We have provided you with Extra and Important Questions from Class 10 Maths Chapter 4 Quadratic Equations.This Extra and Important Questions will help you to score 100% in your Board Exams. These extra questions will be helpful to revise the important topics and concepts.

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**Quadratic Equations** **Class 10 Important Questions with Answers Maths Chapter **4

**Extra Questions for Class 10 Maths Chapter 4 Very Short Answer Type**

**Question: **If a and b are the roots of the equation x² + ax – b = 0, then find a and b.

**Solution:**

Sum of the roots = a + b = – B/A = – a

Product of the roots = ab = B/A = – b

= a + b = – a and ab = -b

⇒ 2a = -b and a = -1

⇒ b = 2 and a = -1

**Question:**

Show that x = – 2 is a solution of 3x² + 13x + 14 = 0.

**Solution:**

Put the value of x in the quadratic equation,

⇒ LHS = 3x^{2} + 13x + 14

⇒ 3(-2)^{2} + 13(-2) + 14

⇒ 12 – 26 + 14 = 0

⇒ RHS Hence, x = -2 is a solution.

**Question:**

Find the discriminant of the quadratic equation 4√2x^{2} + 8x + 2√2 = 0).

**Solution:**

D = 62 – 4ac = (8)^{2} – 4(4√2)(2√2)

⇒ 64 – 64 = 0

**Question:** What will be the nature of roots of quadratic equation 2x² + 4x – n = 0?

**Solution:**

D = b^{2} – 4ac

⇒ 4^{2} – 4 x 2 (-7)

⇒ 16 + 56 = 72 > 0

Hence, roots of quadratic equation are real and unequal.

**Question:**

If 12 is a root of the equation x^{2} + kx – 54 = 0, then find the value of k.

**Solution:**

∴ 12 is a root of quadratic equation.

∴ It must satisfy the quadratic equation.

**Question:** If ax² + bx + c = 0 has equal roots, find the value of c.

**Solution:**

For equal roots D = 0

i.e., b^{2} – 4ac = 0

⇒ b^{2} = 4 ac

⇒ c = b24a

**Question: **State whether the equation (x + 1)(x – 2) + x = 0 has two distinct real roots or not. Justify your answer.

**Solution:**

(x + 1)(x – 2) + x = 0

⇒ x^{2} – x – 2 + x = 0

⇒ x^{2} – 2 = 0

D = b^{2} – 4ac

⇒ (-4(1)(-2) = 8 > 0

∴ Given equation has two distinct real roots.

**Question:**

Is 0.3 a root of the equation x^{2} – 0.9 = 0? Justify.

**Solution:**

∵ 0.3 is a root of the equation x^{2} – 0.9 = 0

∴ x^{2} – 0.9 = (0.3)^{2} – 0.9 = 0.09 – 0.9 ≠ 0

Hence, 0.3 is not a root of given equation.

**Question:**

For what value of k, is 3 a root of the equation 2x^{2} + x + k = 0?

**Solution:**

3 is a root of 2x^{2} + x + k = 0, when

⇒ 2(3)^{2} + 3 + k = 0

⇒ 18+ 3 + k = 0

⇒ k = – 21

**Question:**

Find the values of k for which the quadratic equation 9x^{2} – 3kx + k = 0 has equal roots.

**Solution:**

For equal roots:

D = 0

⇒ b^{2} – 4ac = 0

⇒ (- 3k)^{2} – 4 × 9 × k = 0

⇒ 9k^{2} = 36k

⇒ k = 4

**Question:**

Find the value of k for which the equation x^{2} + k(2x + k – 1)+ 2 = 0 has real and equal roots.

**Solution:**

Given quadratic equation: x^{2} + k(2x + k-1) + 2 = 0)

= x^{2} + 2kx + (k^{2} – k + 2) = 0

For equal roots, b^{2} – 4ac = 0

⇒ 4k^{2} – 4k^{2} + 4k – 8 = 0

⇒ 4k = 8

⇒ k = 2

**Question:**

If -5 is a root of the quadratic equation 2x^{2} + px – 15 = 0 and the quadratic equation p(x^{2} + x) + k = 0 has equal roots, then find the value of k.

