Solve (1)/((x-1)^(2))+(3)/(x-1)-10=0 | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$\frac{ 1 }{ { \left(x-1 \right) }^{ 2 } } + \frac{ 3 }{ x-1 } -10=0$$

Solve for x

$x=\frac{4}{5}=0.8$
$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring By Grouping

Variable $x$ cannot be equal to $1$ since division by zero is not defined. Multiply both sides of the equation by $\left(x-1\right)^{2}$, the least common multiple of $\left(x-1\right)^{2},x-1$.

$$1+\left(x-1\right)\times 3+\left(x-1\right)^{2}\left(-10\right)=0$$

Use the distributive property to multiply $x-1$ by $3$.

$$1+3x-3+\left(x-1\right)^{2}\left(-10\right)=0$$

Subtract $3$ from $1$ to get $-2$.

$$-2+3x+\left(x-1\right)^{2}\left(-10\right)=0$$

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(x-1\right)^{2}$.

$$-2+3x+\left(x^{2}-2x+1\right)\left(-10\right)=0$$

Use the distributive property to multiply $x^{2}-2x+1$ by $-10$.

$$-2+3x-10x^{2}+20x-10=0$$

Combine $3x$ and $20x$ to get $23x$.

$$-2+23x-10x^{2}-10=0$$

Subtract $10$ from $-2$ to get $-12$.

$$-12+23x-10x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-10x^{2}+23x-12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $-10x^{2}+ax+bx-12$. To find $a$ and $b$, set up a system to be solved.

$$a+b=23$$ $$ab=-10\left(-12\right)=120$$

Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is positive, $a$ and $b$ are both positive. List all such integer pairs that give product $120$.

$$1,120$$ $$2,60$$ $$3,40$$ $$4,30$$ $$5,24$$ $$6,20$$ $$8,15$$ $$10,12$$

Calculate the sum for each pair.

$$1+120=121$$ $$2+60=62$$ $$3+40=43$$ $$4+30=34$$ $$5+24=29$$ $$6+20=26$$ $$8+15=23$$ $$10+12=22$$

The solution is the pair that gives sum $23$.

$$a=15$$ $$b=8$$

Rewrite $-10x^{2}+23x-12$ as $\left(-10x^{2}+15x\right)+\left(8x-12\right)$.

$$\left(-10x^{2}+15x\right)+\left(8x-12\right)$$

Factor out $-5x$ in the first and $4$ in the second group.

$$-5x\left(2x-3\right)+4\left(2x-3\right)$$

Factor out common term $2x-3$ by using distributive property.

$$\left(2x-3\right)\left(-5x+4\right)$$

To find equation solutions, solve $2x-3=0$ and $-5x+4=0$.

$$x=\frac{3}{2}$$ $$x=\frac{4}{5}$$

Steps Using the Quadratic Formula

Variable $x$ cannot be equal to $1$ since division by zero is not defined. Multiply both sides of the equation by $\left(x-1\right)^{2}$, the least common multiple of $\left(x-1\right)^{2},x-1$.

$$1+\left(x-1\right)\times 3+\left(x-1\right)^{2}\left(-10\right)=0$$

Use the distributive property to multiply $x-1$ by $3$.

$$1+3x-3+\left(x-1\right)^{2}\left(-10\right)=0$$

Subtract $3$ from $1$ to get $-2$.

$$-2+3x+\left(x-1\right)^{2}\left(-10\right)=0$$

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(x-1\right)^{2}$.

$$-2+3x+\left(x^{2}-2x+1\right)\left(-10\right)=0$$

Use the distributive property to multiply $x^{2}-2x+1$ by $-10$.

$$-2+3x-10x^{2}+20x-10=0$$

Combine $3x$ and $20x$ to get $23x$.

$$-2+23x-10x^{2}-10=0$$

Subtract $10$ from $-2$ to get $-12$.

$$-12+23x-10x^{2}=0$$

All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.

$$-10x^{2}+23x-12=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-10$ for $a$, $23$ for $b$, and $-12$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-23±\sqrt{23^{2}-4\left(-10\right)\left(-12\right)}}{2\left(-10\right)}$$

Square $23$.

$$x=\frac{-23±\sqrt{529-4\left(-10\right)\left(-12\right)}}{2\left(-10\right)}$$

Multiply $-4$ times $-10$.

