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.

Latest Math Topics
Latest Math Problems

A best free mathematics education website for students, teachers and researchers.

Maths Topics

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

Maths Problems

Learn how to solve the maths problems in different methods with understandable steps.

Learn solutions

Subscribe us

You can get the latest updates from us by following to our official page of Math Doubts in one of your favourite social media sites.