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