Question

The equation 124x^2 + bx = 248 has two distinct roots. Find the product of the roots.

Answer 1

$\textbf{Given:}$

$124\,x^2+b\,x=248$

$\textbf{To find:}$

$\text{Product of the roots}$

$\textbf{Solution:}$

$\text{Consider,}$

$124\,x^2+b\,x-248=0$

$\text{We know that,}$

$\textbf{If \bf\alpha and \bf\beta are roots of \bf\,ax^2+bx+c=0 then}$

$\textbf{Product of roots \bf=\alpha\,\beta=\bf\dfrac{c}{a} }$

$\text{Then,}$

$\text{Product of the roots}=\dfrac{c}{a}$

$\text{Product of the roots}=\dfrac{-248}{124}$

$\imples\textbf{Product of the roots=-2}$

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Answer 2

GIVEN :

The equation $124x^2 + bx = 248$ has two distinct roots.

TO FIND :

The product of the roots for the given equation.

SOLUTION :

Given quadratic equation is $124x^2 + bx = 248$ and it has two distinct roots.

Let $\alpha$ and $\beta$ be the two distinct roots.

Given quadratic equation can be written as

$124x^2 + bx-248=0$

It is of the form $ax^2+bx+c=0$

Then the formula for product of the roots is given by

$Product of the roots=\alpha \beta=\frac{c}{a}$

Comparing to the general quadratic equation the values of a=124, b=b and c=-248

Now substituting the values in the formula we get,

$Product of the roots=\alpha \beta=\frac{-248}{124}$

$=-2$

∴ $Product of the roots=\alpha \beta=-2$