Question

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:

Step1: Isolate the square root

For equation \(a\): \(\sqrt{2x + 3}-7 = 0\), add \(7\) to both sides.
\(\sqrt{2x+3}=7\)

Step2: Square both sides

\((\sqrt{2x + 3})^2=7^2\), which simplifies to \(2x+3 = 49\)

Step3: Solve for \(x\)

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

Step4: Check the solution

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

For equation \(b\): \(\sqrt{x}=-4\). Since the square root of a number (\(\sqrt{x}\), where the domain of \(y = \sqrt{x}\) is \(x\geq0\) and the range is \(y\geq0\)), there is no real - number solution because the square root of a non - negative number \(x\) (since \(x\geq0\) for \(y=\sqrt{x}\)) cannot be negative.

For equation \(c\): \(\sqrt{x^{2}-9}-x=-3\)

Step1: Isolate the square root

\(\sqrt{x^{2}-9}=x - 3\)

Step2: Square both sides

\(x^{2}-9=(x - 3)^2\). Expand \((x - 3)^2=x^{2}-6x + 9\). So, \(x^{2}-9=x^{2}-6x + 9\)

Step3: Simplify the equation

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

Step4: Solve for \(x\)

Add \(6x\) to both sides: \(6x-9 = 9\). Add \(9\) to both sides: \(6x=18\). Divide 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 real - number solution
c. \(x = 3\)