CG Board Class 10 Maths Chapter 3 Solutions: Quadratic Equations in One Variable (एक चर का द्विघात समीकरण) (2026-27)

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CG Board (CGBSE / SCERT Chhattisgarh) Class 10 Mathematics Solutions | 2026-27 Edition

📚 Chapter Overview:

  • Board: Chhattisgarh Board of Secondary Education (CGBSE) / CG SCERT Raipur
  • Subject & Unit: Class 10 Mathematics – Unit 1: Algebra (2026-27 Edition)
  • Chapter 3: Quadratic Equations in One Variable (एक चर का द्विघात समीकरण)
  • Medium: Bilingual Step-by-Step Solutions (English Medium + हिंदी माध्यम)
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  • *Quadratic Equation in One Variable

An algebraic equation of degree 2 in one variable is called a quadratic equation.

Standard Form

ax2 + bx + c = 0 (a ≠ 0, a, b, c ∈ R)

  • *Roots / Zeroes of a Quadratic Equation

The values of the variable x for which both sides of the quadratic equation become equal are called its roots.

  • *Discriminant

For a quadratic equation ax2 + bx + c = 0, the expression b2 – 4ac is called its Discriminant (D).

D = b2 – 4ac

  • *Nature of Roots
  • Case 1: If D > 0, roots are real and distinct.
  • Case 2: If D = 0, roots are real and equal.
  • Case 3: If D < 0, roots are imaginary / not real.
  • *Quadratic Formula / Sridharacharya Method

x = (-b ± √(b2 – 4ac))/(2a) = (-b ± √D)/(2a)

  • *Relation between Roots and Coefficients

If α and β are roots of ax2 + bx + c = 0:

  • *Sum of roots

α + β = -b/a = -(Coefficient of x)/(Coefficient of x2)

  • *Product of roots

α · β = c/a = (Constant term)/(Coefficient of x2)

  • *Forming Quadratic Equation from Roots

x2 – (Sum of roots)x + (Product of roots) = 0

x2 – (α + β)x + α β = 0

Q1

Select square quadratic equations from the following equations


📝 Question:

Select square (quadratic) equations from the following equations

💡 Solution:

(i) x2 + 3x – 2 = 0**

Maximum exponent of variable x is 2. It is in standard form ax2 + bx + c = 0.

Quadratic Equation*.

  • (ii) x2 + 1/x = 1

Multiplying both sides by x: x3 + 1 = x ⇒ x3 – x + 1 = 0. Degree is 3.

Not a Quadratic Equation*.

  • (iii) 9x2 – 100x – 20 = 0

Maximum exponent of variable x is 2.

Quadratic Equation*.

  • (iv) x2 – 3√x + 2 = 0

Exponent of x in √x is 1/2, which is not an integer. Thus, it is not a quadratic polynomial/equation.

Not a Quadratic Equation*.

  • (v) x2 – 2x = -x(2 – x)

x2 – 2x = -2x + x2 ⇒ 0 = 0. It is an identity, not a degree 2 equation.)*.

Quadratic Equation*.

  • (vi) x2 – 10/3x + 2 = 0

Maximum exponent of variable x is 2.

Quadratic Equation*.

  • (vii) x2 – 10 = 0

Maximum exponent of variable x is 2.

Quadratic Equation*.

  • (viii) x + y = 10

It contains two variables x and y with degree 1.

Not a Quadratic Equation*.

  • (ix) x + 5 = 7

It is a linear equation in one variable (degree 1).

Not a Quadratic Equation*.

  • (x) x(x – 8) = 0

x2 – 8x = 0. Maximum exponent of variable x is 2.

Quadratic Equation*.

Q1

Check whether given values are roots of the…


📝 Question:

Check whether given values are roots of the given equations

💡 Solution:

(i) (x – 4)(x – 2) = 0; values x = 2, x = 4**

  • For x = 2: (2 – 4)(2 – 2) = (-2)(0) = 0 = RHS.
  • For x = 4: (4 – 4)(4 – 2) = (0)(2) = 0 = RHS.

