For the equation 3x^2 + px + 3 = 0, p 0,if | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the roots be $\alpha$ and $\alpha^2$.From the relation between roots and coefficients for $3x^2 + px + 3 = 0$:Sum of roots: $\alpha + \alpha^2

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) Let the roots be $\alpha$ and $\alpha^2$.From the relation between roots and coefficients for $3x^2 + px + 3 = 0$:Sum of roots: $\alpha + \alpha^2 = -p/3$Product of roots: $\alpha \cdot \alpha^2 = \alpha^3 = 3/3 = 1$Since $\alpha^3 = 1$,the possible values for $\alpha$ are $1, \omega, \omega^2$,where $\omega$ is a complex cube root of unity.If $\alpha = 1$,then $\alpha + \alpha^2 = 1 + 1 = 2$,so $2 = -p/3 \implies p = -6$. But we are given $p > 0$,so this case is rejected.If $\alpha = \omega$ or $\alpha = \omega^2$,then $\alpha + \alpha^2 = \omega + \omega^2 = -1$.Substituting this into the sum of roots equation: $-1 = -p/3 \implies p = 3$.Thus,$p = 3$.

List-$I$	List-$II$
$(i) \alpha = \beta$	$(A) (ac^2)^{1/3} + (a^2c)^{1/3} + b = 0$
$(ii) \alpha = 2\beta$	$(B) 2b^2 = 9ac$
$(iii) \alpha = 3\beta$	$(C) b^2 = 6ac$
$(iv) \alpha = \beta^2$	$(D) 3b^2 = 16ac$
	$(E) b^2 = 4ac$
	$(F) (ac^2)^{1/3} + (a^2c)^{1/3} = b$

For the equation $3x^2 + px + 3 = 0, p > 0$,if one of the roots is the square of the other,then $p$ is equal to:

Similar Questions

If the roots of the equation $x^2 + bx + c = 0$ are reciprocals of each other,then...

If $\alpha, \beta, \gamma$ are the roots of $x^3-6x^2+11x-6=0$,then the equation having the roots $\alpha^2+\beta^2, \beta^2+\gamma^2$ and $\gamma^2+\alpha^2$ is

If $a(p + q)^2 + 2bpq + c = 0$ and $a(p + r)^2 + 2bpr + c = 0$,then $qr$ =

Let $\alpha$ and $\beta$ be the roots of the equation $5x^{2} + 6x - 2 = 0$. If $S_{n} = \alpha^{n} + \beta^{n}$ for $n = 1, 2, 3, \dots$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module