Question

In the following, determine whether the given quadratic equations have real roots and if so, find the roots:
(vii)3x²+2√5x-5=0
(viii)x² − 2x + 1 = 0
(ix)2x²+5√3x+6=0

Answer 1

SOLUTION :  

(vii) Given : 3x² + 2√5x - 5 = 0

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

Here, a = 3 , b = 2√5 , c = - 5

Discriminant , D = b² -  4ac

D = (2√5)² - 4 × 3 × - 5

D = 20 + 60

D = 80

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

x = [- b ± √D]/2a

x= [ −(2√5)  ±√80] / 2 × 3

x = (-2√5  ±√16 × 5) /6

x = (- 2√5 ± 4√5) / 6

x = 2(-√5 ± 2√5) / 6

x = (-√5 ± 2√5) / 3

x = (- √5 + 2√5)/3  or x = (- √5 - 2√5)/3

x = √5/3 or x = -3√5/3 = - √5

Hence, the Roots are  √5/(3) and - √5 .

 

(viii) Given : x² - 2x + 1 = 0

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

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

Discriminant , D = b² -  4ac

D = (-2)² - 4 × 1 × 1

D = 4 - 4  

D = 0

Since, D = 0, so given Quadratic equation has two equal and real roots which are given by

x = [- b ± √D]/2a

x = [( −2)  ± √0] / 2 × 1

x = - 2/2  

x = -1

x = -1  or x  = - 1

Hence, the Roots are - 1  and - 1 .

 

(ix) Given : 2x² + 5√3x + 6 = 0

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

Here, a = 2 , b = 5√3 , c =  6

Discriminant , D = b² -  4ac

D = (5√3)² - 4 × 2 × 6

D = 75 - 48

D = 27

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

x = [- b ± √D]/2a

x= [ −(5√3)  ± √27 ] / 2 × 2

x = [- 5√3 ± √9 × 3 ] / 4

x = [- 5√3 ± 3√3] / 4

x = [- 5√3 + 3√3] / 4  or x =[- 5√3 - 3√3] / 4

x = - 2√3/4 or x = - 8√3/4

x = - √3/2  or x = - 2√3

Hence, the Roots are  - √3/2 and - 2√3 .

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

Solution :

_______________________

Nature of roots of a

quadratic equation

ax²+bx+c=0, a≠0 is

Discreminant (D)

= b²-4ac

If i ) D > 0

Roots are real and distinct.

ii ) D = 0 ,

Roots are real and equal.

iii) D<0

Roots are not real.

_________________________

(vii)3x²+2√5x-5=0 Compare

this with ax²+bx+c=0,

we get

a = 3 , b = 2√5 , c = -5

D = b² - 4ac

= (2√5)² - 4×3×(-5)

= 20 + 60

= 80

D > 0

Therefore,

Roots are real and distinct.

(viii)x² − 2x + 1 = 0

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

D = (-2)² - 4×1××1

= 4 - 4

D = 0

Therefore ,

Roots are real and equal.

(ix)2x²+5√3x+6=0

a = 2 , b = 5√3 , c = 6

D = (5√3)² - 4 × 2 × 6

= 75 - 48

= 27

D >0

Therefore,

Roots are real and distinct.

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