Both x = 2 and x = 4 are roots**

  • (ii) (2x + 3)(3x – 7) = 0; values x = -3/2, x = 7/3
  • For x = -3/2: [2(-3/2) + 3][3(-3/2) – 7] = (-3 + 3)(-9/2 – 7) = 0 = RHS.
  • For x = 7/3: [2(7/3) + 3][3(7/3) – 7] = (14/3 + 3)(7 – 7) = 0 = RHS.

Both x = -3/2 and x = 7/3 are roots**

  • (iii) (x – 7)2 = 0; value x = 7
  • For x = 7: (7 – 7)2 = 02 = 0 = RHS.

x = 7 is a root**

  • (iv) x2 – 11x = 0; values x = 0, x = 11
  • For x = 0: 02 – 11(0) = 0 = RHS.
  • For x = 11: 112 – 11(11) = 121 – 121 = 0 = RHS.

Both x = 0 and x = 11 are roots**

  • (v) (x + 12)2 = 0; value x = -12
  • For x = -12: (-12 + 12)2 = 02 = 0 = RHS.

x = -12 is a root**

  • (vi) x(x + 1) = 0; values x = 0, x = -1
  • For x = 0: 0(0 + 1) = 0 = RHS.
  • For x = -1: -1(-1 + 1) = -1(0) = 0 = RHS.

Both x = 0 and x = -1 are roots**

Q2

Verify whether given values of x are roots…


📝 Question:

Verify whether given values of x are roots of the equations or not

💡 Solution:

(i) x2 – 5x + 6 = 0; values x = 2, x = 3**

  • LHS for x = 2: 22 – 5(2) + 6 = 4 – 10 + 6 = 0 = RHS.
  • LHS for x = 3: 32 – 5(3) + 6 = 9 – 15 + 6 = 0 = RHS.

x = 2, 3 are roots**

  • (ii) x2 + 2x – 4 = 0; values x = 2, x = -3
  • LHS for x = 2: 22 + 2(2) – 4 = 4 + 4 – 4 = 4 ≠ 0.
  • LHS for x = -3: (-3)2 + 2(-3) – 4 = 9 – 6 – 4 = -1 ≠ 0.

Neither x = 2 nor x = -3 are roots**

  • (iii) x2 – 3√3x + 6 = 0; values x = -3, x = 4
  • LHS for x = -3: (-3)2 – 3√3(-3) + 6 = 9 + 9√3 + 6 ≠ 0.
  • LHS for x = 4: 42 – 3√3(4) + 6 = 16 – 12√3 + 6 ≠ 0.

Neither x = -3 nor x = 4 are roots**

  • (iv) 2x2 – 9x + 4 = 0; values x = 1/2, x = 4
  • LHS for x = 1/2: 2(1/2)2 – 9(1/2) + 4 = 2(1/4) – 9/2 + 4 = 1/2 – 9/2 + 4 = -4 + 4 = 0 = RHS.
  • LHS for x = 4: 2(4)2 – 9(4) + 4 = 2(16) – 36 + 4 = 32 – 36 + 4 = 0 = RHS.

Both x = 1/2 and x = 4 are roots**

  • (v) x2 – 3√3x + 6 = 0; values x = √3, x = -2√3
  • LHS for x = √3: (√3)2 – 3√3(√3) + 6 = 3 – 9 + 6 = 0 = RHS.
  • LHS for x = -2√3: (-2√3)2 – 3√3(-2√3) + 6 = 12 + 18 + 6 = 36 ≠ 0.

x = √3 is a root, but x = -2√3 is not a root**

Q3

Is 2 a root of equation xsup2sup 2x…


📝 Question:

Is 2 a root of equation x2 + 2x – 4 = 0?