**Solution:**

Since – 5 is a root of the equation 2x^{2} + px – 15 = 0

∴ 2(-5)^{2} + p(-5) – 15 = 0

⇒ 50 – 5p – 15 = 0 or 5p = 35 or p = 7

Again p(x^{2} + x) + k = 0 or 7x^{2} + 7x + k = 0 has equal roots

∴ D = 0

i.e., b^{2} – 4ac = 0 or 49- 4 × 7k = 0

⇒ k = 4928 = 74

**Extra Questions for Class 10 Maths Chapter 4 Short Answer Type**

**Question:** Find the value of p, for which one root of the quadratic equation px² – 14x + 8 = 0 is 6 times the other.

**Solution:**

Let the roots of the given equation be a and 6α.

Thus the quadratic equation is (x – a) (x – 6α) = 0

⇒ x^{2} – 7αx + 6α^{2} = 0 …(i)

Given equation can be written as

**Question:** If ad ≠ bc, then prove that the equation

(a^{2} + b^{2}) x^{2} + 2(ac + bd)x + (c^{2}+ d^{2}) = 0 has no real roots.

**Solution:**

The given quadratic equation is (a^{2} + b^{2)}x^{2} + 2(ac + bd)x +(c^{2}+ d^{2}) = 0

D = b^{2} – 4ac

= 4(ac + bd)^{2} – 4(a^{2} + b^{2}) (c^{2}+ d^{2})

= -4(a^{2}d^{2} + b^{2}c^{2}– 2abcd) = – 4(ad – bc)^{2}

Since ad ≠ bc

Therefore D < 0

Hence, the equation has no real roots.

**Question:**

Solve for x: √13x^{2} – 2x – 8√3 = 0

**Solution:**

√3x^{2} – 2x – 8√3 = 0

By mid term splitting

⇒ √3x^{2} – 6x + 4x – 8√3 = 0

⇒ √3x(x – 2/3) + 4 (x – 2/3) = 0

⇒ (x – 2√3)(√3x + 4) = 0

⇒ Either (x – 2√3) = 0 or (√3x + 4) = 0

⇒ x = −43√, 2√3

**Question:** If x = 2/3 and x = -3 are roots of the quadratic equation ax² + 7x + b = 0, find the values of a and b.

**Solution:**

Let us assume the quadratic equation be Ax^{2} + Bx + C = 0.

Sum of the roots = –B/A

**Question:**

A two-digit number is four times the sum of the digits. It is also equal to 3 times the product of digits. Find the number.

**Solution: **

Let the ten’s digit be x and unit’s digit = y

Number 10x + y

**Question:**

Find the value of p, for which one root of the quadratic equation px^{2} – 14x + 8 = 0 is 6 times the other.

**Solution:**

Let the roots of the given equation be a and 6α.

Thus the quadratic equation is (x – a) (x – 6α) = 0

⇒ x^{2} – 7αx + 6α^{2} = 0 …(i)

Given equation can be written as

**Question:** Find the roots of the following quadratic equations by factorisation:

(i) √2x² + 7x + 5√2 = 0 (ii) 2x² – x + 18 = 0

**Solution:**

(i) We have, √2x^{2} + 7x + 5√2 = 0

= √2x^{2} + 5x + 2x + 5√2 = 0

x(√2x + 5) + √2 (√2x + 5) = 0

= (√2x + 5)(x + √2) = 0

∴ Either √2x + 5 = 0 or x + √2 = 0

∴ x = – 5/√2 or x = -√2

Hence, the roots are – 5/√2 and -√2.

(ii) We have, 2x^{2} – x + 18 = 0

**Question:** Find the roots of the following quadratic equations, if they exist, by the method of completing the square:

(i) 2x² + x – 4 = 0

(ii) 4x² + 4√3x + 3 = 0

**Solution:**

(i) We have, 2x^{2} + x – 4 = 0

On dividing both sides by 2, we have

(ii) We have, 4x^{2} + 4√3x + 3 = 0

**Question:**

Find the roots of the following quadratic equations by applying the quadratic formula.

(i) 2x^{2} – 7x + 3 = 0

(ii) 4x^{2} + 4√3x + 3 =0

**Solution:**

(i) We have, 2x^{2} – 7x + 3 = 0

Here, a = 2, b = -7 and c = 3

Therefore, D = b^{2} – 4ac

⇒ D = (-7)^{2} – 4 × 2 × 3 = 49 – 24 = 25

∵ D > 0, ∴ roots exist.

So, the roots of given equation are 3 and 12

(ii) We have, 4x^{2} + 4√3x + 3 = 0

Here, a = 4, b = 4√3 and c = 3

Therefore, D = b^{2} – 4ac = (4√3)^{2} – 4 × 4 × 3 = 48 – 48 = 0

∴ D = 0, roots exist and are equal.

