Solve 4x ^ 2 - (1x - 2) * x + 9 = 0 | 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

$$4x^{2}-(1x-2)x+9=0$$

Solve for x (complex solution)

$x=\frac{-1+\sqrt{26}i}{3}\approx -0.333333333+1.699673171i$
$x=\frac{-\sqrt{26}i-1}{3}\approx -0.333333333-1.699673171i$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Use the distributive property to multiply $1x-2$ by $x$.

$$4x^{2}-\left(x^{2}-2x\right)+9=0$$

To find the opposite of $x^{2}-2x$, find the opposite of each term.

$$4x^{2}-x^{2}+2x+9=0$$

Combine $4x^{2}$ and $-x^{2}$ to get $3x^{2}$.

$$3x^{2}+2x+9=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $3$ for $a$, $2$ for $b$, and $9$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-2±\sqrt{2^{2}-4\times 3\times 9}}{2\times 3}$$

Square $2$.

$$x=\frac{-2±\sqrt{4-4\times 3\times 9}}{2\times 3}$$

Multiply $-4$ times $3$.

$$x=\frac{-2±\sqrt{4-12\times 9}}{2\times 3}$$

Multiply $-12$ times $9$.

$$x=\frac{-2±\sqrt{4-108}}{2\times 3}$$

Add $4$ to $-108$.

$$x=\frac{-2±\sqrt{-104}}{2\times 3}$$

Take the square root of $-104$.

$$x=\frac{-2±2\sqrt{26}i}{2\times 3}$$

Multiply $2$ times $3$.

$$x=\frac{-2±2\sqrt{26}i}{6}$$

Now solve the equation $x=\frac{-2±2\sqrt{26}i}{6}$ when $±$ is plus. Add $-2$ to $2i\sqrt{26}$.

$$x=\frac{-2+2\sqrt{26}i}{6}$$

Divide $-2+2i\sqrt{26}$ by $6$.

$$x=\frac{-1+\sqrt{26}i}{3}$$

Now solve the equation $x=\frac{-2±2\sqrt{26}i}{6}$ when $±$ is minus. Subtract $2i\sqrt{26}$ from $-2$.

$$x=\frac{-2\sqrt{26}i-2}{6}$$

Divide $-2-2i\sqrt{26}$ by $6$.

$$x=\frac{-\sqrt{26}i-1}{3}$$

The equation is now solved.

$$x=\frac{-1+\sqrt{26}i}{3}$$ $$x=\frac{-\sqrt{26}i-1}{3}$$

Steps for Completing the Square

Use the distributive property to multiply $1x-2$ by $x$.

$$4x^{2}-\left(x^{2}-2x\right)+9=0$$

To find the opposite of $x^{2}-2x$, find the opposite of each term.

$$4x^{2}-x^{2}+2x+9=0$$

Combine $4x^{2}$ and $-x^{2}$ to get $3x^{2}$.

$$3x^{2}+2x+9=0$$

Subtract $9$ from both sides. Anything subtracted from zero gives its negation.

$$3x^{2}+2x=-9$$

Divide both sides by $3$.

$$\frac{3x^{2}+2x}{3}=-\frac{9}{3}$$

Dividing by $3$ undoes the multiplication by $3$.

$$x^{2}+\frac{2}{3}x=-\frac{9}{3}$$

Divide $-9$ by $3$.

$$x^{2}+\frac{2}{3}x=-3$$

Divide $\frac{2}{3}$, the coefficient of the $x$ term, by $2$ to get $\frac{1}{3}$. Then add the square of $\frac{1}{3}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square $\frac{1}{3}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=-3+\frac{1}{9}$$

Add $-3$ to $\frac{1}{9}$.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=-\frac{26}{9}$$

Factor $x^{2}+\frac{2}{3}x+\frac{1}{9}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+\frac{1}{3}\right)^{2}=-\frac{26}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{3}\right)^{2}}=\sqrt{-\frac{26}{9}}$$

Simplify.

$$x+\frac{1}{3}=\frac{\sqrt{26}i}{3}$$ $$x+\frac{1}{3}=-\frac{\sqrt{26}i}{3}$$

Subtract $\frac{1}{3}$ from both sides of the equation.

$$x=\frac{-1+\sqrt{26}i}{3}$$ $$x=\frac{-\sqrt{26}i-1}{3}$$