Question

How do you solve $\sqrt{x+6}=6-2\sqrt{5-x}$ ? \[\]

Answer 1

Hint: We use the domain of square root function to find the range of values valid for $x$. We take the square roots terms to one side and then square both sides. We square accordingly to eliminate square roots. We use the quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to find the solutions and see if the solutions satisfy valid range of values.

Complete step by step answer:
We know that a square root function takes non-negative real numbers called arguments and returns non-negative real numbers. We are asked in the question to solve
\[\sqrt{x+6}=6-2\sqrt{5-x}.........\left( 1 \right)\]
We see that there are two square roots $\sqrt{x+6},\sqrt{5-x}$. Since the argument for square roots is always positive we have-
\[\begin{align}
  & x+6\ge 0\Rightarrow x\ge -6 \\
 & 5-x\ge 0\Rightarrow 5\ge x \\
\end{align}\]
So the valid range of values for the solution of the given equation is $-6\le x\le 5\Rightarrow x\in \left[ -6,5 \right]$. Since square root function returns non-negative values we have
\[\begin{align}
  & \sqrt{x+6}\ge 0 \\
 & \Rightarrow 6-2\sqrt{5-x}\ge 0 \\
 & \Rightarrow 6\ge 2\sqrt{5-x} \\
 & \Rightarrow 3\ge \sqrt{5-x}....\left( 3 \right) \\
\end{align}\]
Let us rearrange the given equation with square roots at one side.
\[\sqrt{x+6}+2\sqrt{5-x}=6\]
We square both sides of above equation to have
\[\Rightarrow {{\left( \sqrt{x+6}+2\sqrt{5-x} \right)}^{2}}=36\]
We use the algebraic identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ for $a=\sqrt{x+6},b=2\sqrt{5-x}$ in the above step to have
\[\begin{align}
  & \Rightarrow x+6+4\left( 5-x \right)+2\sqrt{6+x}\cdot 2\sqrt{5-x}=36 \\
 & \Rightarrow 26-3x+4\sqrt{6+x}\sqrt{5-x}=36 \\
 & \Rightarrow \sqrt{6+x}\sqrt{5-x}=\dfrac{36-26+3x}{4} \\
 & \Rightarrow \sqrt{6+x}\sqrt{5-x}=\dfrac{10+3x}{4} \\
\end{align}\]
We square both sides of above equation to have;
\[\begin{align}
  & \Rightarrow {{\left( \sqrt{x+6}\sqrt{5-x} \right)}^{2}}=\dfrac{{{\left( 10+3x \right)}^{2}}}{{{4}^{2}}} \\
 & \Rightarrow \left( x+6 \right)\left( 5-x \right)=\dfrac{100+60x+9{{x}^{2}}}{16} \\
\end{align}\]
We multiply the polynomials at the left hand side and then simplify to have;
\[\begin{align}
  & \Rightarrow -{{x}^{2}}-x+30=\dfrac{100+60x+9{{x}^{2}}}{16} \\
 & \Rightarrow -16{{x}^{2}}-16x+480=100+60x+9{{x}^{2}} \\
 & \Rightarrow 25{{x}^{2}}+76x-380=0 \\
\end{align}\]
We note that the above equation is a quadratic equation of the type $a{{x}^{2}}+bx+c=0$. We find the roots of the quadratic equation as
\[\begin{align}
  & x=\dfrac{-76\pm \sqrt{{{\left( -76 \right)}^{2}}-4\cdot 25\cdot \left( -380 \right)}}{2\cdot 25} \\
 & \Rightarrow x=\dfrac{-76\pm \sqrt{43376}}{50} \\
 & \Rightarrow x\simeq \dfrac{-76\pm 208}{50} \\
 & \Rightarrow x\simeq \dfrac{-76+208}{50},\dfrac{-76-208}{50} \\
\end{align}\]
So approximately roots of the equation are
\[\begin{align}
  & \Rightarrow x\simeq \dfrac{132}{50},\dfrac{-284}{50} \\
 & \Rightarrow x\simeq 2.64,-5.84 \\
\end{align}\]
We see that both the obtained roots lie in the interval $\left[ -6,5 \right]$ . Let us check whether they satisfy condition (2) that is $3\ge \sqrt{5-x}$. We have
\[\begin{align}
  & 3\ge \sqrt{5-2.64} \\
 & 3<\sqrt{5-\left( -5\cdot 84 \right)}=\sqrt{10.84} \\
\end{align}\]
So the only solution valid for the equation is approximately $x=2.64$ and exactly is
\[x=\dfrac{-76+\sqrt{43376}}{50}=\dfrac{-76+4\sqrt{2711}}{50}=\dfrac{-38+2\sqrt{2711}}{25}\]

Note:
We note that we get real and unequal roots when discriminant $D={{b}^{2}}-4ac > 0$ and equal roots when $D={{b}^{2}}-4ac=0$. We get rational roots when $D$ is a perfect square. We note that $a\le b\Rightarrow \sqrt{a}\le \sqrt{b}$ and this why in this problem $\sqrt{9} < \sqrt{10.84}\Rightarrow 3 < \sqrt{10.84}$. We can find the square root of 43376 up to one digit in rough for valid range analysis.