write the set value of k for which quadratic equation 2x^2 + kx + 8 =0 has real root
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Answered by
105
Hi ,
2x² + kx + 8 = 0 compare this
quadratic equation with ax²+bx +c = 0
a = 2 , b = k , c = 8
It is given that , the equation has
real roots ,
Discreaminant = D ≥ 0
b² - 4ac ≥ 0
k² - 4 × 2 × 8 ≥ 0
k² ≥ 64
k ≥ ± √64
k ≥ ± 8
I hope this helps you.
: )
2x² + kx + 8 = 0 compare this
quadratic equation with ax²+bx +c = 0
a = 2 , b = k , c = 8
It is given that , the equation has
real roots ,
Discreaminant = D ≥ 0
b² - 4ac ≥ 0
k² - 4 × 2 × 8 ≥ 0
k² ≥ 64
k ≥ ± √64
k ≥ ± 8
I hope this helps you.
: )
Answered by
7
Answer:
k = 8
Step-by-step explanation:
given equation : 2x²+kx+8 which is in the form of ax²+bx+c=0
so, a=2, b=k, c=8
D=√b²-4ac
- k²-√4×2×8
- k² - √ 64
- k ± √64
- k = ± 8
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