$$x=\frac{-23±\sqrt{529+40\left(-12\right)}}{2\left(-10\right)}$$

Multiply $40$ times $-12$.

$$x=\frac{-23±\sqrt{529-480}}{2\left(-10\right)}$$

Add $529$ to $-480$.

$$x=\frac{-23±\sqrt{49}}{2\left(-10\right)}$$

Take the square root of $49$.

$$x=\frac{-23±7}{2\left(-10\right)}$$

Multiply $2$ times $-10$.

$$x=\frac{-23±7}{-20}$$

Now solve the equation $x=\frac{-23±7}{-20}$ when $±$ is plus. Add $-23$ to $7$.

$$x=-\frac{16}{-20}$$

Reduce the fraction $\frac{-16}{-20}$ to lowest terms by extracting and canceling out $4$.

$$x=\frac{4}{5}$$

Now solve the equation $x=\frac{-23±7}{-20}$ when $±$ is minus. Subtract $7$ from $-23$.

$$x=-\frac{30}{-20}$$

Reduce the fraction $\frac{-30}{-20}$ to lowest terms by extracting and canceling out $10$.

$$x=\frac{3}{2}$$

The equation is now solved.

$$x=\frac{4}{5}$$ $$x=\frac{3}{2}$$

Steps for Completing the Square

Variable $x$ cannot be equal to $1$ since division by zero is not defined. Multiply both sides of the equation by $\left(x-1\right)^{2}$, the least common multiple of $\left(x-1\right)^{2},x-1$.

$$1+\left(x-1\right)\times 3+\left(x-1\right)^{2}\left(-10\right)=0$$

Use the distributive property to multiply $x-1$ by $3$.

$$1+3x-3+\left(x-1\right)^{2}\left(-10\right)=0$$

Subtract $3$ from $1$ to get $-2$.

$$-2+3x+\left(x-1\right)^{2}\left(-10\right)=0$$

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(x-1\right)^{2}$.

$$-2+3x+\left(x^{2}-2x+1\right)\left(-10\right)=0$$

Use the distributive property to multiply $x^{2}-2x+1$ by $-10$.

$$-2+3x-10x^{2}+20x-10=0$$

Combine $3x$ and $20x$ to get $23x$.

$$-2+23x-10x^{2}-10=0$$

Subtract $10$ from $-2$ to get $-12$.

$$-12+23x-10x^{2}=0$$

Add $12$ to both sides. Anything plus zero gives itself.

$$23x-10x^{2}=12$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form $x^{2}+bx=c$.

$$-10x^{2}+23x=12$$

Divide both sides by $-10$.

$$\frac{-10x^{2}+23x}{-10}=\frac{12}{-10}$$

Dividing by $-10$ undoes the multiplication by $-10$.

$$x^{2}+\frac{23}{-10}x=\frac{12}{-10}$$

Divide $23$ by $-10$.

$$x^{2}-\frac{23}{10}x=\frac{12}{-10}$$

Reduce the fraction $\frac{12}{-10}$ to lowest terms by extracting and canceling out $2$.

$$x^{2}-\frac{23}{10}x=-\frac{6}{5}$$

Divide $-\frac{23}{10}$, the coefficient of the $x$ term, by $2$ to get $-\frac{23}{20}$. Then add the square of $-\frac{23}{20}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{23}{10}x+\left(-\frac{23}{20}\right)^{2}=-\frac{6}{5}+\left(-\frac{23}{20}\right)^{2}$$

Square $-\frac{23}{20}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{23}{10}x+\frac{529}{400}=-\frac{6}{5}+\frac{529}{400}$$

Add $-\frac{6}{5}$ to $\frac{529}{400}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{23}{10}x+\frac{529}{400}=\frac{49}{400}$$

Factor $x^{2}-\frac{23}{10}x+\frac{529}{400}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{23}{20}\right)^{2}=\frac{49}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{23}{20}\right)^{2}}=\sqrt{\frac{49}{400}}$$

Simplify.

$$x-\frac{23}{20}=\frac{7}{20}$$ $$x-\frac{23}{20}=-\frac{7}{20}$$

Add $\frac{23}{20}$ to both sides of the equation.

$$x=\frac{3}{2}$$ $$x=\frac{4}{5}$$