If one of the roots of the equation x^2+px+q= | vedclass

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

Question

(D) Given equation is $x^2+px+q=0$ ... $(i)$Comparing with $ax^2+bx+c=0$,we get $a=1, b=p, c=q$.Let the roots be $\alpha$ and $\beta$. Given $\alpha =

Vedclass · Accepted Answer

$p(3q-p^2)=q(q+1)$

Vedclass · Answer

(D) Given equation is $x^2+px+q=0$ ... $(i)$Comparing with $ax^2+bx+c=0$,we get $a=1, b=p, c=q$.Let the roots be $\alpha$ and $\beta$. Given $\alpha = \beta^2$.From the relation between roots and coefficients:Sum of roots: $\alpha + \beta = -p \Rightarrow \beta^2 + \beta = -p$ ... (ii)Product of roots: $\alpha \beta = q$ $\Rightarrow \beta^2 \cdot \beta = q$ $\Rightarrow \beta^3 = q$ ... (iii)Cubing equation (ii) on both sides:$(\beta^2 + \beta)^3 = (-p)^3$$\beta^6 + \beta^3 + 3\beta^2 \cdot \beta(\beta^2 + \beta) = -p^3$Substitute $\beta^3 = q$ and $\beta^2 + \beta = -p$:$(q)^2 + q + 3q(-p) = -p^3$$q^2 + q - 3pq = -p^3$$p^3 - 3pq = -q^2 - q$$p(p^2 - 3q) = -q(q+1)$$p(3q - p^2) = q(q+1)$

If one of the roots of the equation $x^2+px+q=0$ is equal to the square of the other,then:

Similar Questions

If $\alpha, \beta$ and $\gamma$ are the roots of the equation $x^3+3x^2-7x+5=0$,then the value of $\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}$ is

If the difference between the roots of the equation $(k - 2)x^2 - (k - 4)x - 2 = 0$ is $3$, find the value of $k$.

Let $\alpha$ and $\beta$ be the roots of $x^{2}-6x-2=0$. If $a_{n}=\alpha^{n}-\beta^{n}$ for $n \geq 1$,then the value of $\frac{a_{10}-2a_{8}}{3a_{9}}$ is ..... .

If $\alpha \ne \beta$ but $\alpha^2 = 5\alpha - 3$ and $\beta^2 = 5\beta - 3$,then the equation whose roots are $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$ is

The coefficient of $x$ in the equation $x^2 + px + q = 0$ was taken as $17$ in place of $13$. Its roots were found to be $-2$ and $-15$. The roots of the original equation are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module