SOLVE QUADRATIC EQUATION USING FORMULA

There are three ways to solve a quadratic equation.

i) Using factoring

ii) Quadratic formula

iii) Completing the square.

Here we are going to see, how to solve quadratic equation using quadratic formula.

General form of quadratic equation is 

ax2 + bx + c = 0

x = -b ± b2 - 4ac2a

Use the quadratic formula to solve exactly for x :

Problem 1 :

x2 - 4x - 3 = 0

Solution :

By comparing the given quadratic equation with general form of quadratic equation,

ax2 + bx + c = 0

a = 1, b = -4 and c = -3

b2 – 4ac = (-4)2 – 4 (1) (-3)

= 16 + 12

= 28

x = -b ± b2 - 4ac2ax = -(-4) ± 282(1)x = 4 ± 272x = 22 ± 72x = 2 ± 7

Problem 2 :

x2 + 6x + 7 = 0

Solution :

By comparing the given quadratic equation with general form of quadratic equation,

ax2 + bx + c = 0

a = 1, b = 6 and c = 7

b2 – 4ac = (6)2 – 4 (1) (7)

= 36 - 28

= 8

x = -b ± b2 - 4ac2ax = -6 ± 82(1)x = -6 ± 222x = 2-3 ± 22x = -3 ± 2

Problem 3 :

x2 + 1 = 4x

Solution :

x2 + 1 = 4x

x2 - 4x + 1 = 0

By comparing the given quadratic equation with general form of quadratic equation,

ax2 + bx + c = 0

a = 1, b = -4 and c = 1

b2 – 4ac = (-4)2 – 4 (1) (1)

= 16 - 4

= 12

x = -b±b2-4ac2ax = -(-4)±122(1)x = 4±232x = 2±3

Problem 4 :

x2 + 4x = 1

Solution : 

x2 + 4x = 1

x2 + 4x – 1 = 0

By comparing the given quadratic equation with general form of quadratic equation,

ax2 + bx + c = 0

a = 1, b = 4 and c = -1

b2 – 4ac = (4)2 – 4 (1) (-1)

= 16 + 4

= 20

x = -b ± b2 - 4ac2ax = -4 ± 202(1)x = -4 ± 252x = 2-2 ± 52x = -2 ± 5

Problem 5 :

x2 – 4x + 2 = 0

Solution :

By comparing the given quadratic equation with general form of quadratic equation,

ax2 + bx + c = 0

a = 1, b = -4 and c = 2

b2 – 4ac = (-4)2 – 4 (1) (2)

= 16 - 8

= 8

x = -b ± b2 - 4ac2ax = -(-4) ± 82(1)x = 4 ± 222x = 22 ± 22x = 2 ± 2

Problem 6 :

2x2 – 2x – 3 = 0

Solution :

By comparing the given quadratic equation with general form of quadratic equation,

ax2 + bx + c = 0

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

b2 – 4ac = (-2)2 – 4 (2) (-3)

= 4 + 24

= 28

x = -b ± b2 - 4ac2ax = -(-2) ± 282(2)x = 2 ± 274x = 21 ± 74x = 1 ± 72x = 12 ± 72

Problem 7 :

x2 – 2√2x + 2 = 0

Solution :

By comparing the given quadratic equation with general form of quadratic equation,

ax2 + bx + c = 0

a = 1, b = -2√2 and c = 2

b2 – 4ac = (-2√2)2 – 4 (1) (2)

= 8 - 8

= 0

x = -b ± b2 - 4ac2ax = --22± 02(1)x = 222x = 2

Problem 8 :

(3x + 1)2 = -2x

Solution :

(3x + 1)2 = -2x

(3x)2 + 12 + 2(3x)(1) = -2x

9x2 + 1 + 6x = -2x

9x2 + 6x + 2x + 1 = 0

9x2 + 8x + 1 = 0

By comparing the given quadratic equation with general form of quadratic equation,

ax2 + bx + c = 0

a = 9, b = 8 and c = 1

b2 – 4ac = (8)2 – 4 (9) (1)

= 64 - 36

= 28

x = -b ± b2 - 4ac2ax = -8 ± 282(9)x = -8 ± 2718x = 2-4 ± 718x = -4 ± 79x = -49 ± 79

Problem 9 :

(x + 3)(2x + 1) = 9

Solution :

(x + 3)(2x + 1) = 9

2x2 + x + 6x + 3 = 9

2x2 + 7x + 3 = 9

2x2 + 7x + 3 – 9 = 0

2x2 + 7x – 6 = 0

By comparing the given quadratic equation with general form of quadratic equation,

ax2 + bx + c = 0

a = 2, b = 7 and c = -6

b2 – 4ac = (7)2 – 4 (2) (-6)

= 49 + 48

= 97

Recent Articles

RSS

  1. Finding Range of Values Inequality Problems

    May 21, 24 08:51 PM

    Finding Range of Values Inequality Problems

    Read More

  2. Solving Two Step Inequality Word Problems

    May 21, 24 08:51 AM

    Solving Two Step Inequality Word Problems

    Read More

  3. Exponential Function Context and Data Modeling

    May 20, 24 10:45 PM

    Exponential Function Context and Data Modeling

    Read More