💡 Solution:

No, 2 is not a root**

Q4

Find the roots of the following quadratic equations…


📝 Question:

Find the roots of the following quadratic equations by factorization

💡 Solution:

(i) 9x2 – 3x – 2 = 0**

Splitting middle term (A · C = -18, sum = -3 ⇒ -6, +3)

9x2 – 6x + 3x – 2 = 0 ⇒ 3x(3x – 2) + 1(3x – 2) = 0

(3x – 2)(3x + 1) = 0 ⇒ 3x – 2 = 0 or 3x + 1 = 0

x = 2/3, -1/3**

  • (ii) 4x2 + 5x = 0

Taking x common

x(4x + 5) = 0 ⇒ x = 0 or 4x + 5 = 0

x = 0, -5/4**

  • (iii) 3x2 – 11x + 10 = 0

Splitting middle term (A · C = 30, sum = -11 ⇒ -6, -5)

3x2 – 6x – 5x + 10 = 0 ⇒ 3x(x – 2) – 5(x – 2) = 0

(x – 2)(3x – 5) = 0 ⇒ x = 2 or x = 5/3

x = 2, 5/3**

  • (iv) 5x2 + 3x – 2 = 0

Splitting middle term (A · C = -10, sum = +3 ⇒ +5, -2)

5x2 + 5x – 2x – 2 = 0 ⇒ 5x(x + 1) – 2(x + 1) = 0

(x + 1)(5x – 2) = 0 ⇒ x = -1 or x = 2/5

x = -1, 2/5**

  • (v) 6x2 + 7x + 2 = 0

Splitting middle term (A · C = 12, sum = 7 ⇒ 4, 3)

6x2 + 4x + 3x + 2 = 0 ⇒ 2x(3x + 2) + 1(3x + 2) = 0

(3x + 2)(2x + 1) = 0 ⇒ x = -2/3 or x = -1/2

x = -2/3, -1/2**

  • (vi) 4√3x2 + 5x – 2√3 = 0

Product A · C = 4√3 × (-2√3) = -24, sum = +5 ⇒ +8, -3

4√3x2 + 8x – 3x – 2√3 = 0 ⇒ 4x(√3x + 2) – √3(√3x + 2) = 0

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

x = √3/4, -2/√3**

  • (vii) 10x – 1/x = 3

Multiply by x: 10x2 – 1 = 3x ⇒ 10x2 – 3x – 1 = 0.

Splitting middle term (A · C = -10, sum = -3 ⇒ -5, +2)

10x2 – 5x + 2x – 1 = 0 ⇒ 5x(2x – 1) + 1(2x – 1) = 0

(2x – 1)(5x + 1) = 0 ⇒ x = 1/2 or x = -1/5

x = 1/2, -1/5**

  • (viii) x2 – 4√2x + 6 = 0

Product = 6, sum = -4√2 ⇒ -3√2, -√2

x2 – 3√2x – √2x + 6 = 0 ⇒ x(x – 3√2) – √2(x – 3√2) = 0

(x – 3√2)(x – √2) = 0 ⇒ x = 3√2 or x = √2

x = 3√2, √2**

  • (ix) abx2 + (b2 – ac)x – bc = 0

abx2 + b2 x – acx – bc = 0 ⇒ bx(ax + b) – c(ax + b) = 0

(ax + b)(bx – c) = 0 ⇒ x = -b/a or x = c/b

x = -b/a, c/b**

  • (x) x+1/x-1 – x-1/x+1 = 5/6

((x+1)2 – (x-1)2)/((x-1)(x+1)) = 5/6 ⇒ (4x)/(x2 – 1) = 5/6

24x = 5x2 – 5 ⇒ 5x2 – 24x – 5 = 0

5x2 – 25x + x – 5 = 0 ⇒ 5x(x – 5) + 1(x – 5) = 0

(x – 5)(5x + 1) = 0 ⇒ x = 5 or x = -1/5

x = 5, -1/5**

Q1

Solve the following quadratic equations by completing the…


📝 Question:

Solve the following quadratic equations by completing the square method

💡 Solution:

(i) 2x2 + x – 4 = 0**

Dividing by 2: x2 + 1/2x – 2 = 0 ⇒ x2 + 2(1/4)x = 2.

