Solve frac{sqrt{3}}{2}sqrt{1-x^2}+1/2x=13/10

$$\frac { \sqrt { 3 } } { 2 } \sqrt { 1 - x ^ { 2 } } + \frac { 1 } { 2 } x = \frac { 13 } { 10 }$$

Solve for x (complex solution)

$$x=\frac{-3\sqrt{23}i+13}{20}\approx 0.65-0.719374728i$$

Solution Steps

Subtract \frac{1}{2}x from both sides of the equation.

$$\frac{\sqrt{3}}{2}\sqrt{1-x^{2}}=\frac{13}{10}-\frac{1}{2}x$$

Multiply both sides of the equation by 10, the least common multiple of 2,10.

$$5\sqrt{3}\sqrt{1-x^{2}}=13-5x$$

Square both sides of the equation.

$$\left(5\sqrt{3}\sqrt{1-x^{2}}\right)^{2}=\left(13-5x\right)^{2}$$

Expand \left(5\sqrt{3}\sqrt{1-x^{2}}\right)^{2}.

$$5^{2}\left(\sqrt{3}\right)^{2}\left(\sqrt{1-x^{2}}\right)^{2}=\left(13-5x\right)^{2}$$

Calculate 5 to the power of 2 and get 25.

$$25\left(\sqrt{3}\right)^{2}\left(\sqrt{1-x^{2}}\right)^{2}=\left(13-5x\right)^{2}$$

The square of \sqrt{3} is 3.

$$25\times 3\left(\sqrt{1-x^{2}}\right)^{2}=\left(13-5x\right)^{2}$$

Multiply 25 and 3 to get 75.

$$75\left(\sqrt{1-x^{2}}\right)^{2}=\left(13-5x\right)^{2}$$

Calculate \sqrt{1-x^{2}} to the power of 2 and get 1-x^{2}.

$$75\left(1-x^{2}\right)=\left(13-5x\right)^{2}$$

Use the distributive property to multiply 75 by 1-x^{2}.

$$75-75x^{2}=\left(13-5x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(13-5x\right)^{2}.

$$75-75x^{2}=169-130x+25x^{2}$$

Subtract 169 from both sides.

$$75-75x^{2}-169=-130x+25x^{2}$$

Subtract 169 from 75 to get -94.

$$-94-75x^{2}=-130x+25x^{2}$$

Add 130x to both sides.

$$-94-75x^{2}+130x=25x^{2}$$

Subtract 25x^{2} from both sides.

$$-94-75x^{2}+130x-25x^{2}=0$$

Combine -75x^{2} and -25x^{2} to get -100x^{2}.

$$-94-100x^{2}+130x=0$$

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.

$$-100x^{2}+130x-94=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -100 for a, 130 for b, and -94 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-130±\sqrt{130^{2}-4\left(-100\right)\left(-94\right)}}{2\left(-100\right)}$$

Square 130.

$$x=\frac{-130±\sqrt{16900-4\left(-100\right)\left(-94\right)}}{2\left(-100\right)}$$

Multiply -4 times -100.

$$x=\frac{-130±\sqrt{16900+400\left(-94\right)}}{2\left(-100\right)}$$

Multiply 400 times -94.

$$x=\frac{-130±\sqrt{16900-37600}}{2\left(-100\right)}$$

Add 16900 to -37600.

$$x=\frac{-130±\sqrt{-20700}}{2\left(-100\right)}$$

Take the square root of -20700.

$$x=\frac{-130±30\sqrt{23}i}{2\left(-100\right)}$$

Multiply 2 times -100.

$$x=\frac{-130±30\sqrt{23}i}{-200}$$

Now solve the equation x=\frac{-130±30\sqrt{23}i}{-200} when ± is plus. Add -130 to 30i\sqrt{23}.

$$x=\frac{-130+30\sqrt{23}i}{-200}$$

Divide -130+30i\sqrt{23} by -200.

$$x=\frac{-3\sqrt{23}i+13}{20}$$

Now solve the equation x=\frac{-130±30\sqrt{23}i}{-200} when ± is minus. Subtract 30i\sqrt{23} from -130.

$$x=\frac{-30\sqrt{23}i-130}{-200}$$

Divide -130-30i\sqrt{23} by -200.

$$x=\frac{13+3\sqrt{23}i}{20}$$

The equation is now solved.

$$x=\frac{-3\sqrt{23}i+13}{20} x=\frac{13+3\sqrt{23}i}{20}$$

Substitute \frac{-3\sqrt{23}i+13}{20} for x in the equation \frac{\sqrt{3}}{2}\sqrt{1-x^{2}}+\frac{1}{2}x=\frac{13}{10}.

$$\frac{\sqrt{3}}{2}\sqrt{1-\left(\frac{-3\sqrt{23}i+13}{20}\right)^{2}}+\frac{1}{2}\times \frac{-3\sqrt{23}i+13}{20}=\frac{13}{10}$$

Simplify. The value x=\frac{-3\sqrt{23}i+13}{20} satisfies the equation.

$$\frac{13}{10}=\frac{13}{10}$$

Substitute \frac{13+3\sqrt{23}i}{20} for x in the equation \frac{\sqrt{3}}{2}\sqrt{1-x^{2}}+\frac{1}{2}x=\frac{13}{10}.

$$\frac{\sqrt{3}}{2}\sqrt{1-\left(\frac{13+3\sqrt{23}i}{20}\right)^{2}}+\frac{1}{2}\times \frac{13+3\sqrt{23}i}{20}=\frac{13}{10}$$

Simplify. The value x=\frac{13+3\sqrt{23}i}{20} does not satisfy the equation.

$$-\frac{13}{20}+\frac{3}{20}i\times 23^{\frac{1}{2}}=\frac{13}{10}$$

Equation 5\sqrt{3}\sqrt{1-x^{2}}=13-5x has a unique solution.

$$x=\frac{-3\sqrt{23}i+13}{20}$$