Solve {left(0.5sqrt{3}xright)}^2+left(5-1.5xright)^2=left(1+xright)^2

$${ \left(0.5 \sqrt{ 3 } x \right) }^{ 2 } + { \left(5-1.5x \right) }^{ 2 } = { \left(1+x \right) }^{ 2 }$$

Solve for x

$$x = \frac{\sqrt{97} + 17}{4} \approx 6.71221445$$

$$x = \frac{17 - \sqrt{97}}{4} \approx 1.78778555$$

Steps Using the Quadratic Formula

Expand \left(0.5\sqrt{3}x\right)^{2}.

$$0.5^{2}\left(\sqrt{3}\right)^{2}x^{2}+\left(5-1.5x\right)^{2}=\left(1+x\right)^{2}$$

Calculate 0.5 to the power of 2 and get 0.25.

$$0.25\left(\sqrt{3}\right)^{2}x^{2}+\left(5-1.5x\right)^{2}=\left(1+x\right)^{2}$$

The square of \sqrt{3} is 3.

$$0.25\times 3x^{2}+\left(5-1.5x\right)^{2}=\left(1+x\right)^{2}$$

Multiply 0.25 and 3 to get 0.75.

$$0.75x^{2}+\left(5-1.5x\right)^{2}=\left(1+x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-1.5x\right)^{2}.

$$0.75x^{2}+25-15x+2.25x^{2}=\left(1+x\right)^{2}$$

Combine 0.75x^{2} and 2.25x^{2} to get 3x^{2}.

$$3x^{2}+25-15x=\left(1+x\right)^{2}$$

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

$$3x^{2}+25-15x=1+2x+x^{2}$$

Subtract 1 from both sides.

$$3x^{2}+25-15x-1=2x+x^{2}$$

Subtract 1 from 25 to get 24.

$$3x^{2}+24-15x=2x+x^{2}$$

Subtract 2x from both sides.

$$3x^{2}+24-15x-2x=x^{2}$$

Combine -15x and -2x to get -17x.

$$3x^{2}+24-17x=x^{2}$$

Subtract x^{2} from both sides.

$$3x^{2}+24-17x-x^{2}=0$$

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

$$2x^{2}+24-17x=0$$

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.

$$2x^{2}-17x+24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -17 for b, and 24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 2\times 24}}{2\times 2}$$

Square -17.

$$x=\frac{-\left(-17\right)±\sqrt{289-4\times 2\times 24}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-17\right)±\sqrt{289-8\times 24}}{2\times 2}$$

Multiply -8 times 24.

$$x=\frac{-\left(-17\right)±\sqrt{289-192}}{2\times 2}$$

Add 289 to -192.

$$x=\frac{-\left(-17\right)±\sqrt{97}}{2\times 2}$$

The opposite of -17 is 17.

$$x=\frac{17±\sqrt{97}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{17±\sqrt{97}}{4}$$

Now solve the equation x=\frac{17±\sqrt{97}}{4} when ± is plus. Add 17 to \sqrt{97}.

$$x=\frac{\sqrt{97}+17}{4}$$

Now solve the equation x=\frac{17±\sqrt{97}}{4} when ± is minus. Subtract \sqrt{97} from 17.

$$x=\frac{17-\sqrt{97}}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{97}+17}{4} x=\frac{17-\sqrt{97}}{4}$$

Steps for Completing the Square

Expand \left(0.5\sqrt{3}x\right)^{2}.

$$0.5^{2}\left(\sqrt{3}\right)^{2}x^{2}+\left(5-1.5x\right)^{2}=\left(1+x\right)^{2}$$

Calculate 0.5 to the power of 2 and get 0.25.

$$0.25\left(\sqrt{3}\right)^{2}x^{2}+\left(5-1.5x\right)^{2}=\left(1+x\right)^{2}$$

The square of \sqrt{3} is 3.

$$0.25\times 3x^{2}+\left(5-1.5x\right)^{2}=\left(1+x\right)^{2}$$

Multiply 0.25 and 3 to get 0.75.

$$0.75x^{2}+\left(5-1.5x\right)^{2}=\left(1+x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-1.5x\right)^{2}.

$$0.75x^{2}+25-15x+2.25x^{2}=\left(1+x\right)^{2}$$

Combine 0.75x^{2} and 2.25x^{2} to get 3x^{2}.

$$3x^{2}+25-15x=\left(1+x\right)^{2}$$

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

$$3x^{2}+25-15x=1+2x+x^{2}$$

Subtract 2x from both sides.

$$3x^{2}+25-15x-2x=1+x^{2}$$

Combine -15x and -2x to get -17x.

$$3x^{2}+25-17x=1+x^{2}$$

Subtract x^{2} from both sides.

$$3x^{2}+25-17x-x^{2}=1$$

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

$$2x^{2}+25-17x=1$$

Subtract 25 from both sides.

$$2x^{2}-17x=1-25$$

Subtract 25 from 1 to get -24.

$$2x^{2}-17x=-24$$

Divide both sides by 2.

$$\frac{2x^{2}-17x}{2}=-\frac{24}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{17}{2}x=-\frac{24}{2}$$

Divide -24 by 2.

$$x^{2}-\frac{17}{2}x=-12$$

Divide -\frac{17}{2}, the coefficient of the x term, by 2 to get -\frac{17}{4}. Then add the square of -\frac{17}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{17}{2}x+\left(-\frac{17}{4}\right)^{2}=-12+\left(-\frac{17}{4}\right)^{2}$$

Square -\frac{17}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{17}{2}x+\frac{289}{16}=-12+\frac{289}{16}$$

Add -12 to \frac{289}{16}.

$$x^{2}-\frac{17}{2}x+\frac{289}{16}=\frac{97}{16}$$

Factor x^{2}-\frac{17}{2}x+\frac{289}{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}.

$$\left(x-\frac{17}{4}\right)^{2}=\frac{97}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{17}{4}\right)^{2}}=\sqrt{\frac{97}{16}}$$

Simplify.

$$x-\frac{17}{4}=\frac{\sqrt{97}}{4} x-\frac{17}{4}=-\frac{\sqrt{97}}{4}$$

Add \frac{17}{4} to both sides of the equation.

$$x=\frac{\sqrt{97}+17}{4} x=\frac{17-\sqrt{97}}{4}$$