In the following, determine whether the given quadratic equations have real roots and if so, find the roots:

$16x^2=24x+1$

Given:

Given quadratic equation is $16x^2=24x+1$.

To do:

We have to determine whether the given quadratic equation has real roots.

Solution:

$16x^2=24x+1$

$16x^2-24x-1=0$

Comparing the given quadratic equation with the standard form of the quadratic equation $ax^2+bx+c=0$, we get,

$a=16, b=-24$ and $c=-1$.

The discriminant of the standard form of the quadratic equation $ax^2+bx+c=0$ is

$D=b^2-4ac$.

Therefore,

$D=(-24)^2-4(16)(-1)=576+64=640$.

As $D>0$, the given quadratic equation has real roots and the roots are

$x=\frac{-b\pm \sqrt{D}}{2a}$

$x=\frac{-(-24)\pm \sqrt{640}}{2(16)}$ 

$x=\frac{24\pm \sqrt{8×8×10}}{32}$ 

$x=\frac{24\pm 8\sqrt{10}}{32}$ 

$x=\frac{8(3\pm \sqrt{10})}{4×8}$ 

$x=\frac{3+\sqrt{10}}{4}$ or $x=\frac{3-\sqrt{10}}{4}$

The roots are $\frac{3+\sqrt10}{4}$ and $\frac{3-\sqrt10}{4}$.

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Updated on: 10-Oct-2022

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