0
$\begingroup$

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?

$\endgroup$
6
  • 2
    $\begingroup$ 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. $\endgroup$ – boojum Apr 8 at 3:22
  • $\begingroup$ Doing it with this method, I got $ 4x^2 -64x +36 = 0 $ $\endgroup$ – Virej Dasani Apr 8 at 3:25
  • $\begingroup$ Solving this still isn't giving me $ x=9, x=1 $ $\endgroup$ – Virej Dasani Apr 8 at 3:25
  • 1
    $\begingroup$ You left out the middle terms again: it's $ \ 4x^2 \ + \ 24x \ + \ 36 \ = \ 64x \ $ . $\endgroup$ – boojum Apr 8 at 3:26
  • 1
    $\begingroup$ Leaving out the middle terms in a "binomial-square" is unfortunately a very common algebra mistake. Beware of that... $\endgroup$ – boojum Apr 8 at 3:28
0
$\begingroup$

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

$\endgroup$
0
1
$\begingroup$

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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.