Solve - x + 4x ^ 2 = 2x ^ 2 + 9 | 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

$$-x+4x^{2}=2x^{2}+9$$

Solve for x

$x = \frac{\sqrt{73} + 1}{4} \approx 2.386000936$
$x=\frac{1-\sqrt{73}}{4}\approx -1.886000936$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

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

$$-x+4x^{2}-2x^{2}=9$$

Combine $4x^{2}$ and $-2x^{2}$ to get $2x^{2}$.

$$-x+2x^{2}=9$$

Subtract $9$ from both sides.

$$-x+2x^{2}-9=0$$

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.

$$2x^{2}-x-9=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $2$ for $a$, $-1$ for $b$, and $-9$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 2\left(-9\right)}}{2\times 2}$$

Multiply $-4$ times $2$.

$$x=\frac{-\left(-1\right)±\sqrt{1-8\left(-9\right)}}{2\times 2}$$

Multiply $-8$ times $-9$.

$$x=\frac{-\left(-1\right)±\sqrt{1+72}}{2\times 2}$$

Add $1$ to $72$.

$$x=\frac{-\left(-1\right)±\sqrt{73}}{2\times 2}$$

The opposite of $-1$ is $1$.

$$x=\frac{1±\sqrt{73}}{2\times 2}$$

Multiply $2$ times $2$.

$$x=\frac{1±\sqrt{73}}{4}$$

Now solve the equation $x=\frac{1±\sqrt{73}}{4}$ when $±$ is plus. Add $1$ to $\sqrt{73}$.

$$x=\frac{\sqrt{73}+1}{4}$$

Now solve the equation $x=\frac{1±\sqrt{73}}{4}$ when $±$ is minus. Subtract $\sqrt{73}$ from $1$.

$$x=\frac{1-\sqrt{73}}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{73}+1}{4}$$ $$x=\frac{1-\sqrt{73}}{4}$$

Steps for Completing the Square

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

$$-x+4x^{2}-2x^{2}=9$$

Combine $4x^{2}$ and $-2x^{2}$ to get $2x^{2}$.

$$-x+2x^{2}=9$$

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

$$2x^{2}-x=9$$

Divide both sides by $2$.

$$\frac{2x^{2}-x}{2}=\frac{9}{2}$$

Dividing by $2$ undoes the multiplication by $2$.

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

Divide $-\frac{1}{2}$, the coefficient of the $x$ term, by $2$ to get $-\frac{1}{4}$. Then add the square of $-\frac{1}{4}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{2}x+\left(-\frac{1}{4}\right)^{2}=\frac{9}{2}+\left(-\frac{1}{4}\right)^{2}$$

Square $-\frac{1}{4}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{9}{2}+\frac{1}{16}$$

Add $\frac{9}{2}$ to $\frac{1}{16}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{73}{16}$$

Factor $x^{2}-\frac{1}{2}x+\frac{1}{16}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{1}{4}\right)^{2}=\frac{73}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{\frac{73}{16}}$$

Simplify.

$$x-\frac{1}{4}=\frac{\sqrt{73}}{4}$$ $$x-\frac{1}{4}=-\frac{\sqrt{73}}{4}$$

Add $\frac{1}{4}$ to both sides of the equation.

$$x=\frac{\sqrt{73}+1}{4}$$ $$x=\frac{1-\sqrt{73}}{4}$$