Adding (1/4)2 = 1/16 to both sides

(x + 1/4)2 = 2 + 1/16 = 33/16 ⇒ x + 1/4 = ± √33/4

x = (-1 + √33)/(4), (-1 – √33)/(4)**

  • (ii) 3x2 + 11x + 10 = 0

Dividing by 3: x2 + 11/3x = -10/3.

Adding (11/6)2 = 121/36 to both sides

(x + 11/6)2 = -10/3 + 121/36 = (-120 + 121)/(36) = 1/36

x + 11/6 = ± 1/6 ⇒ x = -11/6 ± 1/6

x = -5/3, -2**

  • (iii) 5x2 – 6x – 2 = 0

Dividing by 5: x2 – 6/5x = 2/5.

Adding (3/5)2 = 9/25 to both sides

(x – 3/5)2 = 2/5 + 9/25 = 19/25 ⇒ x – 3/5 = ± √19/5

x = (3 + √19)/(5), (3 – √19)/(5)**

  • (iv) x2 – 4√2x + 6 = 0

x2 – 2(2√2)x = -6. Adding (2√2)2 = 8 to both sides

(x – 2√2)2 = -6 + 8 = 2 ⇒ x – 2√2 = ± √2

Taking (+) sign: x = 3√2; Taking (-) sign: x = √2.

x = 3√2, √2**

  • (v) 3x2 + 2x – 1 = 0

Dividing by 3: x2 + 2/3x = 1/3.

Adding (1/3)2 = 1/9 to both sides

(x + 1/3)2 = 1/3 + 1/9 = 4/9 ⇒ x + 1/3 = ± 2/3

Taking (+) sign: x = 1/3; Taking (-) sign: x = -1.

x = 1/3, -1**

  • (vi) x2 – 4x + 3 = 0

x2 – 4x = -3. Adding 22 = 4 to both sides

(x – 2)2 = -3 + 4 = 1 ⇒ x – 2 = ± 1

Taking (+) sign: x = 3; Taking (-) sign: x = 1.

x = 3, 1**

Q2

Find the solution of 7 7 7


📝 Question:

Find the solution of √(7 + √(7 + √(7 + …)))

💡 Solution:

x = (1 + √29)/(2)**

Q3

Find two consecutive positive natural numbers the sum…


📝 Question:

Find two consecutive positive natural numbers, the sum of whose squares is 365

💡 Solution:

Numbers are 13 and 14**

Q4

The product of two consecutive natural numbers is…


📝 Question:

The product of two consecutive natural numbers is 20. Find the numbers

💡 Solution:

Numbers are 4 and 5**

Q5

Find two numbers whose sum is 48 and…


📝 Question:

Find two numbers whose sum is 48 and product is 432

💡 Solution:

Numbers are 36 and 12**

Q6

Area of a rightangled triangle is 165 sq…


📝 Question:

Area of a right-angled triangle is 165 sq m. If the height of the triangle is 7 m more than its base, find the height

💡 Solution:

Base = 15 m, Height = 22 m*

Q7

Perimeter and area of a flower bed are…


📝 Question:

Perimeter and area of a flower bed are 76 m and 357 sq m respectively. Find length and breadth

💡 Solution:

21 meters and 17 meters*

Q8

Cost of sugar increases by Rs 5 per…


📝 Question:

Cost of sugar increases by Rs. 5 per kg, so a person gets 5 kg less sugar for Rs. 250. Find original rate

💡 Solution:

Original rate = Rs. 10 per kg*

Q9

Present age of a man is 4 times…


📝 Question:

Present age of a man is 4 times the square of his son’s age. 9 years hence father’s age will be 2 times son’s age. Find present ages

💡 Solution:

36 years and 3 years*

Q10

Age difference between two brothers is 3 years…


📝 Question:

Age difference between two brothers is 3 years. Product of their ages is 130. Find present ages

💡 Solution:

13 years and 10 years*

Q11

Picnic food budget was Rs 480 8 students…


📝 Question:

Picnic food budget was Rs. 480. 8 students failed to go, so food cost per member increased by Rs. 10. How many attended?

