Question

The system of equations:

$y=x^2-2x$
$y=2x-1$ has two solutions for ($x,y$).

Determine the greater value of $x$.

• A. $2-\sqrt{3}$
• B. $\sqrt{3}$
• C. $2+2\sqrt{3}$
• D. $5$

Answer 1

The correct option is C $2+2\sqrt{3}$

Given, $y=x^2-2x$

$y=2x-1$

Then $x^2-2x=2x-1$

$\Rightarrow x^2-2x-2x+1=0$

$\Rightarrow x^2-4x+1=0$

Manipulate this equation $a{x}^2+bx+c=0$

We know $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\therefore x=\frac{-(-4)\pm \sqrt{(-4{)}^2-4(1\left)\right(1)}}{2\left(1\right)}$

$\Rightarrow x=\frac{4\pm \sqrt{16-4}}{2}$

$\Rightarrow x=\frac{4\pm \sqrt{12}}{2}$

$\Rightarrow x=\frac{4\pm 2\sqrt{3}}{2}$

The greater of the two possible values for $x$ is $x=2\pm 2\sqrt{3}$.

Therefore, the correct answer is (C).