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,$\alpha + \beta = -b$,$\alpha\beta = c$,$\alpha' + \beta' = -q$,and $\alpha'\beta' = r$.Given that the ratio of the roots is the same,$\frac{\alpha}{\beta} = \frac{\alpha'}{\beta'}$.Using the property of componendo and dividendo,$\frac{\alpha + \beta}{\alpha - \beta} = \frac{\alpha' + \beta'}{\alpha' - \beta'}$.Squaring both sides,$\frac{(\alpha + \beta)^2}{(\alpha - \beta)^2} = \frac{(\alpha' + \beta')^2}{(\alpha' - \beta')^2}$.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 $b^2(q^2 - 4r) = q^2(b^2 - 4c)$.$b^2q^2 - 4b^2r = q^2b^2 - 4cq^2$.$-4b^2r = -4cq^2$,which simplifies to $b^2r = cq^2$.

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

Similar Questions

If the roots of the equation $x^2 + 2mx + m^2 - 2m + 6 = 0$ are equal,then the value of $m$ is:

Solve the given two equations and choose the correct option.
$I.$ $\frac{3x}{3x+7} - \frac{3x+7}{3x} = 14$
$II.$ $\frac{y}{18y-5} - \frac{18y-5}{y} = 2$

Solve the given two equations and select the correct answer from the given options.
$I.$ $\sqrt{25} x + \sqrt{16} y = 41$
$II.$ $\sqrt{16} x + \sqrt{25} y = 40$

The equation $e^x - x - 1 = 0$ has

If $8, 2$ are the roots of ${x^2} + ax + \beta = 0$ and $3, 3$ are the roots of ${x^2} + \alpha x + b = 0$,then the roots of ${x^2} + ax + b = 0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module