💡 Solution:

16 students*

Q12

Sum of marks in English and Maths is…


📝 Question:

Sum of marks in English and Maths is 40. If Maths marks increase by 3 and English marks decrease by 4, product is 360. Find marks

💡 Solution:

Maths = 21, English = 19 OR Maths = 12, English = 28*

Q1

Find the discriminant of the following equations


📝 Question:

Find the discriminant of the following equations

💡 Solution:

(i) x2 – 4x + 2 = 0**

a = 1, b = -4, c = 2. D = b2 – 4ac = (-4)2 – 4(1)(2) = 16 – 8 = 8.

D = 8**

  • (ii) (x – 1)(2x – 1) = 0

2x2 – 3x + 1 = 0 ⇒ a = 2, b = -3, c = 1.

D = (-3)2 – 4(2)(1) = 9 – 8 = 1.

D = 1**

  • (iii) √3x2 + 2√2x – 2√3 = 0

a = √3, b = 2√2, c = -2√3.

D = (2√2)2 – 4(√3)(-2√3) = 8 + 24 = 32.

D = 32**

  • (iv) x2 – 4x + a = 0

a’ = 1, b = -4, c = a. D = (-4)2 – 4(1)(a) = 16 – 4a.

D = 16 – 4a**

  • (v) x2 + px + q = 0

a = 1, b = p, c = q. D = p2 – 4(1)(q) = p2 – 4q.

D = p2 – 4q**

Q2

Find the nature of roots of the following…


📝 Question:

Find the nature of roots of the following quadratic equations

💡 Solution:

(i) x2 – 4x + 4 = 0**

D = (-4)2 – 4(1)(4) = 16 – 16 = 0.

Real and equal roots*.

  • (ii) 2x2 + 2x + 2 = 0

D = (2)2 – 4(2)(2) = 4 – 16 = -12 < 0.

Roots are not real / Imaginary*.

  • (iii) 3x2 – 2√6x + 2 = 0

D = (-2√6)2 – 4(3)(2) = 24 – 24 = 0.

Real and equal roots*.

  • (iv) x2 + 2√5x – 1 = 0

D = (2√5)2 – 4(1)(-1) = 20 + 4 = 24 > 0.

Real and distinct roots*.

  • (v) 3/5x2 – 2/3x + 1 = 0

D = (-2/3)2 – 4(3/5)(1) = 4/9 – 12/5 = -88/45 < 0.

Roots are not real / Imaginary*.

Q3

Find the value of k if the roots…


📝 Question:

Find the value of k if the roots of the given equations are real and equal (D = 0)

💡 Solution:

(i) 2x2 – 10x + k = 0**

D = (-10)2 – 4(2)(k) = 0 ⇒ 100 – 8k = 0 ⇒ k = 100/8 = 25/2.

k = 25/2**

  • (ii) kx2 – 5x + k = 0

D = (-5)2 – 4(k)(k) = 0 ⇒ 25 – 4k2 = 0 ⇒ k2 = 25/4 ⇒ k = ± 5/2.

k = ± 5/2**

  • (iii) 2x2 + kx + 9/8 = 0

D = k2 – 4(2)(9/8) = 0 ⇒ k2 – 9 = 0 ⇒ k = ± 3.

k = ± 3**

  • (iv) 9x2 – kx + 16 = 0

D = (-k)2 – 4(9)(16) = 0 ⇒ k2 – 576 = 0 ⇒ k = ± 24.

k = ± 24**

  • (v) kx2 + 4x + 1 = 0

D = 42 – 4(k)(1) = 0 ⇒ 16 – 4k = 0 ⇒ k = 4.

