Solve k ) 5x ^ 2 = 8x + 21 | 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

$$5 x ^ { 2 } = 8 x + 21$$

Solve for x

$x = -\frac{7}{5} = -1\frac{2}{5} = -1.4$
$x=3$

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring By Grouping

Subtract $8x$ from both sides.

$$5x^{2}-8x=21$$

Subtract $21$ from both sides.

$$5x^{2}-8x-21=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $5x^{2}+ax+bx-21$. To find $a$ and $b$, set up a system to be solved.

$$a+b=-8$$ $$ab=5\left(-21\right)=-105$$

Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product $-105$.

$$1,-105$$ $$3,-35$$ $$5,-21$$ $$7,-15$$

Calculate the sum for each pair.

$$1-105=-104$$ $$3-35=-32$$ $$5-21=-16$$ $$7-15=-8$$

The solution is the pair that gives sum $-8$.

$$a=-15$$ $$b=7$$

Rewrite $5x^{2}-8x-21$ as $\left(5x^{2}-15x\right)+\left(7x-21\right)$.

$$\left(5x^{2}-15x\right)+\left(7x-21\right)$$

Factor out $5x$ in the first and $7$ in the second group.

$$5x\left(x-3\right)+7\left(x-3\right)$$

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

$$\left(x-3\right)\left(5x+7\right)$$

To find equation solutions, solve $x-3=0$ and $5x+7=0$.

$$x=3$$ $$x=-\frac{7}{5}$$

Steps Using the Quadratic Formula

Subtract $8x$ from both sides.

$$5x^{2}-8x=21$$

Subtract $21$ from both sides.

$$5x^{2}-8x-21=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $5$ for $a$, $-8$ for $b$, and $-21$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 5\left(-21\right)}}{2\times 5}$$

Square $-8$.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 5\left(-21\right)}}{2\times 5}$$

Multiply $-4$ times $5$.

$$x=\frac{-\left(-8\right)±\sqrt{64-20\left(-21\right)}}{2\times 5}$$

Multiply $-20$ times $-21$.

$$x=\frac{-\left(-8\right)±\sqrt{64+420}}{2\times 5}$$

Add $64$ to $420$.

$$x=\frac{-\left(-8\right)±\sqrt{484}}{2\times 5}$$

Take the square root of $484$.

$$x=\frac{-\left(-8\right)±22}{2\times 5}$$

The opposite of $-8$ is $8$.

$$x=\frac{8±22}{2\times 5}$$

Multiply $2$ times $5$.

$$x=\frac{8±22}{10}$$

Now solve the equation $x=\frac{8±22}{10}$ when $±$ is plus. Add $8$ to $22$.

$$x=\frac{30}{10}$$

Divide $30$ by $10$.

$$x=3$$

Now solve the equation $x=\frac{8±22}{10}$ when $±$ is minus. Subtract $22$ from $8$.

$$x=-\frac{14}{10}$$

Reduce the fraction $\frac{-14}{10}$ to lowest terms by extracting and canceling out $2$.

$$x=-\frac{7}{5}$$

The equation is now solved.

$$x=3$$ $$x=-\frac{7}{5}$$

Steps for Completing the Square

Subtract $8x$ from both sides.

$$5x^{2}-8x=21$$

Divide both sides by $5$.

$$\frac{5x^{2}-8x}{5}=\frac{21}{5}$$

Dividing by $5$ undoes the multiplication by $5$.

$$x^{2}-\frac{8}{5}x=\frac{21}{5}$$

Divide $-\frac{8}{5}$, the coefficient of the $x$ term, by $2$ to get $-\frac{4}{5}$. Then add the square of $-\frac{4}{5}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{8}{5}x+\left(-\frac{4}{5}\right)^{2}=\frac{21}{5}+\left(-\frac{4}{5}\right)^{2}$$

Square $-\frac{4}{5}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{8}{5}x+\frac{16}{25}=\frac{21}{5}+\frac{16}{25}$$

Add $\frac{21}{5}$ to $\frac{16}{25}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{8}{5}x+\frac{16}{25}=\frac{121}{25}$$

Factor $x^{2}-\frac{8}{5}x+\frac{16}{25}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{4}{5}\right)^{2}=\frac{121}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{4}{5}\right)^{2}}=\sqrt{\frac{121}{25}}$$

Simplify.

$$x-\frac{4}{5}=\frac{11}{5}$$ $$x-\frac{4}{5}=-\frac{11}{5}$$

Add $\frac{4}{5}$ to both sides of the equation.

$$x=3$$ $$x=-\frac{7}{5}$$