In this section, you will learn how to solve quadratic equation using quadratic formula.

To use quadratic formula, the quadratic equation must be in the form of

ax2 + bx + c  =  0

We can substitute the values of a, b and c into the formula shown below and solve the quadratic equation given.

(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

-b ± √ (b²-4ac)/2a

Let us see some example problems on solving quadratic equation by using quadratic formula.

Example 1 :

x2 - 2x - 24 = 0

Solution :

By comparing the above equation with the general form of a quadratic equation ax2 + bx + c = 0, we get

a = 1     b = -2    c = -24

 =  -b ± √ (b2- 4ac)/2a  =  -(-2) ± √(4-(-96))/2(1)  =  2 ± √(4+96)/2  =  (2 ± √100)/2  =  (2 ± √(10 x 10))/2  =  (2 ± 10)/2 b2  = (-2)2 ==> 44ac = 4(1)(-24)   =   -96
 =  (2 + 10)/2=  12/2=  6 =  (2 - 10)/2=  -8/2=  -4

Hence the solutions are 6 and -4.

Example 2 :

x2 - 2x - 15  =  0

Solution :

By comparing the above equation with the general form of a quadratic equation ax2 + bx + c = 0, we get

 a = 1     b = -2    c = -15  =  -b ± √ (b2- 4ac)/2ab2- 4ac = 4-(-60) =  4 + 60 =  64  =  -(-2) ± √64/2(1)  =  (2 ± 8)/2 b2  = (-2)2 ==> 44ac = 4(1)(-15)   =   -60
 =  (2 + 8)/2=  10/2=  5 =  (2 - 8)/2=  -4/2=  -2

Hence the solutions are 5 and -2.

Example 3 :

x + (1/x) = 2 ½

Solution :

(x² + 1)/x = 5/2

(x² + 1) = 5 x

2 x² + 2 = 5 x

x² - 5 x + 2 = 0

By comparing the above equation with the general form of a quadratic equation ax2 + bx + c = 0, we get

x = 8/4 , 2/4

x = 2 , 1/2

Hence the solutions are 2 and 1/2

Example 4 :

3 a²x² - ab x - 2b² = 0

Solution :

By comparing the above equation with the general form of a quadratic equation ax2 + bx + c = 0, we get

a = 3 a²   b = - ab   c = - 2b²

x = 6ab/6a²  , -4ab/6a²

x = b/a , x = -2b/a

Hence the solutions are b/a and -2b/a

Example 5 :

a (x² + 1) =  x(a² + 1)

Solution :

a x² + a =  x(a² + 1)

a x² - x(a² + 1) + a = 0

By comparing the above equation with the general form of a quadratic equation ax2 + bx + c = 0, we get

a = a     b = -(a² + 1)   c = a

x = 2a²/2a , 2/2a

= a , 1/a

Hence the solutions are a and 1/a.

Example 6 :

[(x - 1)/(x + 1)] + [(x - 3)/(x - 4)] = 10/3

Solution :

[(x - 1)(x - 4) + (x - 3)(x + 1)]/(x + 1)(x- 4) = 10/3

(x - 1)(x - 4)  =  x² - 5 x + 4  ---(1)

(x - 3)(x + 1)  =  x² - 2 x - 3  -----(2)

(x + 1)(x- 4)  =  (x² - 3 x - 4) -----(3)

(1) + (2) ==>x²-5x+4+x²-2x-3 = 2x²-7x+1  -----(4)

(4)/(3)  ==>  (2x²-7x+1)/(x² - 3 x - 4)  =  10/3

3 (2x² - 7 x + 1) = 10(x² - 3 x - 4)

10 x² - 6 x² - 30 x + 21 x - 40 - 3 = 0

4 x² - 9 x - 43 = 0

By comparing the above equation with the general form of a quadratic equation ax2 + bx + c = 0, we get

a = 4     b = -9    c = -43

 =  -b ± √ (b2- 4ac)/2ab2-4ac = 81 - (-688)= 81 + 688b2-4ac  =  769  =  -(-9) ± √769/2(1)  =  -(-9) ± √769/2(1)  = (9 ± √769)/2 b2 = (-9)2 = 814ac= 4(4)(-43)  =  -688

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

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