Question

Solve the polynomial equation

Answer 1

Hey there!

You can solve this by squaring both sides to first cancel out the square root on the left. Then, you can eliminate each term as your normally would. 

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

$(2 \sqrt{x})^{2} = (x-3)^{2}$

$(4x) = (x-3)(x-3)$

$4x = x^{2} -6x+9$

$x^{2} -10x+9$

Now, you're given the opportunity to find your solution using the quadratic formula. That formula is:

$\frac{-b_-^+ \sqrt{b^2-4ac} }{2a}$

You can find out a, b, and c by comparing your new equation to the standard form. Luckily, we just put our equation into standard form:

$ax^{2} + bx + c$

$x^{2} -10x+9$

$a = 1 \\ b = -10 \\ c = 9$

To find your solution, plug these numbers into the quadratic formula and solve. 

$\frac{-b_-^+ \sqrt{b^2-4ac} }{2a}$

$\frac{10_-^+ \sqrt{-10^2-4(9)}}{2}$

Remember to use PEMDAS when solving these problems. 

$\frac{10_-^+ \sqrt{100-36}}{2}$

$\frac{10_-^+ \sqrt{64}}{2}$

$\frac{10_-^+ 8}{2}$

Now, solve this final bit by adding the top terms then subtracting them for two different answers. 

$\frac{10 + 8}{2} = 9$

$\frac{10- 8}{2} = 1$

Don't assume that both of these are correct. The very last step is to plug both of these into the original problem you were given and see if they make the equation true. 

$2 \sqrt{9} =9-3$
$2*3=6$
$6=6$

$2 \sqrt{1} =1-3$
$2*1=-2$
$2 \neq -2$

So (finally), your answer will be x = 9. 

Hope this helped you out! :-)