Question

What will be the nature of roots of quadratic equation 2x2 + 4x – n = 0​

Answer 1

Answer:

Step-by-step explanation:

Equation is 2$x^{2}$ +4x - n = 0

General equation of quadratic equation is a$x^{2}$+bx +c=0

On comparing two equation we get,

a= 2 , b= 4 and c= - n

As we know that,

D = $b^{2}$ - 4ac

   = $4^{2}$ - 4 × 2 ×(- n)

   = 16 +8 n

If value n >0 then D > 0 then roos are real

If - 2<n<0 then D> o then roots are real

If n< -2 then D < 0 then roots are not real

Answer 2

Nature of roots for 2x² + 4x – n = 0

Roots are real and distinct if  n < 2

Roots are real and equal if n = 2

Roots are imaginary ( not real ) and different if  n > 2

Given:

• 2x² + 4x – n = 0

To Find:

• Nature of roots

Solution:

• Quadratic equation is of the form ax²+bx+c=0  where a  , b and c are real also  a≠0.
• D =  b²-4ac is called discriminant.
• D >0 roots are real and distinct
• D =0 roots are real and equal
• D < 0 roots are imaginary ( not real ) and different

Step 1:

Comparing 2x² + 4x – n = 0​ with ax²+bx+c=0

a = 2

b = 4

c = -n

Step 2:

Calculate Discriminant D

D =  b²-4ac

D = 4²- 4(2)(-n)

D = 16 - 8n

D = 8(2 - n)

Step 3:

Find Nature of roots

Roots are real and distinct if 8(2 - n) > 0  => n < 2

Roots are real and equal if 8(2 - n) = 0  => n = 2

Roots are imaginary ( not real ) and different if  8(2 - n)  < 0 => n > 2