic equation.

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1(iv)(x - 3)(2x + 1) = x(x + 5)Show solution
(x−3)(2x+1)=x(x+5)(x-3)(2x+1)=x(x+5) gives 2x2−5x−3=x2+5x2x^2-5x-3=x^2+5x, so x2−10x−3=0x^2-10x-3=0. This is of the form ax2+bx+c=0ax^2+bx+c=0.

Therefore, it is a quadratic equation.

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1(v)(2x - 1)(x - 3) = (x + 5)(x - 1)Show solution
(2x−1)(x−3)=(x+5)(x−1)(2x-1)(x-3)=(x+5)(x-1) gives 2x2−7x+3=x2+4x−52x^2-7x+3=x^2+4x-5, so x2−11x+8=0x^2-11x+8=0. This is a quadratic equation.

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1(vi)x2+3x+1=(x−2)2x^2 + 3x + 1 = (x - 2)^2Show solution
x2+3x+1=(x−2)2=x2−4x+4x^2+3x+1=(x-2)^2=x^2-4x+4, so 7x−3=07x-3=0. Since the x2x^2 terms cancel, it is not quadratic.

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1(vii)(x+2)3=2x(x2−1)(x + 2)^3 = 2x(x^2 - 1)Show solution
(x+2)3=x3+6x2+12x+8(x+2)^3=x^3+6x^2+12x+8 and 2x(x2−1)=2x3−2x2x(x^2-1)=2x^3-2x. So the equation becomes x3−6x2−14x−8=0x^3-6x^2-14x-8=0, which is degree 3, not degree 2. Hence it is not quadratic.

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1(viii)x3−4x2−x+1=(x−2)3x^3 - 4x^2 - x + 1 = (x - 2)^3Show solution
(x−2)3=x3−6x2+12x−8(x-2)^3=x^3-6x^2+12x-8. So

x3−4x2−x+1=x3−6x2+12x−8x^3-4x^2-x+1=x^3-6x^2+12x-8

which gives 2x2−13x+9=02x^2-13x+9=0. This is a quadratic equation.

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2Represent the following situations in the form of quadratic equations :Show solution
To represent a situation mathematically, let an unknown quantity be xx, express the other quantity in terms of xx, and use the given condition to form an equation. In these problems, after simplification, the equations obtained are quadratic equations.

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2(i)The area of a rectangular plot is 528 m2528 \text{ m}^2. 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.Show solution
Let the breadth be xx m. Then the length is (2x+1)(2x+1) m.

Area =528=528 m2^2, so

(2x+1)x=528(2x+1)x=528

2x2+x−528=02x^2+x-528=0

Thus, the required quadratic equation is **2x2+x−528=02x^2+x-528=0**.

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2(ii)The product of two consecutive positive integers is 306. We need to find the integers.Show solution
Let the smaller integer be xx. Then the next consecutive integer is x+1x+1.

Given product =306=306:

x(x+1)=306x(x+1)=306

x2+x−306=0x^2+x-306=0

So the quadratic equation is **x2+x−306=0x^2+x-306=0**.

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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.Show solution
Let Rohan's present age be xx years. Then his mother's present age is x+26x+26 years.

Three years from now, their ages will be x+3x+3 and x+29x+29.

Given product =360=360:

(x+3)(x+29)=360(x+3)(x+29)=360

x2+32x+87=360x^2+32x+87=360

x2+32x−273=0x^2+32x-273=0

So the required quadratic equation is **x2+32x−273=0x^2+32x-273=0**.

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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.Show solution
Let the speed of the train be xx km/h. Then time taken = 480x\frac{480}{x} h.

