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Question

If a,b,cR and a+b+c=0, then the quadratic equation 4ax2+3bx+2c=0 has

A
one positive and one negative
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B
imaginary roots
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C
real roots
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D
none of these
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Solution

The correct option is C real roots
4ax2+3bx+2c
Discriminant, Δ=(3b)24(4a)(2c)
Δ=9b232ac
Δ=9(a2+c2+2ac)32ac
Δ=9(a2+c2+2ac)32ac
Δ=9a2+9c2+18ac32ac
Δ=9a2+9c214ac
Δ=7(a2+c22ab)+2(a2+c2)
Δ=7(ac)2+2(a2+c2)
clearly, (ac)2>0 & a2+c2>0 a,cR
thus, Δ>0.
Then equation has real roots.

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