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Solving Square Root Equations Quiz 2

mathematics

 Shift $\sqrt{7x-9}$ and 3 on the RHS and LHS respectively and then solve by squaring on both the sides.

  $\begin{align} & \sqrt{7x}-\sqrt{7x-9}=3 \\ & \Rightarrow \sqrt{7x}-3=\sqrt{7x-9} \\ & \Rightarrow 7x+9-6\sqrt{7x}=7x-9\,\,\,\,\,\,\,\,\,\,\text{(squaring on both sides)} \\ & \Rightarrow \sqrt{7x}=3 \\  & \Rightarrow x=\dfrac{9}{7}\,\,\,\,\,\,\,\,\,\,\text{(squaring on both sides)} \end{align}$

 If $x=\sqrt{7}+2$, then find the value of $\sqrt{{{x}^{2}}+2x+49}$. 

 Put the value of $x$ in the given expression and then simplify further. Then, split the terms in such a way that the expression becomes of the form ${{(a+b)}^{2}}$ to remove the square root.

  $\begin{align} & \sqrt{{{x}^{2}}+2x+49} \\ & =\sqrt{{{(\sqrt{7}+2)}^{2}}+2(\sqrt{7}+2)+49} \\ & \Rightarrow \sqrt{7+4+4\sqrt{7}+4+2\sqrt{7}+49} \\ & \Rightarrow \sqrt{63+6\sqrt{7}+1}=\sqrt{{{(3\sqrt{7}+1)}^{2}}} \\ & \Rightarrow 3\sqrt{7}+1\end{align}$

 If $x = 1 – 5y$ and $\sqrt{1+2y}+\sqrt{2y-8}=3$, then find the value of $x$.

 Shift $\sqrt{1+2y}$ on the RHS and then find the value of $y$ by squaring on both the sides. Then put the value of $y$ to find $x$.

 $\begin{align} & \sqrt{1+2y}+\sqrt{2y-8}=3 \\ & \Rightarrow \sqrt{2y-8}=3-\sqrt{1+2y} \,\,\,\,\,\,\,\,\,\,\, \text{(squaring on both sides)} \\ & \Rightarrow 2y-8=9-6\sqrt{1+2y}+1+2y \\ & \Rightarrow 3=\sqrt{1+2y}\,\,\,\,\,\,\,\,\,\,\,\text{(squaring on both sides)} \\ & \Rightarrow 9=1+2y \Rightarrow y=4 \\ & \Rightarrow x=1-5(4)=-19\end{align}$ 

 Simplify: $\sqrt{48-14\sqrt{5}+\sqrt{20+3\sqrt{5}+\sqrt{6+2\sqrt{5}}}}$

 Solve each square root step by step to make it whole square so that we can remove the square root.

 $\begin{align} & \text{Given: }\sqrt{48-14\sqrt{5}+\sqrt{20+3\sqrt{5}+\sqrt{6+2\sqrt{5}}}} \\ & \Rightarrow \sqrt{48-14\sqrt{5}+\sqrt{20+3\sqrt{5}+\sqrt{{{(\sqrt{5}+1)}^{2}}}}} \\  & \Rightarrow \sqrt{48-14\sqrt{5}+\sqrt{20+3\sqrt{5}+\sqrt{5}+1}} \\  & \Rightarrow \sqrt{48-14\sqrt{5}+\sqrt{{{(2\sqrt{5}+1)}^{2}}}} \\  & \Rightarrow \sqrt{45+4-12\sqrt{5}}=\sqrt{{{(3\sqrt{5}-2)}^{2}}} \\  & \Rightarrow 3\sqrt{5}-2 \end{align}$ 

 Solve for x: $\sqrt{4{{x}^{2}}+20x+25}=\sqrt{8{{x}^{2}}+40x+50}-\sqrt{12{{x}^{3}}+60{{x}^{2}}+75x}$

 Convert each square root equation in the form of ${{(a+b)}^{2}}$ and cancel the common factor throughout the equation. Then solve for $x$ by squaring on both the sides.

 $\begin{align} & \sqrt{4{{x}^{2}}+20x+25}=\sqrt{8{{x}^{2}}+40x+50}-\sqrt{12{{x}^{3}}+60{{x}^{2}}+75x} \\ & \Rightarrow \sqrt{{{(2x+5)}^{2}}}=\sqrt{2{{(2x+5)}^{2}}}-\sqrt{3x{{(2x+5)}^{2}}} \\ & \Rightarrow 2x+5=\sqrt{2}(2x+5)-\sqrt{3x}(2x+5) \\ & \Rightarrow \sqrt{3x}=\sqrt{2}-1\,\,\,\,\,\,\,\,\,\text{(now squaring on both sides)} \\ & \Rightarrow 3x=2+1-2\sqrt{2}\Rightarrow x=\dfrac{3-2\sqrt{2}}{3} \\ \end{align}$