Question

find whether following are roots of quadratic equation or not x^2 + 2x -2 ; x= root 3 - 1 & x=root 3 +1

Answer 1

Answer:

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Answer 2

Answer:

⋆Answer:−

(i) Given : 16x² = 24x + 1

16x² – 24x – 1 = 0

On comparing the given equation with, ax² + bx + c = 0

Here, a = 16 , b = - 24 , c = - 1

Discriminant , D = b² - 4ac

D = (-24)² - 4(16)(-1)

D = 576 + 64

D = 640

Since, D ≥ , 0 so given Quadratic equation has distinct real roots which are given by

x = [- b ± √D]/2a

x= −(−24) ±√640/ 2(16)

x =[ 24 ± 8√10] /32

x = 8 [3 ± √10]/32

x= (3 ± √10) /√4

x = (3 + √10) /√4 or x = (3 - √10) /√4

Hence, the Roots are (3 + √10) /√4 and x = (3 - √10) /√4.

(ii)Given : x² + x + 2 = 0

On comparing the given equation with, ax² + bx + c = 0

Here, a = 1 , b = 1 , c = 2

Discriminant , D = b² - 4ac

D = (1)² - 4(1)(2)

D = 1 - 8

D = - 7

Since, D < 0 so given Quadratic equation has no real roots .

(iii) Given : √3x² + 10x - 8√3 = 0

On comparing the given equation with, ax² + bx + c = 0

Here, a = √3 , b = 10 , c = - 8√3

Discriminant , D = b² - 4ac

D = (10)² - 4(√3)(- 8√3)

D = 100 + 32 × 3

D = 100 + 96

D = 196

Since, D ≥ , 0 so given Quadratic equation has distinct real roots which are given by

x = [- b ± √D]/2a

x= −10 ±√196/ 2(√3)

x =( - 10 ± 14)/2√3

x = 2( - 5 ± 7)/2√3

x = ( - 5 ± 7)/√3

x = (-5 + 7)/√3 or x = (- 5 - 7)/√3

x = 2/√3 or x = - 12/√3

x = - 12/√3 = (- 4 ×3)√3/√3×√3 = (- 4 ×3)√3 /3

[By rationalising the denominator]

x = - 4√3

x = 2/√3 or x = - 4√3

Hence, the Roots are 2/√3 and - 4√3 .

Step-by-step explanation:

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