Updated on: 2026-03-31 | Author: Aarti Kulkarni

Algebra (Math 1) Chapter 2 Quadratic Equations Solutions

Q1. Choose the correct alternative answer for the following sub questions and write the correct alphabet. Degree of quadratic equation is always ______

Answer

2

Q2. Choose the correct alternative answer for the following sub questions and write the correct alphabet. If one of the roots of quadratic equation X 2 – kX + 27 = 0 is 3, then find the value of ‘k’

Answer

12

Q3. Choose the correct alternative answer for the following sub questions and write the correct alphabet. If the root of the given quadratic equation are real and equal, then find the value of ‘k’ X 2 + 2X + k = 0

Answer

1

Q4. Choose the correct alternative answer for the following sub questions and write the correct alphabet. What is the value of discriminant for the quadratic equation X 2 – 2X – 3 = 0?

Answer

16

Q5. Choose the correct alternative answer for the following sub questions and write the correct alphabet. Which of the following is a quadratic equation?

Answer

4X2 – 3X – 5 = 0

Q6. Choose the correct alternative answer for the following sub questions and write the correct alphabet. Which of the following is not a quadratic equation?

Answer

X3 – 5X + 3 = 0

Q7. Choose the correct alternative answer for the following sub-questions and write the correct alphabet. Which of the following quadratic equation has roots – 3 and – 5?

Answer

X2 + 8X + 15 = 0

Explanation:

The roots of the quadratic equation x2 + 8x + 15 = 0 are
x2 + 5x + 3x + 15 = 0 ....`((5 xx 3 = + 15),(5 + 3 = + 8))`
x(x + 5) + 3(x + 5) = 0
(x + 5)(x + 3) = 0
∴ x + 5 = 0
x = - 5
∴ x + 3 = 0
x = - 3

Q8. Complete the following activity to find the value of discriminant for quadratic equation 4x 2 – 5x + 3 = 0. Activity: 4x 2 – 5x + 3 = 0 a = 4 , b = ______ , c = 3 b 2 – 4ac = (– 5) 2 – (______) × 4 × 3 = ( ______ ) – 48 b 2 – 4ac = ______

4x2 – 5x + 3 = 0
a = 4 , b = −5, c = 3
b2 – 4ac = (– 5)2 – 4 × 4 × 3

= 25 – 48

∴ b2 – 4ac = − 23

Q9. Complete the following activity to solve the given quadratic equation by factorization method. Activity: x 2 + 8x – 20 = 0 x 2 + ( __ ) – 2x – 20 = 0 x (x + 10) – ( __ ) (x + 10) = 0 (x + 10) ( ____ ) = 0 x = ___ or x = 2

x2 + 8x – 20 = 0
x2 + 10x – 2x – 20 = 0
∴ x (x + 10) – 2 (x + 10) = 0
∴ x (x + 10) – (x – 2) = 0
∴ x + 10 = 0 or x – 2 = 0
x = – 10 or x = 2

Q10. Complete the following activity to solve the given quadratic equation by formula method. 2x 2 + 13x + 15 = 0 Activity: 2x 2 + 13x + 15 = 0 a = (______), b = 13, c = 15 b 2 – 4ac = (13) 2 – 4 × 2 × (______) = 169 – 120 b 2 – 4ac = 49 x = `(-"b" +- sqrt("b"^2 - 4"ac"))/(2"a")` x = `(- ("______") +- sqrt(49))/4` x = `(-13 +- ("______"))/4` x = `(-6)/4` or x = `(-20)/4` x = (______) or x = (______)

2x2 + 13x + 15 = 0
a = 2, b = 13, c = 15
b2 – 4ac = (13)2 – 4 × 2 × 15

= 169 – 120

b2 – 4ac = 49

x = `(-"b" +- sqrt("b"^2 - 4"ac"))/(2"a")`

x = `(- (13) +- sqrt(49))/4`

x = `(-13 +- 7)/4`

x = `(-13 + 7)/4` or x = `(-13 - 7)/4`

x = `(-6)/4` or x = `(-20)/4`

x = `(-3)/2` or x = − 5

Q11. Complete the following activity to solve the given word problem. The Sum of squares of two consecutive even natural numbers is 244, then find those numbers. Activity: Let the first even natural number be x Therefore its consecutive even natural number will be = (______) By the given condition, x 2 + (x + 2) 2 = 244 x 2 + x 2 + 4x + 4 – (______) = 0 2x 2 + 4x – 240 = 0 x 2 + 2x – 120 = 0 x 2 + (______) – (______) – 120 = 0 x(x + 12) – (______) (x + 12) = 0 (x + 12)(x – 10) = 0 x = (______)/x = 10 But natural number cannot be negative, x = – 12 is not possible. Therefore first even natural number is x = 10. Second even consecutive natural number = x + 2 = 10 + 2 = 12.