**Question:**

Using quadratic formula solve the following quadratic equation:

p^{2}x^{2} + (p^{2} – q^{2}) x – q^{2} = 0

**Solution:**

We have, p^{2}x^{2} + (p^{2} – q^{2}) x – q^{2} = 0

Comparing this equation with ax^{2} + bx + c = 0, we have

a = p^{2}, b = p^{2} – q^{2} and c = – q^{2}

∴ D = b^{2} – 4ac

⇒ (p^{2} – q)^{2} – 4 × p^{2} × (-q^{2})

⇒ (p^{2} – q^{2})^{2} + 4p^{2}q^{2}

⇒ (p^{2} + q^{3})^{2} > 0

So, the given equation has real roots given by

**Question:**

Find the roots of the following equation:

**Solution:**

⇒ (x + 3) (x – 6)

⇒ -20 or x^{2} – 3x + 2 = 0

⇒ x^{2} – 2x -x + 2 = 0

⇒ x(x – 2) -1(x – 2) = 0)

⇒ (x – 1) (x – 2) = 0

⇒ x = 1 or x = 2

Both x = 1 and x = 2 are satisfying the given equation. Hence, x = 1, 2 are the solutions of the equation.

**Question:**

Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:

(i) 3x^{2} – 4√3x + 4 = 0) (ii) 2x^{2} – 6x + 3 = 0

**Solution:**

(i) We have, 3x^{2} – 4√3x + 4 = 1

Here, a = 3, b = – 4√3 and c = 4

Therefore,

D = b^{2} – 4ac

⇒ (- 4√3)^{2} – 4 × 3 × 4

⇒ 48 – 48 = 0

Hence, the given quadratic equation has real and equal roots.

(ii) Wehave, 2x^{2} – 6x + 3 = 0

Here, a = 2, b = -6, c = 3

Therefore, D = b^{2} – 4ac

= (-6)^{2} 4 × 2 × 3 = 36 – 24 = 12 > 0

Hence, given quadratic equation has real and distinct roots.

**Question: **Determine the condition for one root of the quadratic equation ax² + bx + c = 0 to be thrice the other.

**Solution:**

Let the roots of the given equation be a and 3α.

**Question:**

Solve for

**Solution:**

⇒ (4x – 2)(2x – 1) – (3x + 9)(x + 3) = 5(x + 3)(2x – 1)

⇒ (8x^{2} – 4x – 4x + 2) – (3x^{2} + 9x + 9x + 27) = 5(2x^{2} – x + 6x – 3)

⇒ 8x^{2} – 8x + 2 – 3x^{2} – 18x – 27 = 10x^{2} + 25x – 15

⇒ 5x^{2} – 26x – 25 = 10x^{2} + 25x – 15

⇒ 5x^{2} + 51x + 10 = 0

⇒ 5x^{2} + 50x + x + 10 = 0

⇒ 5x (x + 10) + 1 (x + 10) = 0

⇒ (5x + 1) (x + 10) = 0

⇒ 5x + 1 = 0 or x + 10 = 0

⇒ x = −15 or x = -10

**Question:**

Solve the equation

**Solution:**

⇒ (4 – 3x) (2x + 3) = 5x ⇒ 8x – 6x^{2} + 12 – 9x = 5x

⇒ 6x^{2} + 6x – 12 = 0

⇒ x^{2} + x – 2 = 0

⇒ x^{2} + 2x + x – 2 = 0

⇒ x(x + 2)-1(x + 2) = 0

⇒ (x – 1)(x + 2) = 0

⇒ x – 1 = 0 or x + 2 = 0

⇒ x = 1 or x = -2

**Question:**

Solve for

**Solution:**

⇒ (16 – x) (x + 1) = 15x

⇒ 16x – x^{2} + 16 – x = 15x

⇒ x^{2} + 15x – 15x – 16 = 0

⇒ x^{2} = 16

⇒ x = ± 4

**Question:**

Solve for x: + 5x – (a^{2} + a – 6) = 0

**Solution:**

⇒ x^{2} + 5x – (a^{2} + a – 6) = 0

⇒ x^{2} + 5x – (a? + 3a – 2a – 6) = 0

⇒ x^{2} + 5x – [a(a + 3) -2 (a + 3)] = 0

⇒ x^{2} + 5x – (a – 2) (a + 3) = 0

∴ x^{2} + (a + 3)x – (a – 2) x – (a – 2) (a + 3) = 0

⇒ x[x + (a + 3)]-(a – 2) [X + (a + 3)] = 0

⇒ [{x + (a + 3)} {x – (a – 2)}] = 0

∴ x = -(a + 3) or x = (a -2)

