Surd Equations Reducible to Quadratics

Surd Equations Reducible to Quadratic for Math Algebra is done squaring both sides for removing surds and radical expressions. Make sure to check whether the solutions are correct by substituting them into the original surd equations.

Question 1

Solve \( x = \sqrt{x+2} \).

\( \begin{aligned} \displaystyle \require{color}
x^2 &= x+2 &\color{red} \text{square both sides} \\
x^2 – x – 2 &= 0 \\
(x-2)(x+1) &= 0 \\
x = 2 &\text{ or } x = -1 \\
\color{green} \text{Check } x=2 \\
2 &= \sqrt{2+2} \\
&= \sqrt{4} \\
&= 2 \\
2 &= 2 \\
\color{green} \text{Check } x=-1 \\
-1 &= \sqrt{-1+2} \\
&= \sqrt{1} \\
&= 1 \\
-1 &\ne 1 \\
\therefore x &= 2 &\color{red} \text{only one solution} \\
\end{aligned} \\ \)

Question 2

Solve \( 3x = \sqrt{x^2+32} \).

\( \begin{aligned} \displaystyle \require{color}
9x^2 &= x^2 + 32 &\color{red} \text{square both sides} \\
8x^2 &= 32 \\
x^2 &= 4 \\
x = 2 &\text{ or } x = -2 \\
\color{green} \text{Check } x=2 \\
3 \times 2 &= \sqrt{2^2 + 32} \\
6 &= \sqrt{36} \\
6 &= 6 \\
\color{green} \text{Check } x=-2 \\
3 \times -2 &= \sqrt{(-2)^2 + 32} \\
-6 &= \sqrt{36} \\
-6 &\ne 6 \\
\therefore x &= 6 &\color{red} \text{only one solution} \\
\end{aligned} \\ \)

Question 3

Solve \( x + 2\sqrt{x+1} = 7 \).

\( \begin{aligned} \displaystyle \require{color}
2 \sqrt{x+1} &= 7 – x \\
4(x+1) &= (7-x)^2 &\color{red} \text{square both sides} \\
4x+4 &= 49-14x+x^2 \\
x^2 – 18x + 45 &= 0 \\
(x-15)(x-3) &= 0 \\
\therefore x &= 15 \text{ or } 3 \\
\end{aligned} \\ \)

Question 4

Solve \( \sqrt{2x+3} – \sqrt{x-2} = 2 \).

\( \begin{aligned} \displaystyle \require{color}
(\sqrt{2x+3} – \sqrt{x-2})^2 &= 2^2 &\color{red} \text{square both sides} \\
2x+3 – 2\sqrt{(2x+3)(x-2)} + x-2 &= 4 \\
– 2\sqrt{(2x+3)(x-2)} &= -3x+3 \\
4(2x^2 – x – 6) &= 9x^2 – 18x + 9 &\color{red} \text{square both sides again} \\
8x^2 – 4x – 24 &= 9x^2 – 18x + 9 \\
x^2 – 14x + 33 &= 0 \\
(x-11)(x-3) &= 0 \\
\therefore x &=11 \text{ or } 3 \\
\end{aligned} \\ \)

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