Question

15xsquare -10 √6x+10 roots for the quadratic equation

Answer 1

$\huge\bf\underline{Solution:-}$

$\sf \bullet 15x{}^{2}-10\sqrt{6}x+10$

• We have to find the roots of quadratic equation

Now by the formula of quadratic equation

$\sf \hookrightarrow ax{}^{2}+bx+c=0$

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

So

$\sf\implies 15x{}^{2}-10\sqrt{6}x+10=0\\ \\ \sf\bullet a=15\:\:\:\:\:\: \bullet b=-10\sqrt{6}\:\:\:\:\:\: \bullet c=10$

Put these values in the formula

$\sf\implies x=\dfrac{-b\pm \sqrt{b{}^{2}-4ac}}{2a}\\ \\ \sf\implies x=\dfrac{-(-10\sqrt{6})\pm \sqrt{(-10\sqrt{6}){}^{2}-4\times 15\times 10}}{2\times 15}\\ \\ \sf\implies x=\dfrac{10\sqrt{6}\pm \sqrt{600-600}}{30}\\ \\ \sf\implies x=\dfrac{10\sqrt{6} \pm 0}{30}$

● So the roots

$\sf\implies taking\:10\: common\\ \\ \sf\implies x=\cancel{\dfrac{10}{10}}\bigg(\dfrac{\sqrt{6}}{3}\bigg)\\ \\ \sf\implies \:\:\: or\:\: x= \cancel{\dfrac{\sqrt{6}}{3}}\\ \\ \sf\implies x=\dfrac{\sqrt{2}}{\sqrt{3}}$

• So the root of this equation is

$\boxed{\sf{x=\dfrac{\sqrt{2}}{\sqrt{3}}}}$and

$\boxed{\sf{x=0}}$

Answer 2

$</p><p>\tt{\purple{\huge{Hello!!! }}}$

$</p><p>\tt{\red{Given \: - }}$

$\tt{ \orange{f(x) = 15 {x}^{2} - \sqrt{6} x + 10 \: }}$