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Find the value of $\dfrac{b^2}{ac}$ if one root is twice the second root of the quadratic equation $ax^2+bx+c = 0$

$ax^2+bx+c = 0$ is a quadratic equation in general form. One root is twice the other root.

01

Relations between Roots

Assume, $\alpha$ and $\beta$ are two roots of the quadratic equation. Summation and product of the both roots are,

$\alpha + \beta = -\dfrac{b}{a}$

$\alpha \beta = \dfrac{c}{a}$

02

Eliminating a Root from equations

It is given that one root is twice the second root. So, $\beta = 2 \alpha$. Now, eliminate $\beta$ from both roots relations by using this condition.

$\alpha + \beta = -\dfrac{b}{a}$

$\implies \alpha + 2\alpha = -\dfrac{b}{a}$

$\implies 3\alpha = -\dfrac{b}{a}$

$\therefore \,\, \alpha = -\dfrac{b}{3a}$

$\alpha \beta = \dfrac{c}{a}$

$\implies \alpha \times 2 \alpha = \dfrac{c}{a}$

$\implies 2 \alpha^2 = \dfrac{c}{a}$

$\therefore \,\, \alpha^2 = \dfrac{c}{2a}$

03

Finding the Required Result

Now, replace the value of $\alpha$, obtained from the summation of the roots of the quadratic equation.

$\implies {\Bigg(-\dfrac{b}{3a}\Bigg)}^2 = \dfrac{c}{2a}$

$\implies \dfrac{b^2}{9a^2} = \dfrac{c}{2a}$

$\implies 2a \times b^2 = 9a^2 \times c$

$\implies 2ab^2 = 9a^2c$

$\require{cancel} \implies 2\cancel{a}b^2 = 9\cancel{a^2}c$

$\implies 2b^2 = 9ac$

$\therefore \,\,\,\,\,\, \dfrac{b^2}{ac} = \dfrac{9}{2} = 4.5$

Therefore, it is the required solution for this quadratic equation.

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