If the ratio of the roots of x^2 + bx + c = 0 | vedclass

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

Question

(C) Let $\alpha, \beta$ be the roots of $x^2 + bx + c = 0$ and $\alpha', \beta'$ be the roots of $x^2 + qx + r = 0$.Then $\alpha + \beta = -b, \alpha\

Vedclass · Accepted Answer

$rb^2 = cq^2$

Vedclass · Answer

(C) Let $\alpha, \beta$ be the roots of $x^2 + bx + c = 0$ and $\alpha', \beta'$ be the roots of $x^2 + qx + r = 0$.Then $\alpha + \beta = -b, \alpha\beta = c$ and $\alpha' + \beta' = -q, \alpha'\beta' = r$.Given that $\frac{\alpha}{\beta} = \frac{\alpha'}{\beta'}$,we apply the property of componendo and dividendo: $\frac{\alpha + \beta}{\alpha - \beta} = \frac{\alpha' + \beta'}{\alpha' - \beta'}$.Squaring both sides,we get $\frac{(\alpha + \beta)^2}{(\alpha - \beta)^2} = \frac{(\alpha' + \beta')^2}{(\alpha' - \beta')^2}$.Since $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$,we have $\frac{b^2}{b^2 - 4c} = \frac{q^2}{q^2 - 4r}$.Cross-multiplying gives $b^2(q^2 - 4r) = q^2(b^2 - 4c) \Rightarrow b^2q^2 - 4rb^2 = q^2b^2 - 4cq^2$.Simplifying,we get $-4rb^2 = -4cq^2$,which implies $rb^2 = cq^2$.

If the ratio of the roots of $x^2 + bx + c = 0$ and $x^2 + qx + r = 0$ is the same,then

Similar Questions

Find the sum of all possible values of $k$ for which the roots of the equation $x^2 + (k + 1)x + \lambda = 0$ are the square of each other.

If $\alpha$ and $\beta$ are the roots of the equation $ax^2 - bx - c = 0$,then $\alpha^2 - \alpha\beta + \beta^2 = .......$

Let $p(x)$ be a quadratic polynomial with constant term $1$. Suppose $p(x)$, when divided by $x-1$, leaves remainder $2$ and when divided by $x+1$, leaves remainder $4$. Then, the sum of the roots of $p(x)=0$ is

If $\alpha, \beta, \gamma$ are roots of the equation $x^3+a x^2+b x+c=0$,then $\alpha^{-1}+\beta^{-1}+\gamma^{-1} = $

If $\alpha, \beta$ and $\gamma$ are the roots of the equation $x^3 - ax^2 + bx - c = 0$,then $\alpha^{-2} + \beta^{-2} + \gamma^{-2} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module