Solve (c)3/4x^2-5/6x^2=1/18x^2

$$( c ) \frac { 3 } { 4 } x ^ { 2 } - \frac { 5 } { 6 } x ^ { 2 } = \frac { 1 } { 18 } x ^ { 2 }$$

Solve for c (complex solution)

$$\left\{\begin{matrix}\\c=\frac{32}{27}\approx 1.185185185\text{, }&\text{unconditionally}\\c\in \mathrm{C}\text{, }&x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Add \frac{5}{6}x^{2} to both sides.

$$c\times \frac{3}{4}x^{2}=\frac{1}{18}x^{2}+\frac{5}{6}x^{2}$$

Combine \frac{1}{18}x^{2} and \frac{5}{6}x^{2} to get \frac{8}{9}x^{2}.

$$c\times \frac{3}{4}x^{2}=\frac{8}{9}x^{2}$$

The equation is in standard form.

$$\frac{3x^{2}}{4}c=\frac{8x^{2}}{9}$$

Divide both sides by \frac{3}{4}x^{2}.

$$\frac{4\times \frac{3x^{2}}{4}c}{3x^{2}}=\frac{8x^{2}}{9\times \frac{3x^{2}}{4}}$$

Dividing by \frac{3}{4}x^{2} undoes the multiplication by \frac{3}{4}x^{2}.

$$c=\frac{8x^{2}}{9\times \frac{3x^{2}}{4}}$$

Divide \frac{8x^{2}}{9} by \frac{3}{4}x^{2}.

$$c=\frac{32}{27}$$

Solve for x (complex solution)

$$\left\{\begin{matrix}\\x=0\text{, }&\text{unconditionally}\\x\in \mathrm{C}\text{, }&c=\frac{32}{27}\end{matrix}\right.$$

Steps by Finding Square Root

Subtract \frac{1}{18}x^{2} from both sides.

$$c\times \frac{3}{4}x^{2}-\frac{5}{6}x^{2}-\frac{1}{18}x^{2}=0$$

Combine -\frac{5}{6}x^{2} and -\frac{1}{18}x^{2} to get -\frac{8}{9}x^{2}.

$$c\times \frac{3}{4}x^{2}-\frac{8}{9}x^{2}=0$$

Combine all terms containing x.

$$\left(c\times \frac{3}{4}-\frac{8}{9}\right)x^{2}=0$$

Dividing by \frac{3}{4}c-\frac{8}{9} undoes the multiplication by \frac{3}{4}c-\frac{8}{9}.

$$x^{2}=\frac{0}{\frac{3c}{4}-\frac{8}{9}}$$

Divide 0 by \frac{3}{4}c-\frac{8}{9}.

$$x^{2}=0$$

Take the square root of both sides of the equation.

$$x=0 x=0$$

The equation is now solved. Solutions are the same.

$$x=0$$

Steps Using the Quadratic Formula

Subtract \frac{1}{18}x^{2} from both sides.

$$c\times \frac{3}{4}x^{2}-\frac{5}{6}x^{2}-\frac{1}{18}x^{2}=0$$

Combine -\frac{5}{6}x^{2} and -\frac{1}{18}x^{2} to get -\frac{8}{9}x^{2}.

$$c\times \frac{3}{4}x^{2}-\frac{8}{9}x^{2}=0$$

Combine all terms containing x.

$$\left(c\times \frac{3}{4}-\frac{8}{9}\right)x^{2}=0$$

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

$$\left(\frac{3c}{4}-\frac{8}{9}\right)x^{2}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{3}{4}c-\frac{8}{9} for a, 0 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}}}{2\left(\frac{3c}{4}-\frac{8}{9}\right)}$$

Take the square root of 0^{2}.

$$x=\frac{0±0}{2\left(\frac{3c}{4}-\frac{8}{9}\right)}$$

Multiply 2 times \frac{3}{4}c-\frac{8}{9}.

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

Divide 0 by \frac{3c}{2}-\frac{16}{9}.

$$x=0$$

Solve for c

$$\left\{\begin{matrix}\\c=\frac{32}{27}\approx 1.185185185\text{, }&\text{unconditionally}\\c\in \mathrm{R}\text{, }&x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Add \frac{5}{6}x^{2} to both sides.

$$c\times \frac{3}{4}x^{2}=\frac{1}{18}x^{2}+\frac{5}{6}x^{2}$$

Combine \frac{1}{18}x^{2} and \frac{5}{6}x^{2} to get \frac{8}{9}x^{2}.

$$c\times \frac{3}{4}x^{2}=\frac{8}{9}x^{2}$$

The equation is in standard form.

$$\frac{3x^{2}}{4}c=\frac{8x^{2}}{9}$$

Divide both sides by \frac{3}{4}x^{2}.

$$\frac{4\times \frac{3x^{2}}{4}c}{3x^{2}}=\frac{8x^{2}}{9\times \frac{3x^{2}}{4}}$$

Dividing by \frac{3}{4}x^{2} undoes the multiplication by \frac{3}{4}x^{2}.

$$c=\frac{8x^{2}}{9\times \frac{3x^{2}}{4}}$$

Divide \frac{8x^{2}}{9} by \frac{3}{4}x^{2}.

$$c=\frac{32}{27}$$

Solve for x

$$\left\{\begin{matrix}\\x=0\text{, }&\text{unconditionally}\\x\in \mathrm{R}\text{, }&c=\frac{32}{27}\end{matrix}\right.$$