Solve 40+5x=14x-((1)^(2))/(2x) | 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

$$40+5x=14x- \frac{ { 1 }^{ 2 } }{ 2x }$$

Solve for x

$x=\frac{\sqrt{1618}}{18}+\frac{20}{9}\approx 4.456909485$
$x=-\frac{\sqrt{1618}}{18}+\frac{20}{9}\approx -0.01246504$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Variable $x$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $2x$.

$$2x\times 40+5x\times 2x=14x\times 2x-1^{2}$$

Multiply $2$ and $40$ to get $80$.

$$80x+5x\times 2x=14x\times 2x-1^{2}$$

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

$$80x+5x^{2}\times 2=14x\times 2x-1^{2}$$

Multiply $5$ and $2$ to get $10$.

$$80x+10x^{2}=14x\times 2x-1^{2}$$

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

$$80x+10x^{2}=14x^{2}\times 2-1^{2}$$

Multiply $14$ and $2$ to get $28$.

$$80x+10x^{2}=28x^{2}-1^{2}$$

Calculate $1$ to the power of $2$ and get $1$.

$$80x+10x^{2}=28x^{2}-1$$

Subtract $28x^{2}$ from both sides.

$$80x+10x^{2}-28x^{2}=-1$$

Combine $10x^{2}$ and $-28x^{2}$ to get $-18x^{2}$.

$$80x-18x^{2}=-1$$

Add $1$ to both sides.

$$80x-18x^{2}+1=0$$

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.

$$-18x^{2}+80x+1=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-18$ for $a$, $80$ for $b$, and $1$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-80±\sqrt{80^{2}-4\left(-18\right)}}{2\left(-18\right)}$$

Square $80$.

$$x=\frac{-80±\sqrt{6400-4\left(-18\right)}}{2\left(-18\right)}$$

Multiply $-4$ times $-18$.

$$x=\frac{-80±\sqrt{6400+72}}{2\left(-18\right)}$$

Add $6400$ to $72$.

$$x=\frac{-80±\sqrt{6472}}{2\left(-18\right)}$$

Take the square root of $6472$.

$$x=\frac{-80±2\sqrt{1618}}{2\left(-18\right)}$$

Multiply $2$ times $-18$.

$$x=\frac{-80±2\sqrt{1618}}{-36}$$

Now solve the equation $x=\frac{-80±2\sqrt{1618}}{-36}$ when $±$ is plus. Add $-80$ to $2\sqrt{1618}$.

$$x=\frac{2\sqrt{1618}-80}{-36}$$

Divide $-80+2\sqrt{1618}$ by $-36$.

$$x=-\frac{\sqrt{1618}}{18}+\frac{20}{9}$$

Now solve the equation $x=\frac{-80±2\sqrt{1618}}{-36}$ when $±$ is minus. Subtract $2\sqrt{1618}$ from $-80$.

$$x=\frac{-2\sqrt{1618}-80}{-36}$$

Divide $-80-2\sqrt{1618}$ by $-36$.

$$x=\frac{\sqrt{1618}}{18}+\frac{20}{9}$$

The equation is now solved.

$$x=-\frac{\sqrt{1618}}{18}+\frac{20}{9}$$ $$x=\frac{\sqrt{1618}}{18}+\frac{20}{9}$$

Steps for Completing the Square

Variable $x$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $2x$.

$$2x\times 40+5x\times 2x=14x\times 2x-1^{2}$$

Multiply $2$ and $40$ to get $80$.

$$80x+5x\times 2x=14x\times 2x-1^{2}$$

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

$$80x+5x^{2}\times 2=14x\times 2x-1^{2}$$

Multiply $5$ and $2$ to get $10$.

$$80x+10x^{2}=14x\times 2x-1^{2}$$

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

$$80x+10x^{2}=14x^{2}\times 2-1^{2}$$

Multiply $14$ and $2$ to get $28$.

$$80x+10x^{2}=28x^{2}-1^{2}$$

Calculate $1$ to the power of $2$ and get $1$.

$$80x+10x^{2}=28x^{2}-1$$

Subtract $28x^{2}$ from both sides.

$$80x+10x^{2}-28x^{2}=-1$$

Combine $10x^{2}$ and $-28x^{2}$ to get $-18x^{2}$.

$$80x-18x^{2}=-1$$

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

$$-18x^{2}+80x=-1$$

Divide both sides by $-18$.

$$\frac{-18x^{2}+80x}{-18}=-\frac{1}{-18}$$

Dividing by $-18$ undoes the multiplication by $-18$.

$$x^{2}+\frac{80}{-18}x=-\frac{1}{-18}$$

Reduce the fraction $\frac{80}{-18}$ to lowest terms by extracting and canceling out $2$.

$$x^{2}-\frac{40}{9}x=-\frac{1}{-18}$$

Divide $-1$ by $-18$.

$$x^{2}-\frac{40}{9}x=\frac{1}{18}$$

Divide $-\frac{40}{9}$, the coefficient of the $x$ term, by $2$ to get $-\frac{20}{9}$. Then add the square of $-\frac{20}{9}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{40}{9}x+\left(-\frac{20}{9}\right)^{2}=\frac{1}{18}+\left(-\frac{20}{9}\right)^{2}$$

Square $-\frac{20}{9}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{40}{9}x+\frac{400}{81}=\frac{1}{18}+\frac{400}{81}$$

Add $\frac{1}{18}$ to $\frac{400}{81}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{40}{9}x+\frac{400}{81}=\frac{809}{162}$$

Factor $x^{2}-\frac{40}{9}x+\frac{400}{81}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{20}{9}\right)^{2}=\frac{809}{162}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{20}{9}\right)^{2}}=\sqrt{\frac{809}{162}}$$

Simplify.

$$x-\frac{20}{9}=\frac{\sqrt{1618}}{18}$$ $$x-\frac{20}{9}=-\frac{\sqrt{1618}}{18}$$

Add $\frac{20}{9}$ to both sides of the equation.

$$x=\frac{\sqrt{1618}}{18}+\frac{20}{9}$$ $$x=-\frac{\sqrt{1618}}{18}+\frac{20}{9}$$