Solve QUESTIONS 1 . Find the value of x / 7 * x - (2 + 5x) = 18 | Equatix - Math Solver

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Example

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a⋅(b-2)=3b

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

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Question

$$\left. \begin{array} { l } { x : 7 x - ( 2 + 5 x ) = 18 } \end{array} \right.$$

Solve for x

$x = \frac{\sqrt{1785} + 35}{2} \approx 38.624630174$
$x=\frac{35-\sqrt{1785}}{2}\approx -3.624630174$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Multiply both sides of the equation by $7$.

$$xx-7\left(2+5x\right)=126$$

Multiply $x$ and $x$ to get $x^{2}$.

$$x^{2}-7\left(2+5x\right)=126$$

Use the distributive property to multiply $-7$ by $2+5x$.

$$x^{2}-14-35x=126$$

Subtract $126$ from both sides.

$$x^{2}-14-35x-126=0$$

Subtract $126$ from $-14$ to get $-140$.

$$x^{2}-140-35x=0$$

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.

$$x^{2}-35x-140=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $-35$ for $b$, and $-140$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-35\right)±\sqrt{\left(-35\right)^{2}-4\left(-140\right)}}{2}$$

Square $-35$.

$$x=\frac{-\left(-35\right)±\sqrt{1225-4\left(-140\right)}}{2}$$

Multiply $-4$ times $-140$.

$$x=\frac{-\left(-35\right)±\sqrt{1225+560}}{2}$$

Add $1225$ to $560$.

$$x=\frac{-\left(-35\right)±\sqrt{1785}}{2}$$

The opposite of $-35$ is $35$.

$$x=\frac{35±\sqrt{1785}}{2}$$

Now solve the equation $x=\frac{35±\sqrt{1785}}{2}$ when $±$ is plus. Add $35$ to $\sqrt{1785}$.

$$x=\frac{\sqrt{1785}+35}{2}$$

Now solve the equation $x=\frac{35±\sqrt{1785}}{2}$ when $±$ is minus. Subtract $\sqrt{1785}$ from $35$.

$$x=\frac{35-\sqrt{1785}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{1785}+35}{2}$$ $$x=\frac{35-\sqrt{1785}}{2}$$

Steps for Completing the Square

Multiply both sides of the equation by $7$.

$$xx-7\left(2+5x\right)=126$$

Multiply $x$ and $x$ to get $x^{2}$.

$$x^{2}-7\left(2+5x\right)=126$$

Use the distributive property to multiply $-7$ by $2+5x$.

$$x^{2}-14-35x=126$$

Add $14$ to both sides.

$$x^{2}-35x=126+14$$

Add $126$ and $14$ to get $140$.

$$x^{2}-35x=140$$

Divide $-35$, the coefficient of the $x$ term, by $2$ to get $-\frac{35}{2}$. Then add the square of $-\frac{35}{2}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-35x+\left(-\frac{35}{2}\right)^{2}=140+\left(-\frac{35}{2}\right)^{2}$$

Square $-\frac{35}{2}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-35x+\frac{1225}{4}=140+\frac{1225}{4}$$

Add $140$ to $\frac{1225}{4}$.

$$x^{2}-35x+\frac{1225}{4}=\frac{1785}{4}$$

Factor $x^{2}-35x+\frac{1225}{4}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{35}{2}\right)^{2}=\frac{1785}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{35}{2}\right)^{2}}=\sqrt{\frac{1785}{4}}$$

Simplify.

$$x-\frac{35}{2}=\frac{\sqrt{1785}}{2}$$ $$x-\frac{35}{2}=-\frac{\sqrt{1785}}{2}$$

Add $\frac{35}{2}$ to both sides of the equation.

$$x=\frac{\sqrt{1785}+35}{2}$$ $$x=\frac{35-\sqrt{1785}}{2}$$