Let the first even natural number be x.

Therefore its consecutive even natural number will be = (x + 2)

By the given condition,

x2 + (x + 2)2 = 244
x2 + x2 + 4x + 4 – 244 = 0
2x2 + 4x – 240 = 0
x2 + 2x – 120 = 0 ........[Dividing both sides by 2]
x2 + 12x – 10x – 120 = 0
x(x + 12) – 10 (x + 12) = 0
(x + 12)(x – 10) = 0
x + 12 = 0 or x - 10 = 0
x = -12/x = 10

But natural number cannot be negative, x = – 12 is not possible.

Therefore first even natural number is x = 10.
Second even consecutive natural number = x + 2 = 10 + 2 = 12.

Q12. Form a quadratic equation if the roots of the quadratic equation are `2 + sqrt(7)` and `2 - sqrt(7)`

Let α = `2 + sqrt(7)` and β = `2 - sqrt(7)`

α + β = `2 + sqrt(7) + 2 - sqrt(7)` = 4

and α × β = `(2 + sqrt(7))(2 - sqrt(7))`

= `(2)^2 - (sqrt(7))^2` ......[(a + b)(a – b) = a2 – b2]

= 4 – 7

= – 3

∴ The required quadratic equation is

x2 – (α + β)x + αβ = 0

∴ x2 – 4x – 3 = 0

Q13. Form a quadratic equation such that one of its roots is 5. Form a quadratic equation for it and write. (For the formation of word problems you can use quantities like age, rupees, or natural numbers.) (Sample solution for the above example is given below students can take another number to form another example) Solution: We need one of the solutions of the quadratic equation as 5. Then we can take another root as any number like a positive or negative number or zero. Here I am taking another root of the quadratic equation as 2. Then we can form a word problem as below, Smita is younger than her sister Mita by 3 years (5 – 2 = 3). If the product of their ages is (5 × 2 = 10). Then find their present ages. Let the age of Mita be x. Therefore age of Smita = x – 3 By the given condition, x(x – 3) = 10 x 2 – 3x – 10 = 0

Answer

Word problem:

The product of two consecutive natural numbers is 30. Find the numbers.

Final answer provided above.

Let the first natural number be x.

∴ The second consecutive natural number = x + 1

According to the given condition,

x(x + 1) = 30
∴ x2 + x = 30
∴ x2 + x – 30 = 0, which is the required quadratic equation.

Q14. If a = 1, b = 4, c = – 5, then find the value of b 2 – 4ac

b2 – 4ac = (4)2 – 4(1)(– 5)

= 16 + 20

= 36

Q15. If b 2 – 4ac > 0 and b 2 – 4ac < 0, then write the nature of roots of the quadratic equation for each given case

Answer

If b2 – 4ac > 0, then the roots are real and unequal.

If b2 – 4ac < 0, then the roots are not real.

Q16. If one of the roots of quadratic equation x 2 – kx – 15 = 0 is – 3, then find the value of ‘k’

– 3 is one of the roots of the equation x2 – kx – 15 = 0.

Putting x = – 3 in the given equation, we get

(– 3)2 – k(– 3) – 15 = 0
∴ 9 + 3k – 15 = 0
∴ 3k – 6 = 0
∴ 3k = 6
∴ k = `6/3`
∴ k = 2

Q17. If one of the roots of quadratic equation x 2 + kx + 54 = 0 is – 6, then complete the following activity to find the value of ‘k’. Activity: One of the roots of the quadratic equation x 2 + kx + 54 = 0 is – 6. Therefore let’s take x = ______ (– 6) 2 + k(– 6) + 54 = 0 (______) – 6k + 54 = 0 – 6k + ______ = 0 k = ______

One of the roots of the quadratic equation x2 + kx + 54 = 0 is – 6.
Therefore let’s take x = − 6
∴ (– 6)2 + k(– 6) + 54 = 0
∴ 36 – 6k + 54 = 0
∴ – 6k + 90 = 0
∴ 6k = 90
∴ k = `90/6`
∴ k = 15

Q18. If roots of a quadratic equation 3y 2 + ky + 12 = 0 are real and equal, then find the value of ‘k’

