Chapter 1, Problem 8T

(a)

To determine

To find: The all real solutions of equation $x+5=14-\frac{1}{2}x$.

Answer to Problem 8T

The real solution of equation $x+5=14-\frac{1}{2}x$ is 6.

Explanation of Solution

Given:

The linear equation is $x+5=14-\frac{1}{2}x$.

Calculation:

Simplify the given equation to find the value of x.

$\begin{array}{c}x+5=14-\frac{1}{2}x\\ x+5-14+\frac{1}{2}x=0\\ \frac{2x+10-28+x}{2}=0\end{array}$

Further simplify the above equation.

$\begin{array}{c}3x-18=0\\ 3x=18\\ x=\frac{18}{3}\\ x=6\end{array}$

Thus, the real solution of equation $x+5=14-\frac{1}{2}x$ is 6.

(b)

To determine

To find: The all real solutions of expression $\frac{2x}{x+1}=\frac{2x-1}{x}$.

Answer to Problem 8T

The real solution of expression $\frac{2x}{x+1}=\frac{2x-1}{x}$ is 1.

Explanation of Solution

Given:

The expression is $\frac{2x}{x+1}=\frac{2x-1}{x}$.

Calculation:

In the given expression eliminate the denominators to multiply each side by the lowest common denominator that is $x\left(x+1\right)$,

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

Thus, the real solution of expression $\frac{2x}{x+1}=\frac{2x-1}{x}$ is 1.

(c)

To determine

To find: The all real solutions of equation $x^2-x-12=0$.

Answer to Problem 8T

The real solutions of equation $x^2-x-12=0$ are $-3$ and 4.

Explanation of Solution

Given:

The equation is $x^2-x-12=0$.

Calculation:

Factor the given equation to get the values of x,

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

Further solve above equation to get values,

$\begin{array}{c}\left(x+3\right)\left(x-4\right)=0\\ x=-3,4\end{array}$

Thus, the real solutions of equation $x^2-x-12=0$ are $-3$ and 4.

(d)

To determine

To find: The all real solutions of equation $2x^2+4x+1=0$.

Answer to Problem 8T

The real solutions of equation $2x^2+4x+1=0$ are $-1\pm \frac{\sqrt{2}}{2}$.

Explanation of Solution

Given:

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

Calculation:

The Quadratic formula to find roots of equation $a{x}^2+bx+c=0$ is,

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

Where,

• $a\ne 0$

Substitute 2 for a, 4 for b and 1 for c from given equation to above formula,

$\begin{array}{c}x=\frac{-4\pm \sqrt{{\left(4\right)}^2-4\cdot 2\cdot 1}}{2\cdot 2}\\ x=\frac{-4\pm \sqrt{16-8}}{4}\\ x=-1\pm \frac{2\sqrt{2}}{4}\\ x=-1\pm \frac{\sqrt{2}}{2}\end{array}$

Thus, the real solutions of equation $2x^2+4x+1=0$ are $-1\pm \frac{\sqrt{2}}{2}$.

(e)

To determine

To find: The all real solutions of expression $\sqrt{3-\sqrt{x+5}}=2$.

Answer to Problem 8T

The expression $\sqrt{3-\sqrt{x+5}}=2$ has no real solution.

Explanation of Solution

Given:

The expression is $\sqrt{3-\sqrt{x+5}}=2$.

Calculation:

To eliminate the square root square both sides of given expression,

$\begin{array}{c}{\left(\sqrt{3-\sqrt{x+5}}\right)}^2={\left(2\right)}^2\\ 3-\sqrt{x+5}=4\end{array}$

The above expression does not define any term of x to get the real solution so that there is no real solution for this expression.

Thus, the expression $\sqrt{3-\sqrt{x+5}}=2$ has no real solution.

(f)

To determine

To find: The all real solutions of equation $x^4-3x^2+2=0$.

Answer to Problem 8T

The real solutions of equation $x^4-3x^2+2=0$ are $\pm 1$ and $\pm \sqrt{2}$.

Explanation of Solution

Given:

The equation is $x^4-3x^2+2=0$.

Calculation:

Factor the given equation to find the value of x.

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

Simplify the above expression.

$\begin{array}{c}{x}^2-1=0,x^2-2=0\\ x^2=1,x^2=2\\ x=\pm 1,x=\pm \sqrt{2}\end{array}$

Thus, the real solutions of equation $x^4-3x^2+2=0$ are $\pm 1$ and $\pm \sqrt{2}$.

(g)

To determine

To find: The all real solutions of equation $3\left|x-4\right|=10$.

Answer to Problem 8T

The real solutions of equation $3\left|x-4\right|=10$ are $\frac{2}{3}$ and $\frac{22}{3}$.

Explanation of Solution

Given:

The equation is $3\left|x-4\right|=10$.

Calculation:

By the definition of absolute value $3\left|x-4\right|=10$ is equivalent to,

$\begin{array}{c}3\left(x-4\right)=10\\ 3x-12=10\\ 3x=22\\ x=\frac{22}{3}\end{array}$

Or,

$\begin{array}{c}3\left(x-4\right)=-10\\ 3x-12=-10\\ 3x=2\\ x=\frac{2}{3}\end{array}$

Thus, the real solutions of equation $3\left|x-4\right|=10$ are $\frac{2}{3}$ and $\frac{22}{3}$.