if the roots of the equation 3x^2+2x+a^2-a=0 are of opposite signs then a lies in
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Step-by-step explanation:
The given quadratic equation is
3x^2+2x+(p+2)(p-1)=03x
2
+2x+(p+2)(p−1)=0
If the roots are\alpha, \betaα,β , the products of the roots is given by
\alpha\beta=\frac{(p+2)(p-1)}{3}αβ=
3
(p+2)(p−1)
If the roots are of opposite sign this procduct should be less than zero
Therefore,
\alpha\beta<0αβ<0
\implies (p+2)(p-1)<0⟹(p+2)(p−1)<0
\implies -2
Therefore, the value of p which is outside this range will be the answer
From the given options this value is -3−3
Hence Option (d) is correct.
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