3y2 + ky + 12 = 0

Comparing the above equation with

ax2 + by + c = 0, we get
a = 3, b = k, c = 12

∆ = b2 – 4ac

= (k)2 – 4 × 3 × 12

= k2 – 144

= k2 – (12)2

∆ = (k + 12)(k – 12) ......[∵ a2 – b2 = (a + b)(a − b)]

Since the roots are real and equal,

∆ = 0

∴ (k + 12)(k – 12) = 0
∴ k + 12 = 0 or k – 12 = 0
∴ k = – 12 or k = 12

Q19. If the roots of a quadratic equation are 4 and – 5, then form the quadratic equation

Let α = 4 and β = – 5

α + β = 4 – 5 = – 1

and α × β = 4 × (– 5) = – 20

∴ The required quadratic equation is

x2 – (α + β)x + αβ = 0

∴ x2 – (– 1) x + (– 20) = 0
∴ x2 + x – 20 = 0

Q20. If the roots of the given quadratic equation are real and equal, then find the value of ‘k’ kx(x – 2) + 6 = 0

kx(x – 2) + 6 = 0
∴ kx2 – 2kx + 6 = 0

Comparing the above equation with

ax2 + bx + c = 0, we get
a = k, b = – 2k, c = 6

∆ = b2 – 4ac

= (–2k)2 – 4 × k × 6

= 4k2 – 24k

∴ ∆ = 4k(k – 6)

Since the roots are real and equal,

∆ = 0

∴ 4k(k – 6) = 0
∴ k(k – 6) = 0
∴ k = 0 or k – 6 = 0
But, if k = 0, then quadratic coefficient becomes zero.
∴ k ≠ 0
∴ k = 6

Q21. If the roots of the given quadratic equation are real and equal, then find the value of ‘m’. (m – 12)x 2 + 2(m – 12)x + 2 = 0

Given quadratic equation is (m – 12)x2 + 2(m – 12)x + 2 = 0
Comparing the above equation with ax2 + bx + c = 0, we get
a = m − 12, b = 2(m − 12), c = 2

∆ = b2 − 4ac

= [2(m − 12)]2 − 4 × (m − 12) × 2

= 4(m – 12)2 – 8(m – 12)

= 4(m2 – 24m + 144) – 8m + 96

= 4m2 – 96m + 576 – 8m + 96

= 4m2 – 104m + 672 ...(1)

∴ m2 – 26m + 168 = 0 ...(dividing by 4 on each side)
∴ m2 – 12m – 14m + 168 = 0 ....`[(168= - 14; -12),(- 14 xx -12 = 168),(- 14 - 12 = - 26)]`
∴ m(m – 12) – 14(m – 12)
∴ (m – 12) (m – 14) = 0
∴ m – 12 = 0 or m – 14 = 0
∴ m = 12 or m = 14
But, m = 12 is invalid because on taking m = 12, the coefficient of x2 = m – 12

= 12 – 12

= 0

Therefore, the given equation will not be a quadratic equation.
∴ m = 14
∴ The value of m is 14.

Q22. In an orchard there are total 200 trees. If the number of trees in each column is more by 10 than the number of trees in each row, then find the number of trees in each row

Let the number of trees in each row be x.

The number of trees in each column is more by 10 than the number of trees in each row.

∴ The number of trees in each column will be x + 10.

According to the given condition,

There is a total of 200 trees.

∴ x(x + 10) = 200
∴ x2 + 10x – 200 = 0
∴ x2 + 20x – 10x – 200 = 0
-200
20  -10
20 × (-10) = -200
20 - 10 = 10
∴ x(x + 20) – 10(x + 20) = 0
∴ (x + 20)(x – 10) = 0

By using the property, if the product of two numbers is zero, then at least one of them is zero, we get

x + 20 = 0 or x – 10 = 0
∴ x = – 20 or x = 10

But, the number of trees cannot be negative.

∴ x = 10
∴ The number of trees in each row is 10.

Q23. Mukund has ₹ 50 more than Sagar. If the product of the amount they have is 15,000, then find the amount each has

Let Sagar possesses ₹ x.

∴ The amount Mukund possesses = ₹ (x + 50)

According to the given condition, the product of the amount they have is ₹ 15,000.

∴ x(x + 50) = 15000
∴ x2 + 50x – 15000 = 0
-15000
150   -100
150 × (-100) = -15000
150 - 100 = 50
∴ x2 + 150x – 100x – 15000 = 0
∴ x(x + 150) – 100(x + 150) = 0
∴ (x + 150)(x – 100) = 0

By using the property, if the product of two numbers is zero, then at least one of them is zero, we get

x + 150 = 0 or x – 100 = 0
∴ x = –150 or x = 100

But, amount cannot be negative.

