Question

if the roots od the quadratic equation 2kx²-3kx+k+1=0 are equal find the value /s of k​

Answer 1

Step-by-step explanation:

Given:-

the roots od the quadratic equation

2kx²-3kx+k+1 =0 are equal.

To find:-

find the value /s of k

Solution:-

Given quadratic equation is 2kx²-3kx+k+1 =0

It can be written as (2k)x²+(-3k)x+(k+1)= 0

On comparing with the standard quadratic equation ax²+bx+c =0

we have ,

a = 2k

b = -3k

c = (k+1)

given that

the roots of the given equation are equal then

=>Discriminant must be equal to zero

We know that

The discriminant of the quadratic equation ax²+bx+c=0 is b²-4ac

=>b²-4ac=0

=>(-3k)²-4(2k)(k+1)=0

=>9k²-8k(k+1)=0

=>9k²-8k²-8k=0

=>(9k²-8k²)-8k=0

=>k²-8k=0

=>k(k-8)=0

=>k=0 or k-8=0

=>k=0 or k=8

If k=0 then the quadratic equation doesn't exist

Therefore k=8

Answer:-

The value of k=8 for the given problem

Check:-

If k=8 then the equation becomes

=2(8)x²-3(8)x+8+1=0

=>16x²-24x+9=0

=>x²-24x/16 +9/16 =0

=>x²-(3/2)x+(9/16)=0

=>x²-2(3/4)x=-9/16

=>x²-2(x)(3/4)+(3/4)²=(3/4)²-(9/16)

=>[x-(3/4)]²=(9/16)-(9/16)

=>[x-(3/4)]²=0

=>x-(3/4)=0

=>x=3/4

The roots are 3/4 and 3/4

They are equal for k=8

Used formula:-

The quadratic equation ax²+bx+c=0 then the discriminant is D=b²-4ac then

• If D>0 then the roots are real and distinct.
• If D=0 then the roots are real and equal.
• If D<0 then the roots are imaginary .i.e. no real

Answer 2

Step-by-step explanation:

a=2I,b=-3k,c=(k+1)

BY USING DISCRIMINANT METHOD,

D=b²- 4ac

(-3k)²-4(2k)(k+1)

9k²-8k(k+1)

9k²-8k(k+1)

9k²-8k²-8k

k²-8k

k(k-8)=0

k=0,k=8

k=8 answer

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