Question

How do you solve \[\sqrt{x+5}+\sqrt{x+15}=\sqrt{9x+40}\]?

Answer 1

Hint: These types of problems are pretty straight forward and are very easy to solve. It is a very simple demonstration of polynomials. In such problems, we square both the sides of the equation until and unless we get rid of the square root terms. After getting rid of the square root, we then try to evaluate the hence formed polynomial equation. In this case we finally need to solve for a quadratic equation and find out the required value for x.

Complete step by step answer:
Now, we start off with the solution by squaring both sides of the given equation and we get,
\[x+5+x+15+2\sqrt{x+5}\sqrt{x+15}=9x+40\]
After this, we rearrange the given equation by bringing the like terms together and we form it as,
\[\begin{align}
  & 2x+20+2\sqrt{x+5}\sqrt{x+15}=9x+40 \\
 & \Rightarrow 2\sqrt{x+5}\sqrt{x+15}=7x+20 \\
\end{align}\]
We now, again square both the sides of the equation, to eliminate the only square root that we have left in this equation. Doing so, we get,
\[\Rightarrow 4\left( x+5 \right)\left( x+15 \right)={{\left( 7x+20 \right)}^{2}}\]
Now, we have no more square roots left in our equation. We now multiply the required factors to get,
\[\begin{align}
  & \Rightarrow 4\left( {{x}^{2}}+20x+75 \right)=49{{x}^{2}}+280x+400 \\
 & \Rightarrow 4{{x}^{2}}+80x+300=49{{x}^{2}}+280x+400 \\
\end{align}\]
Now, taking the left hand terms, to the right hand side of the equation, and then performing the necessary subtraction we get,
\[\Rightarrow 45{{x}^{2}}+200x+100=0\]
Now, we take \[5\] common from this equation and divide both the sides of the equation by \[5\] to get,
\[\Rightarrow 9{{x}^{2}}+40x+20=0\]
Now, we solve for the value of \[x\]. We apply sridharacharya formula to get the value of \[x\] as,
\[\begin{align}
  & \Rightarrow x=\dfrac{-40\pm \sqrt{{{40}^{2}}-4\times 9\times 20}}{2\times 9} \\
 & \Rightarrow x=\dfrac{-40\pm \sqrt{1600-720}}{2\times 9} \\
 & \Rightarrow x=\dfrac{-40\pm \sqrt{880}}{2\times 9} \\
 & \Rightarrow x=\dfrac{-40\pm 29.6648}{2\times 9} \\
 & \Rightarrow x=\dfrac{-40+29.6648}{2\times 9},\dfrac{-40-29.6648}{2\times 9} \\
 & \Rightarrow x=\dfrac{-10.3352}{18},\dfrac{-69.6648}{18} \\
 & \Rightarrow x=\dfrac{-10.3352}{18},\dfrac{-69.6648}{18} \\
 & \Rightarrow x=-0.5742,-3.8703 \\
\end{align}\]
So we get two values of \[x\] , but in the intermediate part we get an equation of \[7x+20\] which should be positive, hence the value \[x=-3.8703\] will not be considered. Thus our answer will be \[x=-0.5742\] .

Note: For these types of problems, we first need to understand the theory of polynomials. We also should never forget the sridharacharya formula to find the value of \[x\] from the polynomials. After finding out the value of \[x\], we need to check whether this particular value is legit or not, by putting it in all the intermediate equations as well as in our original equation.