Question

I really need help quick

Answer 1

Answer:

$\sf x=2+\sqrt{6},\:x=2-\sqrt{6}$

Explanation:

$\sf y = x^2 -4x$

given that y = 2

$\sf x^2 -4x = 2$

$\sf x^2 -4x -2 = 0$

using quadratic formula:

$\sf x = \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{2a} \ \ \ when \ ax^2 + bx + c = 0$

$\rightarrow \sf x_{1,\:2}=\dfrac{-\left(-4\right)\pm \sqrt{\left(-4\right)^2-4\cdot \:1\cdot \left(-2\right)}}{2\cdot \:1}$

$\rightarrow \sf x_1=\dfrac{-\left(-4\right)+2\sqrt{6}}{2\cdot \:1},\:x_2=\dfrac{-\left(-4\right)-2\sqrt{6}}{2\cdot \:1}$

$\rightarrow \sf x=2+\sqrt{6},\:x=2-\sqrt{6}$

in decimals:

$\rightarrow \sf x=4.45 , \ -0.45$