How to find the points of intersection in a curve and a line

Question

The equation of a curve is $$ y=8\sqrt x -2x $$ We have to find the values of $x$ at which the line $y = 6$ meets the curve

I tried equating them and doing using the quadratic formula like this: $$ 8\sqrt x -2x = 6 $$ $$ 64x + 4x^2 = 36 $$ $$ 4x^2 + 64x -36 = 0 $$

The answer to the question is $x=9, x=1$ but after solving this quadratic, I'm getting a completely different answer. What am I doing wrong?

Answer 1

$(a-b)^2=a^2+b^2-2ab$ not $a^2+b^2$ as you have done.

Answer 2

Differently from the previous comments, I propose another way to approach it for the sake of curiosity.

\begin{align*} 8\sqrt{x} - 2x = 6 & \Longleftrightarrow x - 4\sqrt{x} + 3 = 0\\\\ & \Longleftrightarrow (x - \sqrt{x}) - (3\sqrt{x} - 3) = 0\\\\ & \Longleftrightarrow \sqrt{x}(\sqrt{x} - 1) - 3(\sqrt{x} - 1) = 0\\\\ & \Longleftrightarrow (\sqrt{x} - 3)(\sqrt{x} - 1) = 0\\\\ & \Longleftrightarrow (\sqrt{x} = 3)\vee(\sqrt{x} = 1)\\\\ & \Longleftrightarrow (x = 9)\vee(x = 1) \end{align*}

and we are done.

Hopefully this helps!