Frank Solutions for Class 10 Maths Chapter 6 Quadratic Equations

Frank Solutions for Class 10 Maths Chapter 6 Quadratic Equations are given here. We suggest students refer to the Frank Solutions when they get doubtful, while solving the questions from the textbook. These Solutions aim to present a strong and extensive collection of solutions for Maths. The study material is so apt and to the point, that over the years, students have been advised to practise these Frank Solutions to attain good marks in the examinations and reach success.

Chapter 6 – Quadratic Equations is defined as any equation that can be rearranged in standard form. Where x represents an unknown and a, b, c represent known numbers, where a is not equal to 0. If a is equal to 0 then the equation is linear. In this Frank Solutions for Class 10 Maths Chapter 6, students learn about solving different types of Quadratic Equations.

Access answers to Frank Solutions for Class 10 Maths Chapter 6 Quadratic Equations

1. Solve the following equation:

(x – 8) (x + 6) = 0

Solution:-

(x – 8) (x + 6) = 0

Equate both to zero,

(x – 8) = 0, (x + 6) = 0

x = 8 or x = -6

2. Solve the following equation:

(2x + 3) (3x – 7) = 0

Solution:-

(2x + 3) (3x – 7) = 0

Equate both to zero,

(2x + 3) = 0, (3x – 7) = 0

2x = -3, 3x = 7

x = -3/2 or x = 7/3

3. Solve the following equation:

4x2 + 16x = 0

Solution:-

4x2 + 16x = 0

Take out common in each terms,

4x(x + 4) = 0

Equate both to zero,

4x = 0, x + 4 = 0

x = 0, x = – 4

4. Solve the following equation:

2x2 â€“ 3x â€“ 9 = 0

Solution:-

2x2 â€“ 3x â€“ 9 = 0

Divided by 2 for both side of each term we get,

(2x2/2) â€“ (3x/2) â€“ (9/2) = 0/2

x2 â€“ 3x/2 â€“ 9/2 = 0

x2 â€“ 3x + (3/2)x â€“ 9/2 = 0

Take out common in each terms,

x(x – 3) + (3/2) (x – 3) = 0

(x + 3/2) (x – 3) = 0

Equate both to zero,

x + 3/2 = 0, x â€“ 3 = 0

x = -3/2, x = 3

5. Solve the following equation:

2x2 â€“ x â€“ 6 = 0

Solution:-

2x2 â€“ x â€“ 6 = 0

Divided by 2 for both side of each term we get,

2x2/2 â€“ x/2 â€“ 6/2 = 0

x2 â€“ x/2 â€“ 3 = 0

x2 â€“ 2x + (3/2)x â€“ 3 = 0

Take out common in each terms,

x(x – 2) +3/2 (x – 2) = 0

(x – 2) + (x + (3/2)) = 0

Equate both to zero,

x â€“ 2 = 0, x + 3/2 = 0

x = 2, x = – 3/2

6. Solve the following question:

5x2 â€“ 11x + 2 = 0

Solution:-

5x2 â€“ 11x + 2 = 0

5x2 â€“ 10x â€“ x + 2 = 0

Take out common in each terms,

5x(x – 2) â€“ 1(x – 2) = 0

(5x – 1) (x – 2) = 0

Equate both to zero,

5x â€“ 1 = 0, x â€“ 2 = 0

5x = 1, x = 2

x = 1/5, x = 2

7. Solve the following equation:

4x2 â€“ 13x â€“ 12 = 0

Solution:-

4x2 -13x â€“ 12 = 0

Divided by 4 for both side of each term we get,

4x2/4 â€“ 13x/4 â€“ 12/4 = 0/4

x2 â€“ 13x/4 â€“ 3 = 0

x2 â€“ 4x + 3x/4 â€“ 3 = 0

Take out common in each terms,

x(x – 4) + Â¾(x – 4) = 0

(x – 4) (x + Â¾) = 0

Equate both to zero,

(x – 4) = 0, x + Â¾ = 0

X = 4, x = -Â¾

8. Solve the following:

3x2 + 25x + 42 = 0

Solution:-

3x2 + 25x + 42 = 0

Divided by 3 for both side of each term we get,