∴ x = 100 and x + 50 = 100 + 50 = 150
∴ The amount possessed by Sagar and Mukund are `100 and `150 respectively.

Q24. Present age of mother of Manish is 1 year more than 5 times the present age of Manish. Four years before, if the product of their ages was 22, then find the present age of Manish and his mother

Let the present age of Manish be x years.

The present age of the mother of Manish is 1 year more than 5 times the present age of Manish.

∴ The present age of the mother of Manish will be (5x + 1) years.

Four years before,

Manish’s age was (x – 4) years and

Mother’s age was (5x + 1 – 4) years i.e., (5x – 3) years

According to the given condition,

Four years before, the product of their ages was 22.

∴ (x – 4)(5x – 3) = 22
∴ 5x2 – 3x – 20x + 12 – 22 = 0
∴ 5x2 – 23x – 10 = 0
5 × (-10) =  -50
  +25  +2
  (-25) × 2 = -50
  -25 + 2 = -23
∴ 5x2 – 25x + 2x – 10 = 0
∴ 5x(x – 5) + 2(x – 5) = 0
∴ (x – 5)(5x + 2) = 0

By using the property, if the product of two numbers is zero, then at least one of them is zero, we get

x – 5 = 0 or 5x + 2 = 0
∴ x = 5 or x = `(-2)/5`

But, age cannot be negative.

∴ x = 5
∴ Manish’s present age = x = 5 years,
Mother’s present age = 5x + 1

= 5(5) + 1

= 25 + 1

= 26 years

Q25. Roots of a quadratic equation are 5 and – 4, then form the quadratic equation

Let α = 5 and β = – 4

α + β = 5 – 4 = 1

and α × β = 5 × (– 4) = – 20

∴ The required quadratic equation is

x2 – (α + β)x + αβ = 0

∴ x2 – (1) x + (– 20) = 0
∴ x2 – x – 20 = 0

Q26. Solve the following quadratic equation. `1/(4 - "p") - 1/(2 + "p") = 1/4`

`1/(4 - "p") - 1/(2 + "p") = 1/4`
∴ `(2 + "p" - (4 - "p"))/((4 - "p")(2 + "p")) = 1/4`
∴ `(2 + "p" - 4 + "p")/(8 + 4"p" - 2"p" - "p"^2) = 1/4`
∴ `(2"p" - 2)/(8 + 2"p" - "p"^2) = 1/4`
∴ 4(2p – 2) = 8 + 2p – p2
∴ 8p – 8 = 8 + 2p – p2
∴ p2 – 2p + 8p – 8 – 8 = 0
∴ p2 + 6p – 16 = 0
-16
8  -2
8 × (-2) = -16
8 - 2 = 6
∴ p2 + 8p – 2p – 16 = 0
∴ p(p + 8) – 2(p + 8) = 0
∴ (p + 8)(p – 2) = 0

By using the property, if the product of two numbers is zero, then at least one of them is zero, we get

p + 8 = 0 or p – 2 = 0
∴ p = – 8 or p = 2
∴ The roots of the given equation are – 8 and 2.

Q27. Solve the following quadratic equation. `sqrt(3) x^2 + sqrt(2)x - 2sqrt(3)` = 0

`sqrt(3) x^2 + sqrt(2)x - 2sqrt(3)` = 0

Comparing the above equation with ax2 + bx + c = 0, we get

a = `sqrt(3)`, b = `sqrt(2)`, c = `-2sqrt(3)`

∴ b2 – 4ac = `(sqrt(2))^2 - 4 xx sqrt(3) xx (-2sqrt(3))`

= 2 + 24

= 26

x = `(-"b" +- sqrt("b"^2 - 4"ac"))/(2"a")`

= `(-sqrt(2) +- sqrt(26))/(2sqrt(3))`

= `(sqrt(2)(-1 +- sqrt(13)))/(2sqrt(3))`

= `(sqrt(2)(-1 +- sqrt(13)))/(2sqrt(3)) xx sqrt(2)/(sqrt(2)`

= `(2(-1 +- sqrt(13)))/(2sqrt(6))`

∴ x = `(-1 +- sqrt(13))/sqrt(6)`
∴ x = `(-1 + sqrt(13))/sqrt(6)` or x = `(-1 - sqrt(13))/sqrt(6)`
∴ The roots of the given equation are `(-1 + sqrt(13))/sqrt(6)` and x = `(-1 - sqrt(13))/sqrt(6)`.

