Question

pre - calculus 11
solving radical equations
name:
1 solve
a) $\sqrt{3x + 1}=7$ b) $\sqrt{2x + 7}=5$
c) $2\sqrt{x}=8$ d) $12\sqrt{x}=30$

1. solve

a) $\sqrt{x + 2}-5 = 0$ b) $\sqrt{x - 4}-7 = 0$
c) $\sqrt{7x - 3}-2 = 3$ d) $\sqrt{3x - 1}+7 = 10$

Explanation:

Step1: Isolate the radical

For part a) $\sqrt{3x + 1}=7$, the radical is already isolated.
For part b) $\sqrt{2x + 7}=5$, the radical is already isolated.
For part c) $2\sqrt{x}=8$, divide both sides by 2: $\sqrt{x}=\frac{8}{2}=4$.
For part d) $12\sqrt{x}=30$, divide both sides by 12: $\sqrt{x}=\frac{30}{12}=\frac{5}{2}$.
For part 2a) $\sqrt{x + 2}-5 = 0$, add 5 to both sides: $\sqrt{x + 2}=5$.
For part 2b) $\sqrt{x - 4}-7 = 0$, add 7 to both sides: $\sqrt{x - 4}=7$.
For part 2c) $\sqrt{7x - 3}-2 = 3$, add 2 to both sides: $\sqrt{7x - 3}=5$.
For part 2d) $\sqrt{3x - 1}+7 = 10$, subtract 7 from both sides: $\sqrt{3x - 1}=3$.

Step2: Square both sides

For part a): $(\sqrt{3x + 1})^2=7^2$, so $3x + 1 = 49$.
For part b): $(\sqrt{2x + 7})^2=5^2$, so $2x + 7 = 25$.
For part c): $(\sqrt{x})^2=4^2$, so $x = 16$.
For part d): $(\sqrt{x})^2=(\frac{5}{2})^2$, so $x=\frac{25}{4}$.
For part 2a): $(\sqrt{x + 2})^2=5^2$, so $x + 2 = 25$.
For part 2b): $(\sqrt{x - 4})^2=7^2$, so $x - 4 = 49$.
For part 2c): $(\sqrt{7x - 3})^2=5^2$, so $7x - 3 = 25$.
For part 2d): $(\sqrt{3x - 1})^2=3^2$, so $3x - 1 = 9$.

Step3: Solve for \(x\)

For part a):

$$\begin{align*} 3x+1&=49\\ 3x&=49 - 1\\ 3x&=48\\ x&=\frac{48}{3}\\ x& = 16 \end{align*}$$

For part b):

$$\begin{align*} 2x+7&=25\\ 2x&=25 - 7\\ 2x&=18\\ x&=\frac{18}{2}\\ x&=9 \end{align*}$$

For part c): \(x = 16\) (already found in Step2).
For part d): \(x=\frac{25}{4}\) (already found in Step2).
For part 2a):

$$\begin{align*} x+2&=25\\ x&=25 - 2\\ x&=23 \end{align*}$$

For part 2b):

$$\begin{align*} x - 4&=49\\ x&=49+4\\ x&=53 \end{align*}$$

For part 2c):

$$\begin{align*} 7x-3&=25\\ 7x&=25 + 3\\ 7x&=28\\ x&=\frac{28}{7}\\ x&=4 \end{align*}$$

For part 2d):

$$\begin{align*} 3x-1&=9\\ 3x&=9 + 1\\ 3x&=10\\ x&=\frac{10}{3} \end{align*}$$

Answer:

a) \(x = 16\)
b) \(x = 9\)
c) \(x = 16\)
d) \(x=\frac{25}{4}\)
2a) \(x = 23\)
2b) \(x = 53\)
2c) \(x = 4\)
2d) \(x=\frac{10}{3}\)