Answer:
[tex]x=\pm \sqrt{\frac{3}{10}}[/tex]
Step-by-step explanation:
[tex]-10x^2+12-9=0\\\\-10x^2+3=0\\\\-10x^2=-3\\\\x^2=\frac{-3}{-10}\\\\x^2=\frac{3}{10}\\\\[/tex]
here we have to cases
case 1
[tex]x=\sqrt{\frac{3}{10}}\\[/tex]
case 2
[tex]x=-\sqrt{\frac{3}{10}}[/tex]
Given:
Consider the given equation is
[tex]-10x^2+12x-9=0[/tex]
To find:
The roots of x in the given equation.
Solution:
We have,
[tex]-10x^2+12x-9=0[/tex]
On comparing this equation with [tex]ax^2+bx+c=0[/tex], we get
[tex]a=-10,b=12,c=-9[/tex]
Using quadratic formula, we get
[tex]x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]x=\dfrac{-(12)\pm \sqrt{(12)^2-4(-10)(-9)}}{2(-10)}[/tex]
[tex]x=\dfrac{-12\pm \sqrt{144-360}}{-20}[/tex]
[tex]x=\dfrac{-12\pm \sqrt{-216}}{-20}[/tex]
We know that, [tex]\sqrt{-1}=i[/tex].
[tex]x=\dfrac{-12\pm 6\sqrt{6}i}{-20}[/tex]
[tex]x=\dfrac{-2(6\mp 3\sqrt{6}i)}{-20}[/tex]
[tex]x=\dfrac{6\mp 3\sqrt{6}i}{10}[/tex]
[tex]x=\dfrac{6+3\sqrt{6}i}{10}[/tex] and [tex]x=\dfrac{6-3\sqrt{6}i}{10}[/tex]
Therefore, the roots of the given equation are [tex]x=\dfrac{6+3\sqrt{6}i}{10}[/tex] and [tex]x=\dfrac{6-3\sqrt{6}i}{10}[/tex].
