find the value of k if the given set has real roots 4x^2-3kx+1=0
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for real root
D ≥0
b²-4ac ≥ 0
(-3k)²-4*4*1≥0
9k²≥16
k ² ≥9/16
k = ≥3/4
D ≥0
b²-4ac ≥ 0
(-3k)²-4*4*1≥0
9k²≥16
k ² ≥9/16
k = ≥3/4
deepanshu3333:
but l have only answer not solution
u got solutions na
hmm
there's no need that the ans given in ur book match perfectly
ok
9k^2≥16 and k^2≥4/3
aapka ye kaise aa gaya ki k^2≥3/4
dekho 9 k²≥16 hai ,then 9is multiply here on replacement 9 on other side it divide 16
then u got k²≥9/16 then u take underoot
at last u got k ≥3/4
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