Question

PLZ HELP IM 100% LOST ON THIS

Answer 1

The equation $f(x)=g(x)$ is

$|x-4|-11=-\sqrt{5x}$

Some observations:

• $\sqrt{5x}$ is defined only as long as $5x\ge0$, or $x\ge0$
• wherever $\sqrt{5x}$ is defined, its value must be non-negative, so that $-\sqrt{5x}$ is never positive
• by the definition of absolute value, we have $|x-4|=x-4$ if $x\ge4$, and $|x-4|=-(x-4)=4-x$ if $x$. Then

$|x-4|-11=\begin{cases}x-15&\text{for }x\ge4\\-x-7&\text{for }x$

If $x$, the equation becomes

$-x-7=-\sqrt{5x}\implies x+7=\sqrt{5x}$

Taking the square of both sides gives

$(x+7)^2=\left(\sqrt{5x}\right)^2\implies x^2+14x+49=5x\implies x^2+9x+49=0$

but since the discriminant is $9^2-4\cdot1\cdot49$, there are no real solutions.

If $x\ge4$, then

$x-15=-\sqrt{5x}$

Taking squares gives

$(x-15)^2=\left(-\sqrt{5x}\right)^2\implies x^2-30x+225=5x$

and solving by the quadratic formula gives two potential solutions,

$x=\dfrac{35\pm5\sqrt{13}}2$

which have approximate values of 8.49 and 26.51.

We know for any value of $x$ that $g(x)\le0$. We have $f(8.49)\approx-6.51$ and $f(26.51)\approx11.51$, so only the first solution 8.49 is valid.