SOLVING QUADRATIC EQUATIONS BY FORMULA METHOD 

Solving Quadratic Equations by Formula Method :

In this section, how we solve quadratic equation using formula.

How to use quadratic formula to solve quadratic equation ?

(i) Compare the given quadratic equation with the general form of quadratic equation ax² + bx + c = 0

(ii) Here the coefficient of x² is a, coefficient of x is b and the constant term is c.

(iii) Then apply those values in the formula

Solving Quadratic Equations by Formula Method Examples

Question 1 :

Solve by using quadratic formula

x² – 7 x + 12  =  0

Solution :

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

ax² + bx + c  =  0

a  =  1, b  =  -7 and c  =  12

b² – 4ac  =  (-7)² - 4(1) (12)

  =  49 - 48

  =  1

x  =  -b ± √(b² – 4ac)/2a

x  =  [ 7 ± √1 ] / 2(1)

x  =  [ 7 ± √1 ] / 2

Hence the solution is { 3, 4 }.

Question 2 :

Solve by using quadratic formula

15x² – 11 x + 2  =  0

Solution :

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

ax² + bx + c  =  0

a  =  15, b  =  -11 and c  =  2

b² – 4ac  =  (-11)² - 4(15) (2)

  =  121 - 120

  =  1

x  =  -b ± √(b² – 4ac)/2a

x  =  [ 11 ± √1 ] / 2(15)

x  =  [ 11 ± √1 ] / 30

Hence the solution is { 1/3, 2/5 }.

Question 3 :

Solve by using quadratic formula

x + (1/x) = 2  1/2

Solution :

(x² + 1)/x  =  5/2

2(x² + 1)  =  5x

2x² + 2  =  5x

2x² – 5x + 2  =  0

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

ax² + bx + c  =  0

a  =  2, b  =  -5 and c  =  2

b² – 4ac  =  (-5)² - 4(2) (2)

  =  25 - 16

  =  9

x  =  -b ± √(b² – 4ac)/2a

x  =  [ 5 ± √9 ] / 2(4)

x  =  [ 5 ± 3 ] / 8

Hence the solution is { 1, 1/4 }.

Question 4 :

Solve by using quadratic formula

3a²x² – abx  - 2b²  =  0

Solution :

3 a²x² – a b x - 2b²  =  0

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

ax² + bx + c  =  0

a  =  3a², b  =  -ab and c  =  -2b²

b² – 4ac  =  (-ab)² - 4(3a²) (- 2b²)

  =  a²b² + 24a²b²

  =  25a²b²

x  =  -b ± √(b² – 4ac)/2a

x  =  [ ab ± √(25a²b²) ] / 2(3a²)

x  =  [ ab ± 5ab ] / 6a²

After having gone through the stuff given above, we hope that the students would have understood how to solve quadratic equations using quadratic formula.

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