Question

Find all real values of $x$ such that $f\left(x\right)=g\left(x\right)$ where $f$ and $g$ are functions given by
$f\left(x\right)=3x+\sqrt{x}$ and $g\left(x\right)=2x+6$

• A. $4$
• B. $9$
• C. $3$
• D. $6$

Answer 1

Answer: (A), (B)

The correct options are
A $4$
B $9$

Since $f\left(x\right)=g\left(x\right)$ therefore,

$3x+\sqrt{x}=2x+6$

$\sqrt{x}=2x+6-3x$

$\sqrt{x}=-x+6$

Squaring both sides, we get,

$x=(-x+6{)}^2$ that is

$x=x^2+36-12x$

$x^2+36-12x-x=0$

$x^2+36-13x=0$ or

$x^2-13x+36=0$

Now find the roots of the above equation as follows:

$x^2-13x+36=0$

$x^2-9x-4x+36=0$

$x(x-9)-4(x-9)=0$ that is

$(x-9)=0$ and $(x-4)=0$

$x=9$ and $x=4$