Question

If the roots of equation 3x2+2x+(p+2)(p-1)=0 are of opposite sign then which of thefollowing cannot be the value of p?a)0 b)-1 c)12d)-3

Answer 1

Option (d) -3

Step-by-step explanation:

The given quadratic equation is

$3x^2+2x+(p+2)(p-1)=0$

If the roots are$\alpha, \beta$, the products of the roots is given by

$\alpha\beta=\frac{(p+2)(p-1)}{3}$

If the roots are of opposite sign this procduct should be less than zero

Therefore,

$\alpha\beta<0$

$\implies (p+2)(p-1)<0$

$\implies -2<p<1$

Therefore, the value of p which is outside this range will be the answer

From the given options this value is $-3$

Hence Option (d) is correct.

Hope this answer was helpful.

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Answer 2

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.

Hope this answer was helpful