Question

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

Answer 1

`x=\pm \sqrt{(3)/(10)}`

Explanation:

`-10x^2+12-9=0\\\\-10x^2+3=0\\\\-10x^2=-3\\\\x^2=(-3)/(-10)\\\\x^2=(3)/(10)\\\\`

here we have to cases

case 1

`x=\sqrt{(3)/(10)}\\`

case 2

`x=-\sqrt{(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=(-b\pm √(b^2-4ac))/(2a)`

`x=(-(12)\pm √((12)^2-4(-10)(-9)))/(2(-10))`

`x=(-12\pm √(144-360))/(-20)`

`x=(-12\pm √(-216))/(-20)`

We know that,
`√(-1)=i`.

`x=(-12\pm 6√(6)i)/(-20)`

`x=(-2(6\mp 3√(6)i))/(-20)`

`x=(6\mp 3√(6)i)/(10)`

`x=(6+3√(6)i)/(10)` and
`x=(6-3√(6)i)/(10)`

Therefore, the roots of the given equation are
`x=(6+3√(6)i)/(10)` and
`x=(6-3√(6)i)/(10)`.