3x2/3 + 25x/3 + 42/3 = 0/3

x2 + 25x/3 + 14 = 0

x2 + 6x + 7x/3 + 14 = 0

Take out common in each terms,

x(x + 6) + 7/3(x + 6) = 0

(x + 6) (x + 7/3) = 0

Equate both to zero,

x + 6 = 0, x = -7/3

9. Solve the following equation:

25x(x + 1) = -4

Solution:-

25x(x + 1) = -4

25x2 + 25x = – 4

Divided by 25 for both side of each term we get,

25x2/25 + 25x/25 = – 4/25

x2 + x = – 4/25

x2 + x + 4/25 = 0

x2 + (1/5)x + (4/5)x + 4/25 = 0

Take out common in each terms,

x(x + 1/5) +4/5(x + 1/5) = 0

(x + 4/5) (x + 1/5) = 0

Equate both to zero,

x + 4/5 = 0, x + 1/5 = 0

x = -4/5, x = -1/5

10. Solve the following equation:

10x â€“ 1/x = 3

Solution:-

10x â€“ 1/x = 3

(10x2 – 1)/x = 3

By cross multiplication we get,

10x2 â€“ 1 = 3x

10x2 â€“ 3x -1 = 0

Divided by 10 for both side of each term we get,

10x2/10 â€“ 3x/10 â€“ 1/10 = 0/10

x2 â€“ 3x/10 â€“ 1/10 = 0

x2 â€“ 1x/5 = Â½x â€“ 1/10 = 0

Take out common in each terms,

x(x + 1/5) â€“ Â½(x + 1/5) = 0

(x â€“ Â½) (x + 1/5) = 0

Equate both to zero,

x â€“ Â½ = 0, x + 1/5 = 0

x = Â½, x = -1/5

11. Solve the following equation:

2/x2 â€“ 5/x + 2 = 0

Solution:-

2/x2 â€“ 5/x + 2 = 0

Multiply by x2 for both side of each term we get,

2x2/x2 â€“ 5x2/x + 2x2 = 0

2 â€“ 5x + 2x2 = 0

Above equation can be written as,

2x2 â€“ 5x + 2 = 0

Divided by 2 for both side of each term we get,

2x2/2 â€“ 5x/2 + 2/2 = 0/2

x2 â€“ 5x/2 + 1 = 0

x2 â€“ 2x â€“ Â½x + 1 = 0

Take out common in each terms,

x(x – 2) â€“ Â½(x – 2) = 0

(x â€“ Â½) (x â€“ 2) = 0

Equate both to zero,

x â€“ Â½ = 0, x â€“ 2 = 0

x = Â½, x = 2

12. Solve the following equation:

âˆš2x2 â€“ 3x – 2âˆš2 = 0

Solution:-

âˆš2x2Â â€“ 3x – 2âˆš2 = 0

Divided by âˆš2 for both side of each term we get,

âˆš2x2/âˆš2 â€“ 3x/âˆš2 – 2âˆš2/âˆš2 = 0

x2 â€“ 3x/âˆš2 â€“ 2 = 0

x2 +(1/âˆš2)x – 2âˆš2x â€“ 2 = 0

Take out common in each terms,

x(x + 1/âˆš2) – 2âˆš2(x + 1/âˆš2) = 0

(x + 1/âˆš2) (x – 2âˆš2) = 0

Equate both to zero,

x + 1/âˆš2 = 0, x – 2âˆš2 = 0

x = -1/âˆš2, x = 2âˆš2

13. Solve the following equation:

a2x2 â€“ 3abx + 2b2 = 0

Solution:-

a2x2 â€“ 3abx + 2b2 = 0

Divided by a2 for both side of each term we get,

a2x2/a2 â€“ 3abx/a2 + 2b2/a2 = 0

x2 â€“ 3bx/a + 2(b/a)2 = 0

x2 â€“ (b/a)x â€“ 2(b/a)x + 2(b/a)2 = 0

Take out common in each terms,

x(x â€“ b/a) â€“ 2(b/a) (x â€“ b/a) = 0

(x â€“ 2(b/a)) (x â€“ b/a) = 0

Equate both to zero,

x â€“ 2(b/a) = 0, x â€“ b/a = 0

x = 2(b/a), x = b/a

14. Solve the following equation:

x2 â€“ (âˆš2 + 1)x + âˆš2 = 0

Solution:-

x2 â€“ (âˆš2 + 1)x + âˆš2 = 0

x2 â€“ x – âˆš2x + âˆš2 = 0

Take out common in each terms,

x(x – 1) – âˆš2(x – 1) = 0

(x – 1) (x – âˆš2) = 0

Equate both to zero,

x â€“ 1 = 0, x – âˆš2 = 0

x = 1, x = âˆš2

15. Solve the following equation:

x2 â€“ (âˆš3 + 1)x + âˆš3 = 0

Solution:-

x2 â€“ (âˆš3 + 1)x + âˆš3 = 0

x2 – âˆš3x â€“ x + âˆš3 = 0

Take out common in each terms,

x(x – âˆš3) â€“ 1(x – âˆš3) = 0

(x – âˆš3) (x – 1) = 0

Equate both to zero,

x – âˆš3 = 0, x â€“ 1 = 0

x = âˆš3, x = 1