Q28. Solve the following quadratic equation by factorization method. 3p 2 + 8p + 5 = 0

3p2 + 8p + 5 = 0
∴ 3p2 + 5p + 3p + 5 = 0
3 × 5 =  15
  5 3
  5 × 3 = 15
5 + 3 = 8
∴ p(3p + 5) + 1(3p + 5) = 0
∴ (3p + 5) (p + 1) = 0

By using the property, if the product of two numbers is zero, then at least one of them is zero, we get

3p + 5 = 0 or p + 1 = 0
∴ 3p = – 5 or p = – 1
∴ p = `(-5)/3` or p = – 1
∴ The roots of the given quadratic equation are `(-5)/3` and – 1

Q29. Solve the following quadratic equation by the formula method: `y^2 + 1/3y = 2`

`y^2 + 1/3y` = 2

∴ 3y2 + y = 6 ......[Multiplying both sides by 3]
∴ 3y2 + y – 6 = 0
Comparing the above equation with ay2 + by + c = 0, we get
a = 3, b = 1, c = – 6
∴ b2 – 4ac = (1)2 – 4 × 3 × (– 6)

= 1 + 72

= 73

y = `(-"b" +- sqrt("b"^2 - 4"ac"))/(2"a")`

= `(-1 +- sqrt(73))/(2(3))`

∴ y = `(-1 +- sqrt(73))/6`
∴ y = `(-1 + sqrt(73))/6` or y = `(-1 - sqrt(73))/6`
∴ The roots of the given quadratic equation are `(-1 + sqrt(73))/6` and `(-1 - sqrt(73))/6`.

Q30. Solve the following quadratic equations by formula method. 5m 2 – 4m – 2 = 0

5m2 – 4m – 2 = 0
Comparing the above equation with am2 + bm + c = 0, we get
a = 5, b = – 4, c = – 2
∴ b2 – 4ac = (– 4)2 − 4 × 5 × (– 2)

= 16 + 40

= 56

m = `(-"b" +- sqrt("b"^2 - 4"ac"))/(2"a")`

= `((-4) +- sqrt(56))/(2(5))`

= `(4 +- sqrt(4 xx 14))/10`

= `(4 +- 2sqrt(14))/10`

= `(2(2 +- sqrt(14)))/10`

∴ m = `(2 +- sqrt(14))/5`
∴ m = `(2 + sqrt(14))/5` or m = `(2 - sqrt(14))/5`
∴ The roots of the given quadratic equation are `(2 + sqrt(14))/5` and `(2 - sqrt(14))/5`

Q31. Sum of the roots of the quadratic equation is 5 and sum of their cubes is 35, then find the quadratic equation

Let α and β be the roots of the quadratic equation.

According to the given conditions,

α + β = 5 and α3 + β3 = 35

Now, (α + β)3 = α3 + 3α2β + 3αβ2 + β3

∴ (α + β)3 = α3 + β3 + 3αβ(α + β)
∴ (5)3 = 35 + 3αβ(5)
∴ 125 = 35 + 15αβ
∴ 125 – 35 = 15αβ
∴ 15αβ = 90
∴ αβ = `90/15`
∴ αβ = 6
∴ The required quadratic equation is

x2 − (α + β)x + αβ = 0

∴ x2 − 5x + 6 = 0

Q32. To decide whether 1 is a root of quadratic equation x 2 + 4x – 5 = 0 or not, complete the following activity. Activity: When x = (______) L.H.S. = 1 2 + 4(______) – 5 = 1 + 4 – 5 = (______) – 5 = ______ = R.H.S Therefore x = 1 is a root of quadratic equation x 2 + 4x – 5 = 0

When x = 1,

L.H.S. = 12 + 4(1) – 5

= 1 + 4 – 5

= 5 – 5

= 0

= R.H.S.

Therefore x = 1 is a root of quadratic equation x2 + 4x – 5 = 0.

Q33. Write the given quadratic equation in standard form and also write the values of a, b and c 4y 2 – 3y = – 7

Answer

Given equation is 4y2 – 3y = – 7

∴ 4y2 – 3y + 7 = 0
Comparing the above equation with ay2 + by + c = 0, we get
a = 4, b = – 3, c = 7

Q34. Write the given quadratic equation in standard form. m (m – 6) = 9

Answer

m (m – 6) = 9
∴ m2 – 6m = 9
∴ m2 – 6m – 9 = 0

Q35. Write the roots of following quadratic equation. (p – 5) (p + 3) = 0

Answer

(p – 5) (p + 3) = 0
∴ p – 5 = 0 or p + 3 = 0
∴ p = 5 or p = – 3
∴ The roots of the given equation are 5 and – 3.

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