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Question

If the equation 4x3+5x+k=0(k R) has a negative real root then k will be

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Solution

Let f(x)=4x3+5x+k
f(x)=12x2+5>0xR
f(x) is strictly increasing on R.
So, for f(x)=0 to have a negative real root, f(0)>0k>0.
1111500_993027_ans_7a5918bc87fd457d9d1cdd9b116975c9.png

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