k = 4**

Q4

Solve the following quadratic equations with the help…


📝 Question:

Solve the following quadratic equations with the help of formula

💡 Solution:

(i) 9x2 + 7x – 2 = 0**

a = 9, b = 7, c = -2. D = 72 – 4(9)(-2) = 49 + 72 = 121.

x = (-7 ± √(121))/(2(9)) = (-7 ± 11)/(18) ⇒ x = 4/18 = 2/9 or x = -18/18 = -1

x = 2/9, -1**

  • (ii) 6x2 + x – 2 = 0

a = 6, b = 1, c = -2. D = 12 – 4(6)(-2) = 1 + 48 = 49.

x = (-1 ± 7)/(12) ⇒ x = 6/12 = 1/2 or x = -8/12 = -2/3

x = 1/2, -2/3**

  • (iii) 6x2 + 7x – 10 = 0

a = 6, b = 7, c = -10. D = 49 – 4(6)(-10) = 49 + 240 = 289.

x = (-7 ± 17)/(12) ⇒ x = 10/12 = 5/6 or x = -24/12 = -2

x = 5/6, -2**

  • (iv) 2x2 – 9x + 7 = 0

a = 2, b = -9, c = 7. D = (-9)2 – 4(2)(7) = 81 – 56 = 25.

x = (9 ± 5)/(4) ⇒ x = 14/4 = 7/2 or x = 4/4 = 1

x = 7/2, 1**

  • (v) x2 – 7x – 5 = 0

a = 1, b = -7, c = -5. D = (-7)2 – 4(1)(-5) = 49 + 20 = 69.

x = (7 + √69)/(2), (7 – √69)/(2)**

  • (vi) 4x2 – 11x – 3 = 0

a = 4, b = -11, c = -3. D = (-11)2 – 4(4)(-3) = 121 + 48 = 169.

x = (11 ± 13)/(8) ⇒ x = 24/8 = 3 or x = -2/8 = -1/4

x = 3, -1/4**

  • (vii) 9x2 – 4 = 0

a = 9, b = 0, c = -4. D = 0 – 4(9)(-4) = 144.

x = (0 ± 12)/(18) ⇒ x = ± 2/3

x = 2/3, -2/3**

  • (viii) 3x2 – 10x – 8√3 = 0 (Textbook printing variant: √3x2 – 10x + 3√3 = 0 or 3x2 – 10x – 8√3 = 0)

For 3x2 – 10x – 8√3 = 0: a = 3, b = -10, c = -8√3.

x = 4√3, -2/√3 (12/√3, -2/√3)**

  • (ix) 2x2 + x – 6 = 0

a = 2, b = 1, c = -6. D = 12 – 4(2)(-6) = 1 + 48 = 49.

x = (-1 ± 7)/(4) ⇒ x = 6/4 = 3/2 or x = -8/4 = -2

x = 3/2, -2**

  • (x) 2x2 – 2√6x + 3 = 0

a = 2, b = -2√6, c = 3. D = (-2√6)2 – 4(2)(3) = 24 – 24 = 0.

x = (2√6 ± 0)/(4) = √6/2

x = √6/2, √6/2**

Q1

Form the quadratic equation whose sum and product…


📝 Question:

Form the quadratic equation whose sum and product of roots are given below: Formula*: x2 – (Sum)x + (Product) = 0

💡 Solution:

(i) Sum = -4, Product = 12**

x2 – (-4)x + 12 = 0 ⇒ x2 + 4x + 12 = 0

x2 + 4x + 12 = 0**

  • (ii) Sum = 6, Product = -9

x2 – (6)x + (-9) = 0 ⇒ x2 – 6x – 9 = 0

x2 – 6x – 9 = 0**

  • (iii) Sum = 2√7, Product = 8

x2 – (2√7)x + 8 = 0 ⇒ x2 – 2√7x + 8 = 0

x2 – 2√7x + 8 = 0**

  • (iv) Sum = 4/9, Product = 1

x2 – 4/9x + 1 = 0 ⇒ 9x2 – 4x + 9 = 0

9x2 – 4x + 9 = 0**

Q2

Find the quadratic equation whose roots are given…


📝 Question:

