Question

square root functions and equations
student activity sheet 4; exploring \solving square root equations algebraically\ page 5 of 5

1. reinforce solve each equation algebraically. check numerically or graphically.

a. $\sqrt{2x + 3}-7 = 0$
b. $\sqrt{x}=-4$
c. $\sqrt{x^{2}-9}-x=-3$

Explanation:

Part a

Step1: Isolate the square root

Add \(7\) to both sides of the equation \(\sqrt{2x + 3}-7 = 0\).
\(\sqrt{2x+3}=7\)

Step2: Eliminate the square root

Square both sides of the equation. Using the property \((\sqrt{a})^2=a\) (\(a\geq0\)), we get \((\sqrt{2x + 3})^2=7^2\).
\(2x+3 = 49\)

Step3: Solve for \(x\)

Subtract \(3\) from both sides: \(2x=49 - 3=46\). Then divide both sides by \(2\): \(x=\frac{46}{2}=23\).

Step4: Check the solution

Substitute \(x = 23\) into the original equation: \(\sqrt{2\times23+3}-7=\sqrt{46 + 3}-7=\sqrt{49}-7=7 - 7=0\).

Part b

The square root function \(y=\sqrt{x}\) has a range \(y\geq0\) (by definition, \(\sqrt{x}\) represents the non - negative square root of \(x\) where \(x\geq0\)). In the equation \(\sqrt{x}=-4\), since the left - hand side \(\sqrt{x}\geq0\) for all \(x\) in its domain (\(x\geq0\)) and the right - hand side is \(-4<0\), there is no solution.

Part c

Step1: Isolate the square root

Add \(x\) to both sides of the equation \(\sqrt{x^{2}-9}-x=-3\).
\(\sqrt{x^{2}-9}=x - 3\)

Step2: Eliminate the square root

Square both sides: \((\sqrt{x^{2}-9})^2=(x - 3)^2\). Using the formula \((a - b)^2=a^{2}-2ab + b^{2}\), we have \(x^{2}-9=x^{2}-6x + 9\).

Step3: Simplify the equation

Subtract \(x^{2}\) from both sides: \(-9=-6x + 9\).

Step4: Solve for \(x\)

Add \(6x\) to both sides: \(6x-9 = 9\). Then add \(9\) to both sides: \(6x=9 + 9=18\). Divide both sides by \(6\): \(x = 3\).

Step5: Check the solution

Substitute \(x = 3\) into the original equation: \(\sqrt{3^{2}-9}-3=\sqrt{9 - 9}-3=0-3=-3\).

Answer:

a. \(x = 23\)
b. No solution
c. \(x = 3\)