HaCkEd By KaSpEr511

Worked Problems (Another Type Equations)

September 14th, 2009

Find all real solutions of the equation.
1. $x^4+64x=0$

Solution

$x(x^3+64)=0$ . Hence $x=0$ or $x^3+64=0$ . The last equation is equivalent to $x^3=-64\Leftrightarrow x=-4$ .
So the solutions are $x=0,-4$ .

2. $x^4-x^3-6x^2=0$

Solution

$\begin{align*}x^4-x^3-6x^2&=0\\x^2(x^2-x-6)&=0\\x^2(x-3)(x+2)&=0\end{align*}$
Hence $x=0,3,-2$ .

3. $2x^3+x^2-18x-9=0$

Solution

$\begin{align*}2x^3+x^2-18x-9&=0\\(2x^3+x^2)-(18x+9)&=0\\x^2(2x+1)-9(2x+1)&=0\\(2x+1)(x^2-9)&=0\\(2x+1)(x+3)(x-3)&=0\end{align*}$
Hence $\textstyle x=-\frac{1}{2},-3,3$ .

4. $\frac{1}{x-1}+\frac{1}{x+2}=\frac{5}{4}$

Solution

Multiply both sides by $4(x-1)(x+2)$ to get
$\begin{align*} {\color{Red} 4(x-1)(x+2)}\left(\frac{1}{x-1}+\frac{1}{x+2}\right)&={\color{Red}4(x-1)(x+2)}\frac{5}{4}\\ 4(x+2)+4(x-1)&=5(x-1)(x+2)\\ 4x+8+4x-4&=5(x^2+x-2)\\ 8x+4&=5x^2+5x-10\\ 5x^2-3x-14&=0\\ 5x^2-10x+7x-14&=0\\ (5x^2-10x)+(7x-14)&=0\\ 5x(x-2)+7(x-2)&=0\\ (5x+7)(x-2)&=0 \end{align*}$
Hence $\textstyle x=-\frac{7}{5},2$ .

5. $(x+5)^2-3(x+5)-10=0$

Solution

Let $z=x+5$ then the equation is equivalent to
$\begin{align*}z^2-3z-10&=0\\(z-5)(z+2)&=0\end{align*}$
Hence $z=5$ or $z=-2$ . It follows that $x+5=-2$ or $x+5=5$ . Solving these equations we have $x=-7$ or $x=0$ .

6. $\left(\frac{x+1}{x}\right)^2+4\left(\frac{x}{x+1}\right)+3=0$

Solution

Let $x=\frac{x+1}{x}$ , then the equation is equivalent to
$\begin{align*} z^2+4z+3&=0\\ (z+1)(z+3)&=0 \end{align*}$
Hence $z=-1$ or $z=-3$ . It follows that $\frac{x}{x+1}=-1$ or $\frac{x}{x+1}=-3$ .
From the first equation we have $x=-(x+1)$ . It follows that $2x=-1$ and hence $x=-1/2$ .
From the second equation we have $x=-3(x+1)$ . It follows that $4x=-3$ and hence $x=-3/4$ .
Therefore $x=-1/2$ or $x=-3/4$ are the solution of the original equation.

7. $x^6-26x^3-27=0$

Solution

Let $z=x^3$ , then the equation is equivalent to
$\begin{align*} z^2-26z-27&=0\\ (z-27)(z+1)&=0 \end{align*}$
Hence $z=27$ or $z=-1$ . It follows that $x^3=27$ or $x^3=-1$ and these implies that $x=3$ or $x=-1$ .

8. $2x+\sqrt{x+1}=8$

Solution

Rewrite the equation as $\sqrt{x+1}=8-2x$ . Square both sides, we have
$\begin{align*} x+1&=(8-2x)^2\\ x+1&=64-32x+4x^2\\ 0&=4x^2-33x+63\\ 0&=4x^2-12x-21x+63\\ 0&=4x(x-3)-21(x-3)\\ 0&=(4x-21)(x-3) \end{align*}$
Hence $x=21/4$ or $x=3$
One can check that $x=21/4$ is not a solution of the original equation and $x=2$ is a solution.
Therefore $x=2$ is the only solution.

9. $x+2\sqrt{x-7}=10$

Solution

Rewrite the equation as $2\sqrt{x-7}=10-x$ . Square both sides of the equation to have
$\begin{align*} 4(x-7)&=100-20x+x^2\\ 4x-28&=x^2-2x+100\\ 0&=x^2-6x+128 \end{align*}$
The discriminant of the last equation is $6^2-4\cdot1\cdot 128=36-512<0$ . Hence the equation has no solution.

10. $\sqrt[3]{4x^2-4x}=x$

Solution

Raise both sides to the third power to get $4x^2-4x=x^3$ . Now this equation is equivalent to
$\begin{align*} x^3-4x^2+4x&=0\\ x(x^2-4x+4)&=0\\ x(x-2)^2&=0 \end{align*}$
Hence $x=0$ or $x=2$ and one can easily verify that these two numbers are solutions of the original equation.

Categories: Algebra, MA109

Tags: Equation of Quadratic Type, Equation with Radicals Leave a comment

Kirthi Raman October 23rd, 2009

The solution of Problem 9
x+ 2 sqrt(x-7) = 10

is wrong.

There is a solution x=8

watchmath October 23rd, 2009

Yes you are right. I didn’t copy the second line correctly. After moving $x$ to the left and then squaring both sides we have
$\begin{align*}4(x-7)&=100-20x+x^2\\4x-28&=100-20x+x^2\\0&=x^2-24x+128\\0&=(x-8)(x-16) \end{align*}$
So $x=8$ or $x=16$ . By checking to the original equation one can see that $x=16$ is an extraneous solution. Therefore $x=8$ is the only solution.

Thanks again for the correction

Worked Problems (Another Type Equations)

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