Find the quadratic equation whose roots are given below

💡 Solution:

(i) 7, 4**

Sum = 7 + 4 = 11, Product = 7 × 4 = 28.

x2 – 11x + 28 = 0

x2 – 11x + 28 = 0**

  • (ii) -5, -11

Sum = (-5) + (-11) = -16, Product = (-5)(-11) = 55.

x2 – (-16)x + 55 = 0 ⇒ x2 + 16x + 55 = 0

x2 + 16x + 55 = 0**

  • (iii) -2, 4

Sum = -2 + 4 = 2, Product = (-2)(4) = -8.

x2 – 2x – 8 = 0

x2 – 2x – 8 = 0**

  • (iv) 12, -24

Sum = 12 + (-24) = -12, Product = 12(-24) = -288.

x2 – (-12)x – 288 = 0 ⇒ x2 + 12x – 288 = 0

x2 + 12x – 288 = 0**

  • (v) 4/5, -3/5

Sum = 4/5 – 3/5 = 1/5, Product = (4/5)(-3/5) = -12/25.

x2 – 1/5x – 12/25 = 0 ⇒ 25x2 – 5x – 12 = 0

x2 – 1/5x – 12/25 = 0 (or 25x2 – 5x – 12 = 0)**

  • (vi) 4, 4

Sum = 4 + 4 = 8, Product = 4 × 4 = 16.

x2 – 8x + 16 = 0

x2 – 8x + 16 = 0**

  • (vii) -1/3, 2/5

Sum = -1/3 + 2/5 = 1/15, Product = (-1/3)(2/5) = -2/15.

x2 – 1/15x – 2/15 = 0 ⇒ 15x2 – x – 2 = 0

15x2 – x – 2 = 0**

  • (viii) 8, 3

Sum = 8 + 3 = 11, Product = 8 × 3 = 24.

x2 – 11x + 24 = 0

x2 – 11x + 24 = 0**

  • (ix) 3 – √7, 3 + √7

Sum = (3 – √7) + (3 + √7) = 6.

Product = (3 – √7)(3 + √7) = 32 – (√7)2 = 9 – 7 = 2.

x2 – 6x + 2 = 0

x2 – 6x + 2 = 0**

  • (x) 6 + √5, 6 – √5

Sum = (6 + √5) + (6 – √5) = 12.

Product = (6 + √5)(6 – √5) = 36 – 5 = 31.

x2 – 12x + 31 = 0

x2 – 12x + 31 = 0**

Q3

Find the sum and product of roots of…


📝 Question:

Find the sum and product of roots of the following quadratic equations: Formula*: For ax2 + bx + c = 0, Sum (α + β) = -b/a, Product (α β) = c/a

💡 Solution:

(i) 3x2 + 7x + 1 = 0**

a = 3, b = 7, c = 1.

Sum = -7/3, Product = 1/3**

  • (ii) 2x2 – 2x + 3 = 0

a = 2, b = -2, c = 3.

Sum = -(-2)/2 = 1, Product = 3/2.

Sum = 1, Product = 3/2**

  • (iii) 3x2 – 5x – 2 = 0

a = 3, b = -5, c = -2.

Sum = -(-5)/3 = 5/3, Product = -2/3.

Sum = 5/3, Product = -2/3**

  • (iv) 2x2 – 2√6x + 3 = 0

a = 2, b = -2√6, c = 3.

Sum = -(-2√6)/2 = √6, Product = 3/2.

Sum = √6, Product = 3/2**

  • (v) x2 + 6x – 6 = 0

a = 1, b = 6, c = -6.

Sum = -6/1 = -6, Product = -6/1 = -6.

Sum = -6, Product = -6**

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