Solve (x)/(2)\times(3x)/(8)+9=x | 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

$$\frac{ x }{ 2 } \times \frac{ 3x }{ 8 } +9=x$$

Solve for x (complex solution)

$x=\frac{8+4\sqrt{23}i}{3}\approx 2.666666667+6.394442031i$
$x=\frac{-4\sqrt{23}i+8}{3}\approx 2.666666667-6.394442031i$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Multiply $\frac{x}{2}$ times $\frac{3x}{8}$ by multiplying numerator times numerator and denominator times denominator.

$$\frac{x\times 3x}{2\times 8}+9=x$$

To add or subtract expressions, expand them to make their denominators the same. Multiply $9$ times $\frac{2\times 8}{2\times 8}$.

$$\frac{x\times 3x}{2\times 8}+\frac{9\times 2\times 8}{2\times 8}=x$$

Since $\frac{x\times 3x}{2\times 8}$ and $\frac{9\times 2\times 8}{2\times 8}$ have the same denominator, add them by adding their numerators.

$$\frac{x\times 3x+9\times 2\times 8}{2\times 8}=x$$

Do the multiplications in $x\times 3x+9\times 2\times 8$.

$$\frac{3x^{2}+144}{2\times 8}=x$$

Multiply $2$ and $8$ to get $16$.

$$\frac{3x^{2}+144}{16}=x$$

Divide each term of $3x^{2}+144$ by $16$ to get $\frac{3}{16}x^{2}+9$.

$$\frac{3}{16}x^{2}+9=x$$

Subtract $x$ from both sides.

$$\frac{3}{16}x^{2}+9-x=0$$

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.

$$\frac{3}{16}x^{2}-x+9=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $\frac{3}{16}$ for $a$, $-1$ for $b$, and $9$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times \frac{3}{16}\times 9}}{2\times \frac{3}{16}}$$

Multiply $-4$ times $\frac{3}{16}$.

$$x=\frac{-\left(-1\right)±\sqrt{1-\frac{3}{4}\times 9}}{2\times \frac{3}{16}}$$

Multiply $-\frac{3}{4}$ times $9$.

$$x=\frac{-\left(-1\right)±\sqrt{1-\frac{27}{4}}}{2\times \frac{3}{16}}$$

Add $1$ to $-\frac{27}{4}$.

$$x=\frac{-\left(-1\right)±\sqrt{-\frac{23}{4}}}{2\times \frac{3}{16}}$$

Take the square root of $-\frac{23}{4}$.

$$x=\frac{-\left(-1\right)±\frac{\sqrt{23}i}{2}}{2\times \frac{3}{16}}$$

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

$$x=\frac{1±\frac{\sqrt{23}i}{2}}{2\times \frac{3}{16}}$$

Multiply $2$ times $\frac{3}{16}$.

$$x=\frac{1±\frac{\sqrt{23}i}{2}}{\frac{3}{8}}$$

Now solve the equation $x=\frac{1±\frac{\sqrt{23}i}{2}}{\frac{3}{8}}$ when $±$ is plus. Add $1$ to $\frac{i\sqrt{23}}{2}$.

$$x=\frac{\frac{\sqrt{23}i}{2}+1}{\frac{3}{8}}$$

Divide $1+\frac{i\sqrt{23}}{2}$ by $\frac{3}{8}$ by multiplying $1+\frac{i\sqrt{23}}{2}$ by the reciprocal of $\frac{3}{8}$.

$$x=\frac{8+4\sqrt{23}i}{3}$$

Now solve the equation $x=\frac{1±\frac{\sqrt{23}i}{2}}{\frac{3}{8}}$ when $±$ is minus. Subtract $\frac{i\sqrt{23}}{2}$ from $1$.

$$x=\frac{-\frac{\sqrt{23}i}{2}+1}{\frac{3}{8}}$$

Divide $1-\frac{i\sqrt{23}}{2}$ by $\frac{3}{8}$ by multiplying $1-\frac{i\sqrt{23}}{2}$ by the reciprocal of $\frac{3}{8}$.

$$x=\frac{-4\sqrt{23}i+8}{3}$$

The equation is now solved.

$$x=\frac{8+4\sqrt{23}i}{3}$$ $$x=\frac{-4\sqrt{23}i+8}{3}$$

Steps for Completing the Square

Multiply $\frac{x}{2}$ times $\frac{3x}{8}$ by multiplying numerator times numerator and denominator times denominator.

$$\frac{x\times 3x}{2\times 8}+9=x$$

To add or subtract expressions, expand them to make their denominators the same. Multiply $9$ times $\frac{2\times 8}{2\times 8}$.

$$\frac{x\times 3x}{2\times 8}+\frac{9\times 2\times 8}{2\times 8}=x$$

Since $\frac{x\times 3x}{2\times 8}$ and $\frac{9\times 2\times 8}{2\times 8}$ have the same denominator, add them by adding their numerators.

$$\frac{x\times 3x+9\times 2\times 8}{2\times 8}=x$$

Do the multiplications in $x\times 3x+9\times 2\times 8$.

$$\frac{3x^{2}+144}{2\times 8}=x$$

Multiply $2$ and $8$ to get $16$.

$$\frac{3x^{2}+144}{16}=x$$

Divide each term of $3x^{2}+144$ by $16$ to get $\frac{3}{16}x^{2}+9$.

$$\frac{3}{16}x^{2}+9=x$$

Subtract $x$ from both sides.

$$\frac{3}{16}x^{2}+9-x=0$$

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

$$\frac{3}{16}x^{2}-x=-9$$

Divide both sides of the equation by $\frac{3}{16}$, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{3}{16}x^{2}-x}{\frac{3}{16}}=-\frac{9}{\frac{3}{16}}$$

Dividing by $\frac{3}{16}$ undoes the multiplication by $\frac{3}{16}$.

$$x^{2}+\left(-\frac{1}{\frac{3}{16}}\right)x=-\frac{9}{\frac{3}{16}}$$

Divide $-1$ by $\frac{3}{16}$ by multiplying $-1$ by the reciprocal of $\frac{3}{16}$.

$$x^{2}-\frac{16}{3}x=-\frac{9}{\frac{3}{16}}$$

Divide $-9$ by $\frac{3}{16}$ by multiplying $-9$ by the reciprocal of $\frac{3}{16}$.

$$x^{2}-\frac{16}{3}x=-48$$

Divide $-\frac{16}{3}$, the coefficient of the $x$ term, by $2$ to get $-\frac{8}{3}$. Then add the square of $-\frac{8}{3}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{16}{3}x+\left(-\frac{8}{3}\right)^{2}=-48+\left(-\frac{8}{3}\right)^{2}$$

Square $-\frac{8}{3}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{16}{3}x+\frac{64}{9}=-48+\frac{64}{9}$$

Add $-48$ to $\frac{64}{9}$.

$$x^{2}-\frac{16}{3}x+\frac{64}{9}=-\frac{368}{9}$$

Factor $x^{2}-\frac{16}{3}x+\frac{64}{9}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{8}{3}\right)^{2}=-\frac{368}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{8}{3}\right)^{2}}=\sqrt{-\frac{368}{9}}$$

Simplify.

$$x-\frac{8}{3}=\frac{4\sqrt{23}i}{3}$$ $$x-\frac{8}{3}=-\frac{4\sqrt{23}i}{3}$$

Add $\frac{8}{3}$ to both sides of the equation.

$$x=\frac{8+4\sqrt{23}i}{3}$$ $$x=\frac{-4\sqrt{23}i+8}{3}$$