If speed were 88 km/h less, speed would be x−8x-8 km/h and time would be 480x−8\frac{480}{x-8} h. Given this is 3 hours more:

480x−8=480x+3\frac{480}{x-8}=\frac{480}{x}+3

This simplifies to a quadratic equation:

480x=480(x−8)+3x(x−8)480x=480(x-8)+3x(x-8)

480x=480x−3840+3x2−24x480x=480x-3840+3x^2-24x

3x2−24x−3840=03x^2-24x-3840=0

Dividing by 3:

**x2−8x−1280=0x^2-8x-1280=0**

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EXERCISE 4.2

1Find the roots of the following quadratic equations by factorisation:Show solution
The roots are found by factorisation after simplifying each equation.

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1(i)x2−3x−10=0x^2 - 3x - 10 = 0Show solution
Factorise:

x2−3x−10=x2−5x+2x−10x^2-3x-10=x^2-5x+2x-10

=x(x−5)+2(x−5)=x(x-5)+2(x-5)

=(x−5)(x+2)=(x-5)(x+2)

So, (x−5)(x+2)=0(x-5)(x+2)=0.

Hence x=5x=5 or x=−2x=-2.

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1(ii)2x2+x−6=02x^2 + x - 6 = 0Show solution
2x2+x−6=02x^2+x-6=0

Split the middle term:

2x2+4x−3x−6=02x^2+4x-3x-6=0

=2x(x+2)−3(x+2)=0=2x(x+2)-3(x+2)=0

=(2x−3)(x+2)=0=(2x-3)(x+2)=0

So, 2x−3=02x-3=0 or x+2=0x+2=0.

Hence x=32x=\frac{3}{2} or x=−2x=-2.

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1(iii)2x2+7x+52=0\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0Show solution
sqrt2x2+7x+52=0sqrt{2}x^2+7x+5\sqrt{2}=0.

Split the middle term:

2x2+5x+2x+52=0\sqrt{2}x^2+5x+2x+5\sqrt{2}=0 is not the right split. Better factorise by grouping:

2x2+7x+52=sqrt2x2+5x+2x+52\sqrt{2}x^2+7x+5\sqrt{2}=sqrt{2}x^2+5x+2x+5\sqrt{2}

=x(2x+5)+2(x+5)=x(\sqrt{2}x+5)+\sqrt{2}(x+5) is not matching.

Using the chapter's factorisation:

2x2+7x+52=(2x+5)( ?)\sqrt{2}x^2+7x+5\sqrt{2}=(\sqrt{2}x+\sqrt{5})(\,?) not convenient.

Actually, we can verify by inspection that it factorises as

(2x+5)(2x+5)(\sqrt{2}x+\sqrt{5})(\sqrt{2}x+\sqrt{5})? That gives 2x2+210x+52x^2+2\sqrt{10}x+5, not correct.

So the correct roots are obtained by factoring as

(2x+52)(x+2)=0(\sqrt{2}x+\frac{5}{\sqrt{2}})(x+\sqrt{2})=0 after matching coefficients, giving

x=−2x=-\sqrt{2} and x=−52x=-\frac{5}{\sqrt{2}}.

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1(iv)2x2−x+18=02x^2 - x + \frac{1}{8} = 0
1(v)100x2−20x+1=0100x^2 - 20x + 1 = 0
2Solve the problems given in Example 1.
3Find two numbers whose sum is 27 and product is 182.
4Find two consecutive positive integers, sum of whose squares is 365.
5The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
6A 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.

EXERCISE 4.3

1Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
1(i)2x2−3x+5=02x^2 - 3x + 5 = 0
1(ii)3x2−43x+4=03x^2 - 4\sqrt{3}x + 4 = 0
1(iii)2x2−6x+3=02x^2 - 6x + 3 = 0
2Find the values of kk for each of the following quadratic equations, so that they have two equal roots.
2(i)2x2+kx+3=02x^2 + kx + 3 = 0
2(ii)kx(x - 2) + 6 = 0
3Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2^2? If so, find its length and breadth.
4Is 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.
5Is it possible to design a rectangular park of perimeter 80 m and area 400 m2^2? If so, find its length and breadth.

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