Chapter T, Problem 8ADT

Solve the equation. (Find only the real solutions.)

1. (a) $x+5=14-\frac{1}{2}x$
2. (b) $\frac{2x}{x+1}=\frac{2x-1}{x}$
3. (c) x² − x − 12 = 0
4. (d) 2x² + 4x + 1 = 0
5. (e) x⁴ − 3x² + 2 = 0
6. (f) 3|x − 4| = 10
7. (g) $2x{(4-x)}^{-1/2}-3\sqrt{4-x}=0$

To determine

To solve: The equation $x+5=14-\frac{1}{2}x$.

Answer to Problem 8ADT

The value of x is 6.

Explanation of Solution

Consider the equation $x+5=14-\frac{1}{2}x$.

Simplify the above expression as follows,

$\begin{array}{c}x+5=14-\frac{1}{2}x\\ x+\frac{1}{2}x=14-5\\ \frac{3}{2}x=9\\ x=9\times \frac{2}{3}\end{array}$

On further simplification the above equation becomes,

$\begin{array}{c}x=3\times 2\\ =6\end{array}$

Thus, the value of x is 6.

To determine

To solve: The equation $\frac{2x}{x+1}=\frac{2x-1}{x}$.

Answer to Problem 8ADT

The value of x is 1.

Explanation of Solution

Consider the equation $\frac{2x}{x+1}=\frac{2x-1}{x}$.

Simplify the above expression as follows,

$\begin{array}{c}\frac{2x}{x+1}=\frac{2x-1}{x}\\ 2x\times x=\left(x+1\right)\left(2x-1\right)\\ 2x^2=2x^2+2x-x-1\\ x-1=0\end{array}$

$x=1$

Thus, the value of x is 1.

To determine

To solve: The equation $x^2-x-12=0$.

Answer to Problem 8ADT

The values of x are $4,-3$.

Explanation of Solution

Consider the equation $x^2-x-12=0$.

Simplify the above expression as follows,

$\begin{array}{l}{x}^2-x-12=0\\ x^2-4x+3x-12=0\\ x\left(x-4\right)+3\left(x-4\right)=0\\ \left(x-4\right)\left(x+3\right)=0\end{array}$

$x=4,-3$

Thus, the values of x are $4,-3$.

To determine

To solve: The equation $2x^2+4x+1=0$.

Answer to Problem 8ADT

The values of x are $-1\pm \frac{1}{2}\sqrt{2}$.

Explanation of Solution

Formula used:

The quadratic equation is $a{x}^2+bx+c=0$.

The value is $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

Calculation:

Consider the equation $x^4-3x^2+2=0$.

Use the above mentioned Formula and simplifies the above expression as follows,

The quadratic equation is $2x^2+4x+1=0$

The value is

$\begin{array}{c}x=\frac{-4\pm \sqrt{4^2-4\times 2\times 1}}{2\times 2}\\ =\frac{-4\pm \sqrt{16-8}}{4}\\ =\frac{-4\pm \sqrt{8}}{4}\\ =\frac{-4\pm 2\sqrt{2}}{4}\end{array}$

$=-1\pm \frac{1}{2}\sqrt{2}$

Thus, the values of x are $-1\pm \frac{1}{2}\sqrt{2}$.

Thus, the values of x are $-1\pm \frac{1}{2}\sqrt{2}$.

To determine

To solve: The equation $x^4-3x^2+2=0$.

Answer to Problem 8ADT

The values of x are $\pm \sqrt{2},\pm 1$.

Explanation of Solution

Consider the equation $x^4-3x^2+2=0$.

Simplify the above expression as follows,

$\begin{array}{l}{x}^4-3x^2+2=0\\ x^4-2x^2-x^2+2=0\\ x^2\left(x^2-2\right)-1\left(x^2-2\right)=0\\ \left(x^2-2\right)\left(x^2-1\right)=0\end{array}$

Either,

$\begin{array}{l}{x}^2-2=0\\ x^2=2\\ x=\pm \sqrt{2}\end{array}$

Or

$\begin{array}{l}{x}^2-1=0\\ x^2=1\\ x=\pm 1\end{array}$

Thus, the values of x are $\pm \sqrt{2},\pm 1$.

To determine

To solve: The equation $3\left|x-4\right|=10$.

Answer to Problem 8ADT

The values of x are $\frac{22}{3},\frac{2}{3}$.

Explanation of Solution

Consider the equation $3\left|x-4\right|=10$.

Simplify the above expression as follows,

$\begin{array}{l}3\left|x-4\right|=10\\ \left|x-4\right|=\frac{10}{3}\\ x-4=\pm \frac{10}{3}\end{array}$

Take positive,

$\begin{array}{c}x-4=\frac{10}{3}\\ x=4+\frac{10}{3}\\ =\frac{22}{3}\end{array}$

Take negative,

$\begin{array}{c}x-4=-\frac{10}{3}\\ x=4-\frac{10}{3}\\ =\frac{2}{3}\end{array}$

Thus, the values of x are $\frac{22}{3},\frac{2}{3}$.

To determine

To solve: The equation $2x{\left(4-x\right)}^{-\frac{1}{2}}-3\sqrt{4-x}=0$.

Answer to Problem 8ADT

The value of x is $\frac{12}{5}$.

Explanation of Solution

Consider the equation $2x{\left(4-x\right)}^{-\frac{1}{2}}-3\sqrt{4-x}=0$.

Simplify the above expression as follows,

$\begin{array}{l}2x{\left(4-x\right)}^{-\frac{1}{2}}-3\sqrt{4-x}=0\\ {\left(4-x\right)}^{-\frac{1}{2}}\left(2x-3\left(4-x\right)\right)=0\\ {\left(4-x\right)}^{-\frac{1}{2}}\left(2x-12+3x\right)=0\\ {\left(4-x\right)}^{-\frac{1}{2}}\left(5x-12\right)=0\end{array}$

On further simplification the above equation becomes,

$\begin{array}{l}\left(5x-12\right)=0\\ 5x=12\\ x=\frac{12}{5}\end{array}$

Thus, the value of x is $\frac{12}{5}$.