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

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?

• When you squared the left side of your equation, you left out the "middle terms": $\ (a-b)^2 \ \neq \ a^2 + b^2 \ , \ \text{but} \ a^2 - 2ab + b^2 \ .$ You will have an easier time getting rid of the square-root if you write $\ 8 \sqrt{x} \ = \ 6 + 2x \$ , square both sides and then simplify the equation before solving it. – boojum Apr 8 at 3:22
• Doing it with this method, I got $4x^2 -64x +36 = 0$ – Virej Dasani Apr 8 at 3:25
• Solving this still isn't giving me $x=9, x=1$ – Virej Dasani Apr 8 at 3:25
• You left out the middle terms again: it's $\ 4x^2 \ + \ 24x \ + \ 36 \ = \ 64x \$ . – boojum Apr 8 at 3:26
• Leaving out the middle terms in a "binomial-square" is unfortunately a very common algebra mistake. Beware of that... – boojum Apr 8 at 3:28

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