Answer
The real solutions to this equation are $x=\dfrac{1}{2}\pm\dfrac{\sqrt{13}}{2}$
Work Step by Step
$x^{2}\sqrt{x+3}=(x+3)^{3/2}$
Square both sides:
$(x^{2}\sqrt{x+3})^{2}=[(x+3)^{3/2}]^{2}$
$x^{4}(x+3)=(x+3)^{3}$
Take $(x+3)$ to divide the right side and simplify:
$x^{4}=\dfrac{(x+3)^{3}}{x+3}$
$x^{4}=(x+3)^{2}$
Evaluate the power on the right side:
$x^{4}=x^{2}+6x+9$
Take all terms to the left:
$x^{4}-x^{2}-6x-9=0$
Factor the left side:
$(x^{2}+x+3)(x^{2}-x-3)=0$
Set both factor equal to $0$ and solve each individual equation for $x$:
$x^{2}+x+3=0$
Solve using the quadratic formula:
$x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}=\dfrac{-1\pm\sqrt{1^{2}-4(1)(3)}}{2(1)}=...$
$...=\dfrac{-1\pm\sqrt{1-12}}{2}=\dfrac{-1\pm\sqrt{-11}}{2}$
Since the number inside the square root is a negative number, solving this equation will yield two complex solutions. Since this problem asks for the real solutions of the equation, move on to the next equation
$x^{2}-x-3=0$
Solve using the quadratic formula:
$x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}=\dfrac{-(-1)\pm\sqrt{(-1)^{2}-4(1)(-3)}}{2(1)}=...$
$...=\dfrac{1\pm\sqrt{1+12}}{2}=\dfrac{1\pm\sqrt{13}}{2}=\dfrac{1}{2}\pm\dfrac{\sqrt{13}}{2}$
The real solutions to this equation are $x=\dfrac{1}{2}\pm\dfrac{\sqrt{13}}{2}$
