## Solution 41 Page 60 Math 10 SBT – Kite>

Topic

Solve the following equation:

a) $$\sqrt {7 – 2x} + x = 2$$

b) $$\sqrt { – 2{x^2} + 7x + 1} + 3x = 7$$

Solution method – See details

Step 1: Return the PT to the form $$\sqrt {f\left( x \right)} = g\left( x \right)$$

Step 2: $$\sqrt {f\left( x \right)} = g\left( x \right) \Leftrightarrow \left\{ \begin{array}{l}g\left( x \right) \ge 0\\f\left( x \right) = {\left[ {g\left( x \right)} \right]^2}\end{array} \right.$$

Detailed explanation

a) $$\sqrt {7 – 2x} + x = 2 \Leftrightarrow \sqrt {7 – 2x} = 2 – x$$

$$\begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l}2 – x \ge 0\\7 – 2x = {\left( {2 – x} \right)^2 }\end{array} \right \Leftrightarrow \left\{ \begin{array}{l}x \le 2\\7 – 2x = {x^2} – 4x + 4\end{array} \right. \\ \Leftrightarrow \left\{ \begin{array}{l}x \le 2\\{x^2} – 2x – 3 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{ array}{l}x \le 2\\\left[\begin{array}{l}x=3\;(L)\\x=–1\;\end{array}\right\end{array}\right\end{array}$$[\begin{array}{l}x=3\;(L)\\x= –1\;\end{array}\right\end{array}\right\end{array}\)

So $$S = \left\{ { – 1} \right\}$$

b) $$\sqrt { – 2{x^2} + 7x + 1} + 3x = 7 \Leftrightarrow \sqrt { – 2{x^2} + 7x + 1} = 7 – 3x$$

$$\begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l}7 – 3x \ge 0\\ – 2{x^2} + 7x + 1 = {\left( {7 – 3x} \right)^2}\end{array} \right \Leftrightarrow \left\{ \begin{array}{l}x \le \frac{7}{3}\\ – 2{x^2 } + 7x + 1 = 9{x^2} – 42x + 49\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x \le \frac{7}{3 }\\11{x^2} – 49x + 48 = 0\end{array} \right \Leftrightarrow \left\{ \begin{array}{l}x \le \frac{7}{3}\\ \left[\begin{array}{l}x=3\;(L)\\x=\frac{{16}}{{11}}\;\end{array}\right\quad\end{array}\right\Leftrightarrowx=\frac{{16}}{{11}}\;\end{array}$$[\begin{array}{l}x=3\;(L)\x=\frac{{16}}{{11}}\;\end{array}\right\quad\end{array}\right\Leftrightarrowx=\frac{{16}}{{11}}\;\end{array}\)

So $$S = \left\{ {\frac{{16}}{{11}}} \right\}$$