Question

Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula:
i. 2x² + x – 4 = 0 ii. 4x² + 4 3x + 3 = 0
iii. 5x² − 7x − 6 = 0 iv. x² + 5 = −6x

Answer 1

if a quadratic equation ax² + bx + c = 0 is given then, x = $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

i. 2x² + x - 4 = 0
here , a = 2 , b = 1 and c = -4
then, x = {- 1 ± √(1² - 4 × 2 × -4)}/2 × 2
x = {-1 ± √(1 + 32)}/4
x = (-1 + √33)/4 and (-1 - √33)/4

ii. 4x² + 4√3x + 3 = 0
here, a = 4 , b = 4√3 and c = 3
so, x = [-4√3 ± √{(4√3)² - 4(3)(4)}]/2.4
x = [-4√3 ± √{48 - 48}]/8
x = -4√3/8 = -√3/2
hence, both roots are equal e.g., -√3/2

iii. 5x² - 7x - 6 = 0
here, a = 5 , b = -7 and c = -6
so, x = [7 ± √{(-7)² - 4(-6)(5)}]/2.5
x = [7 ± √{49 + 120}]/10
x = (7 ± 13)/10
x = -6/10 , 2

iv. x² + 5 = 6x
x² - 6x + 5 = 0
here, a = 1 , b = -6 and c = 5
so, x = [6 ± √{(-6)²-4(5)(1)}]/2
x = [6 ± √{36 - 20}]/2
x = [6 ± 4]/2
x = 5 , 1

Answer 2

GIVEN :

i. 2x² + x – 4 = 0  

ii. 4x² + 4 3x + 3 = 0

iii. 5x² − 7x − 6 = 0  

iv. x² + 5 = −6x

SOLUTION :

i. 2x² + x - 4 = 0

On comparing with ax² + bx + c = 0 , We get

a = 2 , b = 1 and c = - 4  

On substituting the values of a, b and c in the quadratic formula :  

x = -b ±√b² - 4ac / 2a

x = - 1 ± √(1² - 4 × 2 × - 4) / (2 × 2)  

x = -1 ± √(1 + 32) / 4

x =  (√33 -1)/4 and  (- √33 -1)/4  

Hence,the roots of the quadratic equation are x =  (√33 -1)/4 and  (- √33 -1)/4

ii. 4x² + 4√3x + 3 = 0

On comparing with ax² + bx + c = 0 , We get

a = 4 , b = 4√3 and c = 3

On substituting the values of a, b and c in the quadratic formula :  

x = -b ±√b² - 4ac / 2a

x = - 4√3 ± √(4√3)² - 4×3×4 /(2 ×4)

x = - 4√3 ± √48 - 48 / 8  

x = - 4√3/8 = -√3/2  

Hence,the roots of the quadratic equation x = -√3/2(both roots are equal )

iii. 5x² - 7x - 6 = 0

On comparing with ax² + bx + c = 0 , We get

a = 5 , b = -7 and c = -6

On substituting the values of a, b and c in the quadratic formula :  

x = -b ±√b² - 4ac / 2a

so, x = [-(-7) ± √(-7)² - 4 × -6 × 5] / (2 ×5)

x =[ 7 ± √49 + 120] / 10

x = 7 ± √169  / 10

x = 7 ± 13 /10  

x = (7 - 13) / 10 = -6/10= -⅗

x = (7 + 13) / 10 = 20/10= 2

x = -3/5 , 2  

Hence,the roots of the quadratic equation are x = -3/5 , 2

iv. x² + 5 = 6x  

x² - 6x + 5 = 0  

On comparing with ax² + bx + c = 0 , We get

a = 1 , b = -6 and c = 5

On substituting the values of a, b and c in the quadratic formula :  

x = -b ±√b² - 4ac / 2a

so, x = [-(-6) ± √(-6)²- 4 ×5 ×1] /(2× 1)

x = [6 ± √{36 - 20}]/2

x = [6 ± 4]/2  

x = (6 + 4) /2 = 10/2 = 5

x = (6 -  4) /2 = 2 /2 = 1

x = 5 , 1

Hence,the roots of the quadratic equation are x = 5 , 1

HOPE THIS ANSWER WILL HELP YOU..