Question

Find all real solutions to the equation with absolute value
$|2x-1|=x^2$

• A. $-1-\sqrt{2},-1+\sqrt{2},1$
• B. $1$
• C. $-1,1$
• D. $-1-\sqrt{2},-1+\sqrt{2}$
• E. no solutions

Answer 1

The correct option is A $-1-\sqrt{2},-1+\sqrt{2},1$

Since $|x|=x$ and $-x$, therefore,

$|2x-1|=x^2$ may be calculated as:

$2x-1=x^2$ and $-(2x-1)=x^2$

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

$\Rightarrow x^2-2x+1=0$..........(1) and

$\Rightarrow x^2+2x-1=0$......... (2)

Finding the roots of equation (1) is as shown below:

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

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

$x(x-1)-1(x-1)=0$

$x-1=0$ and $x-1=0$

$x=1$

Finding the roots of equation (2) is as shown below:

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

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

$=\frac{-2\pm \sqrt{8}}{2}$

$=\frac{-2\pm 2\sqrt{2}}{2}$

$=-1\pm \sqrt{2}$

Therefore, $x=1$ and $x=-1\pm \sqrt{2}$.

Hence, option A is correct.