⇒ -(a + 3), (a – 2)

Alternative method

x^{2} + 5x -(a^{2} + a – 6) = 0

**Question:**

Solve for

**Solution:**

⇒ 2x(2x + 3) + (x – 3) + (3x + 9) = 0

⇒ 4x^{2} + 10x + 6 = 0

⇒ 2x^{2} + 5x + 3 = 0

⇒ (x + 1) (2x + 3) = 0

⇒ x = -1, x = – 32

But x ≠ – 32

∴ x = -1

**Extra Questions for Class 10 Maths Chapter 4 Long Answer Type**

**Question: **Using quadratic formula, solve the following equation for x:

abx² + (b2 – ac) x – bc = 0

**Solution:**

We have, abx^{2} + (b^{2} – ac) x – bc = 0

Here, A = ab, B = b^{2} – ac, C = – bc

**Question:**

Find the value of p for which the quadratic equation

(2p + 1)x^{2} – (7p + 2)x + (7p – 3) = 0 has equal roots. Also find these roots.

**Solution:**

Since the quadratic equation has equal roots, D = 0

i.e., b^{2} – 4ac = 0

In (2p + 1 )x^{2} – (7p + 2)x + (7p – 3) = 0

Here, a = (2p + 1), b = -(7p + 2), c = (7p – 3)

**Question:**

Solve for

**Solution:**

**Question:**

The sum of the reciprocals of Rehman’s age (in years) 3 years ago and 5 years from now is Find his present age.

**Solution:**

Let the present age of Rehman be x years.

So, 3 years ago, Rehman’s age = (x – 3) years

And 5 years from now, Rehman’s age = (x + 5) years

Now, according to question, we have

But x ≠ -3 (age cannot be negative)

Therefore, present age of Rehman = 7 years.

**Question: **A train travels 360 km at a uniform speed. If the speed has been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.

**Solution:**

Let the uniform speed of the train be x km/h.

But x cannot be negative, so x ≠ – 45

therefore, x = 40

Hence, the uniform speed of train is 40 km/h.

**Question:**

The sum of the areas of two squares is 468 m^{2}. If the difference of their perimeters is 24 m, find the sides of the two squares.

**Solution:**

Let x be the length of the side of first square and y be the length of side of the second square.

Then, x^{2} + y^{2} = 468 …(i)

Let x be the length of the side of the bigger square.

4x – 4y = 24

⇒ x – y = 6 or x = y + 6 …(ii)

Putting the value of x in terms of y from equation (ii), in equation (i), we get

(y + 6)^{2} + y^{2} = 468

⇒ y^{2} + 12y + 36 + y^{2} = 468 or 232 + 12y – 432 = 0

⇒ y^{2} + 6y – 216 = 0

⇒ y^{2} + 18y – 12y – 216 = 0

⇒ y(y + 18) – 12(y + 18) = 0

⇒ (y + 18)(y – 12) = 0

Either y + 18 = 0 or y – 12 = 0

⇒ y = -18 or y = 12

But, sides cannot be negative, so y = 12

Therefore, x = 12 + 6 = 18

Hence, sides of two squares are 18 m and 12 m.

**Question:**

Seven years ago Varun’s age was five times the square of Swati’s age. Three years hence, Swati’s age will be two-fifth of Varun’s age. Find their present ages.

**Solution:**

Seven years ago, let Swati’s age be x years. Then, seven years ago Varun’s age was 5x^{2} years.

∴ Swati’s present age = (x + 7) years

Varun’s present age = (5x^{2} + 7) years

Three years hence,

Swati’s age = (x + 7 + 3) years = (x + 10) years

Varun’s age (5x^{2} + 7 + 3) years = (5x^{2} + 10) years

According to the question,

Hence, Swati’s present age = (2 + 7) years = 9 years

and Varun’s present age = (5 × 2^{2} + 7) years = 27 years

**Question:**

A train covers a distance of 300 km at a uniform speed. If the speed of the train is increased by 5 km/hour, it takes 2 hours less in the journey. Find the original speed of the train.

**Solution:**

Let the original speed of the train = x km/h.

Therefore, the usual speed of the train = 25 km/h.

**Question:**

A two digit number is such that the product of its digits is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number.

**Solution:**

Let the digit at tens place be x.

**Question: **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 her marks would have been 210. Find her marks in the two subjects.

**Solution:**

Let Shefali’s marks in Mathematics be x.

Therefore, Shefali’s marks in English is (30 – x).

