Solve 5(x+3)-5((x)^(2)-1)=(x)^(2)+7(3) | 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

$$5(x+3)-5( { x }^{ 2 } -1)= { x }^{ 2 } +7(3)$$

Solve for x

$x=\frac{1}{3}\approx 0.333333333$
$x=\frac{1}{2}=0.5$

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring By Grouping

Use the distributive property to multiply $5$ by $x+3$.

$$5x+15-5\left(x^{2}-1\right)=x^{2}+7\times 3$$

Use the distributive property to multiply $-5$ by $x^{2}-1$.

$$5x+15-5x^{2}+5=x^{2}+7\times 3$$

Add $15$ and $5$ to get $20$.

$$5x+20-5x^{2}=x^{2}+7\times 3$$

Multiply $7$ and $3$ to get $21$.

$$5x+20-5x^{2}=x^{2}+21$$

Subtract $x^{2}$ from both sides.

$$5x+20-5x^{2}-x^{2}=21$$

Combine $-5x^{2}$ and $-x^{2}$ to get $-6x^{2}$.

$$5x+20-6x^{2}=21$$

Subtract $21$ from both sides.

$$5x+20-6x^{2}-21=0$$

Subtract $21$ from $20$ to get $-1$.

$$5x-1-6x^{2}=0$$

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

$$-6x^{2}+5x-1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $-6x^{2}+ax+bx-1$. To find $a$ and $b$, set up a system to be solved.

$$a+b=5$$ $$ab=-6\left(-1\right)=6$$

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 $6$.

$$1,6$$ $$2,3$$

Calculate the sum for each pair.

$$1+6=7$$ $$2+3=5$$

The solution is the pair that gives sum $5$.

$$a=3$$ $$b=2$$

Rewrite $-6x^{2}+5x-1$ as $\left(-6x^{2}+3x\right)+\left(2x-1\right)$.

$$\left(-6x^{2}+3x\right)+\left(2x-1\right)$$

Factor out $-3x$ in $-6x^{2}+3x$.

$$-3x\left(2x-1\right)+2x-1$$

Factor out common term $2x-1$ by using distributive property.

$$\left(2x-1\right)\left(-3x+1\right)$$

To find equation solutions, solve $2x-1=0$ and $-3x+1=0$.

$$x=\frac{1}{2}$$ $$x=\frac{1}{3}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply $5$ by $x+3$.

$$5x+15-5\left(x^{2}-1\right)=x^{2}+7\times 3$$

Use the distributive property to multiply $-5$ by $x^{2}-1$.

$$5x+15-5x^{2}+5=x^{2}+7\times 3$$

Add $15$ and $5$ to get $20$.

$$5x+20-5x^{2}=x^{2}+7\times 3$$

Multiply $7$ and $3$ to get $21$.

$$5x+20-5x^{2}=x^{2}+21$$

Subtract $x^{2}$ from both sides.

$$5x+20-5x^{2}-x^{2}=21$$

Combine $-5x^{2}$ and $-x^{2}$ to get $-6x^{2}$.

$$5x+20-6x^{2}=21$$

Subtract $21$ from both sides.

$$5x+20-6x^{2}-21=0$$

Subtract $21$ from $20$ to get $-1$.

$$5x-1-6x^{2}=0$$

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.

$$-6x^{2}+5x-1=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-6$ for $a$, $5$ for $b$, and $-1$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-5±\sqrt{5^{2}-4\left(-6\right)\left(-1\right)}}{2\left(-6\right)}$$

Square $5$.

$$x=\frac{-5±\sqrt{25-4\left(-6\right)\left(-1\right)}}{2\left(-6\right)}$$

Multiply $-4$ times $-6$.

$$x=\frac{-5±\sqrt{25+24\left(-1\right)}}{2\left(-6\right)}$$

Multiply $24$ times $-1$.

$$x=\frac{-5±\sqrt{25-24}}{2\left(-6\right)}$$

Add $25$ to $-24$.

$$x=\frac{-5±\sqrt{1}}{2\left(-6\right)}$$

Take the square root of $1$.

$$x=\frac{-5±1}{2\left(-6\right)}$$

Multiply $2$ times $-6$.

$$x=\frac{-5±1}{-12}$$

Now solve the equation $x=\frac{-5±1}{-12}$ when $±$ is plus. Add $-5$ to $1$.

$$x=-\frac{4}{-12}$$

Reduce the fraction $\frac{-4}{-12}$ to lowest terms by extracting and canceling out $4$.

$$x=\frac{1}{3}$$

Now solve the equation $x=\frac{-5±1}{-12}$ when $±$ is minus. Subtract $1$ from $-5$.

$$x=-\frac{6}{-12}$$

Reduce the fraction $\frac{-6}{-12}$ to lowest terms by extracting and canceling out $6$.

$$x=\frac{1}{2}$$

The equation is now solved.

$$x=\frac{1}{3}$$ $$x=\frac{1}{2}$$

Steps for Completing the Square

Use the distributive property to multiply $5$ by $x+3$.

$$5x+15-5\left(x^{2}-1\right)=x^{2}+7\times 3$$

Use the distributive property to multiply $-5$ by $x^{2}-1$.

$$5x+15-5x^{2}+5=x^{2}+7\times 3$$

Add $15$ and $5$ to get $20$.

$$5x+20-5x^{2}=x^{2}+7\times 3$$

Multiply $7$ and $3$ to get $21$.

$$5x+20-5x^{2}=x^{2}+21$$

Subtract $x^{2}$ from both sides.

$$5x+20-5x^{2}-x^{2}=21$$

Combine $-5x^{2}$ and $-x^{2}$ to get $-6x^{2}$.

$$5x+20-6x^{2}=21$$

Subtract $20$ from both sides.

$$5x-6x^{2}=21-20$$

Subtract $20$ from $21$ to get $1$.

$$5x-6x^{2}=1$$

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 $x^{2}+bx=c$.

$$-6x^{2}+5x=1$$

Divide both sides by $-6$.

$$\frac{-6x^{2}+5x}{-6}=\frac{1}{-6}$$

Dividing by $-6$ undoes the multiplication by $-6$.

$$x^{2}+\frac{5}{-6}x=\frac{1}{-6}$$

Divide $5$ by $-6$.

$$x^{2}-\frac{5}{6}x=\frac{1}{-6}$$

Divide $1$ by $-6$.

$$x^{2}-\frac{5}{6}x=-\frac{1}{6}$$

Divide $-\frac{5}{6}$, the coefficient of the $x$ term, by $2$ to get $-\frac{5}{12}$. Then add the square of $-\frac{5}{12}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{5}{6}x+\left(-\frac{5}{12}\right)^{2}=-\frac{1}{6}+\left(-\frac{5}{12}\right)^{2}$$

Square $-\frac{5}{12}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{6}x+\frac{25}{144}=-\frac{1}{6}+\frac{25}{144}$$

Add $-\frac{1}{6}$ to $\frac{25}{144}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{5}{6}x+\frac{25}{144}=\frac{1}{144}$$

Factor $x^{2}-\frac{5}{6}x+\frac{25}{144}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{5}{12}\right)^{2}=\frac{1}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{12}\right)^{2}}=\sqrt{\frac{1}{144}}$$

Simplify.

$$x-\frac{5}{12}=\frac{1}{12}$$ $$x-\frac{5}{12}=-\frac{1}{12}$$

Add $\frac{5}{12}$ to both sides of the equation.

$$x=\frac{1}{2}$$ $$x=\frac{1}{3}$$