Question

In the following, determine whether the given quadratic equations have real roots and if so, find the roots:
(x) √2x²+7x+5√2=0
(xi)2x²-2√2x+1=0
(xii)3x² − 5x + 2 = 0

Answer 1

SOLUTION :  

(x) Given : √2x² + 7x + 5√2 = 0

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

Here, a = √2 , b = 7, c =  5√2

Discriminant , D = b² -  4ac

D = (7)² - 4 × √2 × 5√2

D = 49 - 40

D = 9

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

x = [- b ± √D]/2a

x= [ −7  ± √9] / 2 × √2

x = [- 7 ±  3 ] / 2√2

x =  [- 7 +  3 ] / 2√2  or x = [- 7 -  3 ] / 2√2

x =  - 4 /2√2  or  x = - 10/2√2

x = - 2/√2  or  x = - 5/√2

x = (- 2 × √2) / (√2 × √2)   or x = - 5/√2

[By rationalising the denominator]

x = - 2√2 / 2 or   x = - 5/√2

x = - √2  or    x = - 5/√2

Hence, the Roots are  - √2 and - 5√2 .

 

(xi) Given : 2x² - 2√2x + 1 = 0

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

Here, a = 2 , b = -  2√2 , c =  1

Discriminant , D = b² -  4ac

D = (-2√2)² - 4 × 2 × 1

D = 8 - 8

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√2)  ± √0 ] / 2 × 2

x = 2√2 /4

x = √2/2

x = (√2)/ (√2 × √2)

x = 1/√2  

x = 1/√2  or x = 1/√2

Hence, the Roots are  1/√2 and 1/√2

 

(xii) Given : 3x² - 5x + 2 = 0

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

Here, a = 3 , b = - 5 , c = 2

Discriminant , D = b² -  4ac

D = (- 5)² - 4 × 3 × 2

D = 25  - 24

D = 1

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

x = [- b ± √D]/2a

x = [- (−5)  ± √1] / 2 × 3

x = (5  ± 1) /6

x =  (5  + 1) /6  or x =  (5  - 1) /6

x =  6/6 or x = 4/6

x = 1 or x = 2/3

Hence, the Roots are  1  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 ,

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.

____________________

Now,

x) Compare given

quadratic equation

√2x²+7x+5√2=0 with

ax²+bx+c = 0 ,we get

a = √2 ,b = 7 ,c = 5√2

D = 7² - 4×√2×5√2

= 49 - 40

= 9

= 3²

D > 0

Therefore ,

Roots are real and distinct.

(xi)2x²-2√2x+1=0,

a = 2 , b = -2√2 , c = 1

D = (-2√2)²-4×2×1

= 8 - 8

= 0

D = 0

Therefore ,

Roots are real and equal.

(xii)3x² − 5x + 2 = 0

a = 3 , b = -5 , c = 2

D = (-5)² - 4×3×2

= 25 - 24

= 1

D > 0

Therefore ,

Roots are real and distinct.

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