Solve {left(05sqrt{3}xright)}^2+left(5-15xright)^2=left(1+xright)^2

$${ \left(05 \sqrt{ 3 } x \right) }^{ 2 } + { \left(5-15x \right) }^{ 2 } = { \left(1+x \right) }^{ 2 }$$

Solve for x

$$x=\frac{1}{4}=0.25$$

$$x=\frac{3}{7}\approx 0.428571429$$

Steps Using the Quadratic Formula

Multiply 0 and 5 to get 0.

$$\left(0\sqrt{3}x\right)^{2}+\left(5-15x\right)^{2}=\left(1+x\right)^{2}$$

Anything times zero gives zero.

$$0^{2}+\left(5-15x\right)^{2}=\left(1+x\right)^{2}$$

Calculate 0 to the power of 2 and get 0.

$$0+\left(5-15x\right)^{2}=\left(1+x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-15x\right)^{2}.

$$0+25-150x+225x^{2}=\left(1+x\right)^{2}$$

Add 0 and 25 to get 25.

$$25-150x+225x^{2}=\left(1+x\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+x\right)^{2}.

$$25-150x+225x^{2}=1+2x+x^{2}$$

Subtract 1 from both sides.

$$25-150x+225x^{2}-1=2x+x^{2}$$

Subtract 1 from 25 to get 24.

$$24-150x+225x^{2}=2x+x^{2}$$

Subtract 2x from both sides.

$$24-150x+225x^{2}-2x=x^{2}$$

Combine -150x and -2x to get -152x.

$$24-152x+225x^{2}=x^{2}$$

Subtract x^{2} from both sides.

$$24-152x+225x^{2}-x^{2}=0$$

Combine 225x^{2} and -x^{2} to get 224x^{2}.

$$24-152x+224x^{2}=0$$

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.

$$224x^{2}-152x+24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 224 for a, -152 for b, and 24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-152\right)±\sqrt{\left(-152\right)^{2}-4\times 224\times 24}}{2\times 224}$$

Square -152.

$$x=\frac{-\left(-152\right)±\sqrt{23104-4\times 224\times 24}}{2\times 224}$$

Multiply -4 times 224.

$$x=\frac{-\left(-152\right)±\sqrt{23104-896\times 24}}{2\times 224}$$

Multiply -896 times 24.

$$x=\frac{-\left(-152\right)±\sqrt{23104-21504}}{2\times 224}$$

Add 23104 to -21504.

$$x=\frac{-\left(-152\right)±\sqrt{1600}}{2\times 224}$$

Take the square root of 1600.

$$x=\frac{-\left(-152\right)±40}{2\times 224}$$

The opposite of -152 is 152.

$$x=\frac{152±40}{2\times 224}$$

Multiply 2 times 224.

$$x=\frac{152±40}{448}$$

Now solve the equation x=\frac{152±40}{448} when ± is plus. Add 152 to 40.

$$x=\frac{192}{448}$$

Reduce the fraction \frac{192}{448} to lowest terms by extracting and canceling out 64.

$$x=\frac{3}{7}$$

Now solve the equation x=\frac{152±40}{448} when ± is minus. Subtract 40 from 152.

$$x=\frac{112}{448}$$

Reduce the fraction \frac{112}{448} to lowest terms by extracting and canceling out 112.

$$x=\frac{1}{4}$$

The equation is now solved.

$$x=\frac{3}{7} x=\frac{1}{4}$$

Steps for Completing the Square

Multiply 0 and 5 to get 0.

$$\left(0\sqrt{3}x\right)^{2}+\left(5-15x\right)^{2}=\left(1+x\right)^{2}$$

Anything times zero gives zero.

$$0^{2}+\left(5-15x\right)^{2}=\left(1+x\right)^{2}$$

Calculate 0 to the power of 2 and get 0.

$$0+\left(5-15x\right)^{2}=\left(1+x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-15x\right)^{2}.

$$0+25-150x+225x^{2}=\left(1+x\right)^{2}$$

Add 0 and 25 to get 25.

$$25-150x+225x^{2}=\left(1+x\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+x\right)^{2}.

$$25-150x+225x^{2}=1+2x+x^{2}$$

Subtract 2x from both sides.

$$25-150x+225x^{2}-2x=1+x^{2}$$

Combine -150x and -2x to get -152x.

$$25-152x+225x^{2}=1+x^{2}$$

Subtract x^{2} from both sides.

$$25-152x+225x^{2}-x^{2}=1$$

Combine 225x^{2} and -x^{2} to get 224x^{2}.

$$25-152x+224x^{2}=1$$

Subtract 25 from both sides.

$$-152x+224x^{2}=1-25$$

Subtract 25 from 1 to get -24.

$$-152x+224x^{2}=-24$$

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.

$$224x^{2}-152x=-24$$

Divide both sides by 224.

$$\frac{224x^{2}-152x}{224}=-\frac{24}{224}$$

Dividing by 224 undoes the multiplication by 224.

$$x^{2}+\left(-\frac{152}{224}\right)x=-\frac{24}{224}$$

Reduce the fraction \frac{-152}{224} to lowest terms by extracting and canceling out 8.

$$x^{2}-\frac{19}{28}x=-\frac{24}{224}$$

Reduce the fraction \frac{-24}{224} to lowest terms by extracting and canceling out 8.

$$x^{2}-\frac{19}{28}x=-\frac{3}{28}$$

Divide -\frac{19}{28}, the coefficient of the x term, by 2 to get -\frac{19}{56}. Then add the square of -\frac{19}{56} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{19}{28}x+\left(-\frac{19}{56}\right)^{2}=-\frac{3}{28}+\left(-\frac{19}{56}\right)^{2}$$

Square -\frac{19}{56} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{19}{28}x+\frac{361}{3136}=-\frac{3}{28}+\frac{361}{3136}$$

Add -\frac{3}{28} to \frac{361}{3136} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{19}{28}x+\frac{361}{3136}=\frac{25}{3136}$$

Factor x^{2}-\frac{19}{28}x+\frac{361}{3136}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{19}{56}\right)^{2}=\frac{25}{3136}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{19}{56}\right)^{2}}=\sqrt{\frac{25}{3136}}$$

Simplify.

$$x-\frac{19}{56}=\frac{5}{56} x-\frac{19}{56}=-\frac{5}{56}$$

Add \frac{19}{56} to both sides of the equation.

$$x=\frac{3}{7} x=\frac{1}{4}$$