Question

What are the roots of x in -10xsquare+12-9=0

Answer 1

Answer:

$x=\pm \sqrt{\frac{3}{10}}$

Step-by-step explanation:

$-10x^2+12-9=0\\\\-10x^2+3=0\\\\-10x^2=-3\\\\x^2=\frac{-3}{-10}\\\\x^2=\frac{3}{10}$

here we have to cases

case 1

$x=\sqrt{\frac{3}{10}}$

case 2

$x=-\sqrt{\frac{3}{10}}$

Answer 2

Given:

Consider the given equation is

$-10x^2+12x-9=0$

To find:

The roots of x in the given equation.

Solution:

We have,

$-10x^2+12x-9=0$

On comparing this equation with $ax^2+bx+c=0$, we get

$a=-10,b=12,c=-9$

Using quadratic formula, we get

$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\dfrac{-(12)\pm \sqrt{(12)^2-4(-10)(-9)}}{2(-10)}$

$x=\dfrac{-12\pm \sqrt{144-360}}{-20}$

$x=\dfrac{-12\pm \sqrt{-216}}{-20}$

We know that, $\sqrt{-1}=i$.

$x=\dfrac{-12\pm 6\sqrt{6}i}{-20}$

$x=\dfrac{-2(6\mp 3\sqrt{6}i)}{-20}$

$x=\dfrac{6\mp 3\sqrt{6}i}{10}$

$x=\dfrac{6+3\sqrt{6}i}{10}$ and $x=\dfrac{6-3\sqrt{6}i}{10}$

Therefore, the roots of the given equation are $x=\dfrac{6+3\sqrt{6}i}{10}$ and $x=\dfrac{6-3\sqrt{6}i}{10}$.