Now, according to question,

⇒ (x + 2) (30 – x – 3) = 210

⇒ (x + 2) (27 – x) = 210

⇒ 27x – x^{2} + 54 – 2x = 210

⇒ 25x – x^{2} + 54 – 210 = 0

⇒ 25x – x^{2} – 156 = 0

⇒ -(x^{2} – 25x + 156) = 0

⇒ x^{2} – 25x + 156 = 0

= x^{2} – 13x – 12x + 156 = 0

⇒ x(x – 13) – 12(x – 13) = 0

⇒ (x – 13) (x – 12) = 0

Either x – 13 or x – 12 = 0

∴ x = 13 or x = 12

Therefore, Shefali’s marks in Mathematics = 13

Marks in English = 30 – 13 = 17

or Shefali’s marks in Mathematics = 12

marks in English = 30 – 12 = 18.

**Question: **Find the positive value(s) of k for which both quadratic equations x^{2} + kx + 64 = 0 and x^{2} – 8x + k = 0 will have real roots.

**Solution: **

(i) For x^{2} + kx + 64 = 0 to have real roots

⇒ k^{2} – 4(1)(64) ≥ 0 i.e., k^{2} – 256 ≥ 0

⇒ k ≥ ± 16

(ii) For x^{2} – 8x + k = 0 to have real roots

⇒ (-8)^{2} – 4(k) ≥ 0 i.e., 64 – 4k ≥ 0

⇒ k ≤ ± 16

For (i) and (ii) to hold simultaneously k = 16

**Question:**

Speed of a boat in still water is 15 km/h. It goes 30 km upstream and returns back at the same point in 4 hours 30 minutes. Find the speed of the stream.

**Solution: **

Let the speed of stream be x km/h.

∴ Speed of boat upstream = (15 – x) km/h.

Speed of boat downstream = (15 + x) km/h.

According to question,

⇒ 30 × 2 × 30 = 9(225 – x^{2})

⇒ 100 × 2 = 225 – x^{2}

⇒ 200 = 225 – x^{2}

⇒ x^{2} = 25

⇒ x = ±5

⇒ x = 5 (Rejecting – 5)

∴ Speed of stream = 5 km/h

**Question:**

Ram takes 6 days less than Bhagat to finish a piece of work. If both of them together can finish the work in 4 days, in how many days Bhagat alone can finish the work.

**Solution:**

Let Bhagat alone can do the work in x number of days

∴ Ram takes (x – 6) number of days

Work done by Bhagat in 1 day = 1x

Work done by Ram in 1 day = 1x−6

According to the question,

**Question: **In a rectangular park of dimensions 50 m × 40 m, a rectangular pond is constructed so that the area of grass strip of uniform width surrounding the pond would be 1184 m2. Find the length and breadth of the pond.

**Solution: **

Let ABCD be rectangular lawn and EFGH be rectangular pond. Let x m be the width of grass area, which is same around the pond.

Given, Length of lawn = 50 m

Width of lawn = 40 m

Length of pond = (50 – 2x)m

Breadth of pond = (40 – 2x)m

Also given,

Area of grass surrounding the pond = 1184 m^{2}

⇒ Area of rectangular lawn – Area of pond = 1184 m^{2}

⇒ 50 × 40 – {(50 – 2x) × (40 – 2x)} = 1184

⇒ 2000 – (2000 – 80x – 100x + 4x^{2}) = 1184

⇒ 2000 – 2000 + 180x – 4x^{2} = 1184

⇒ 4x^{2} – 180x + 1184 = 0

⇒ x^{2} – 45x + 296 = 0

⇒ x^{2} – 37x – 8x + 296 = 0

⇒ x(x – 37) – 8(x – 37) = 0

⇒ (x – 37) (x – 8) = 0

⇒ x- 37 = 0 or x – 8 = 0)

⇒ x = 37 or x = 8

x = 37 is not possible as in this case length of pond becomes 50 – 2 × 37 = -24 (not possible) Hence, x = 8 is acceptable

∴ Length of pond = 50 – 2 × 8 = 34 m

Breadth of pond = 40 – 2 × 8 = 24 m

**Question:**

A car covers a distance of 2592 km with uniform speed. The number of hours taken for the journey is one-half the number representing the speed, in km/hour. Find the time taken to cover the distance.

**Solution:**

Let speed of the car be x km/h

According to question

Time taken = x2 h.

⇒ x = 72 km/h [Taking square root both sides]

∴ Time taken = x/2 = 36 hours.