8. Find the value of k for which the equation x2 + k (2x + k – 1) + 2 = 0 has real and equal roots.
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1
Answer:
k=2
Explanation:
Given, x
2
+k(2x+k−1)+2=0
Simplify above equation:
x
2
+2kx+(k
2
−k+2)=0
Compare given equation with the general form of quadratic equation, which ax
2
+bx+c=0
here, a=1,b=2k,c=(k
2
−k+2)
Find discriminant:
D=b
2
−4ac
=(2k)
2
−4×1×(k
2
−k+2)
=4k
2
−4k
2
+4k−8
=4k−8
Since roots are real and equal (given)
Put D=0
4k−8=0
k=2
Hence, the value of k is 2.
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