for what value of k,(4-k)x²+(2k+4)x+(8k+1)=0 is a perfect square.
Answers
Answered by
30
hello dear,
b2-4ac=0(given)
(2k+4)2-4(4-k)(8k+1)=0
4k2+16k+16-4(32k+4-8k2-k)=0
4k2+16k+16-128k-16+32k2+4k=0
36k2-108k=0 (Dividing the equation by 4)
9k2-27k=0
9k2=27k
9k=27k/k
9k=27
K=27/9
K=3
hope it will will help you. :-)
b2-4ac=0(given)
(2k+4)2-4(4-k)(8k+1)=0
4k2+16k+16-4(32k+4-8k2-k)=0
4k2+16k+16-128k-16+32k2+4k=0
36k2-108k=0 (Dividing the equation by 4)
9k2-27k=0
9k2=27k
9k=27k/k
9k=27
K=27/9
K=3
hope it will will help you. :-)
Rahulnegi1:
thanks
Answered by
5
Given, (4 – k)x² + (2k + 4)x + (8k + 1) = 0
It is in the form of ax² + bx + c = 0
Where, a = 4 – k, b = 2k + 4, c = 8k + 1
Calculating the discriminant,
D = b²– 4ac
= (2k + 4)² – 4(4 – k)(8k + 1)
= 4k² + 16 + 4k – 4(32 + 4 – 8k² – k)
= 4(k² + 4 + k – 32 – 4 + 8k² + k)
= 4(9k² – 27k)
As the given equation is a perfect square, then D = 0
⇒ 4(9k² – 27k) = 0
⇒ (9k² – 27k) = 0
⇒ 3k(k – 3) = 0
Thus, 3k = 0
⇒ k = 0 Or k-3 = 0
⇒k = 3
Hence, the value of k should be 0 or 3 for the given to be perfect square.
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