Solve 5-5x=frac{5sqrt{3(x^2+1)}}{2}

$$5-5x= \frac{ 5 \sqrt{ 3( { x }^{ 2 } +1) } }{ 2 }$$

Solve for x

$$x=4-\sqrt{15}\approx 0.127016654$$

Solution Steps

Multiply both sides of the equation by 2.

$$10-10x=5\sqrt{3\left(x^{2}+1\right)}$$

Use the distributive property to multiply 3 by x^{2}+1.

$$10-10x=5\sqrt{3x^{2}+3}$$

Subtract 5\sqrt{3x^{2}+3} from both sides.

$$10-10x-5\sqrt{3x^{2}+3}=0$$

Subtract 10-10x from both sides of the equation.

$$-5\sqrt{3x^{2}+3}=-\left(10-10x\right)$$

To find the opposite of 10-10x, find the opposite of each term.

$$-5\sqrt{3x^{2}+3}=-10+10x$$

Square both sides of the equation.

$$\left(-5\sqrt{3x^{2}+3}\right)^{2}=\left(-10+10x\right)^{2}$$

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

$$\left(-5\right)^{2}\left(\sqrt{3x^{2}+3}\right)^{2}=\left(-10+10x\right)^{2}$$

Calculate -5 to the power of 2 and get 25.

$$25\left(\sqrt{3x^{2}+3}\right)^{2}=\left(-10+10x\right)^{2}$$

Calculate \sqrt{3x^{2}+3} to the power of 2 and get 3x^{2}+3.

$$25\left(3x^{2}+3\right)=\left(-10+10x\right)^{2}$$

Use the distributive property to multiply 25 by 3x^{2}+3.

$$75x^{2}+75=\left(-10+10x\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-10+10x\right)^{2}.

$$75x^{2}+75=100-200x+100x^{2}$$

Subtract 100 from both sides.

$$75x^{2}+75-100=-200x+100x^{2}$$

Subtract 100 from 75 to get -25.

$$75x^{2}-25=-200x+100x^{2}$$

Add 200x to both sides.

$$75x^{2}-25+200x=100x^{2}$$

Subtract 100x^{2} from both sides.

$$75x^{2}-25+200x-100x^{2}=0$$

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

$$-25x^{2}-25+200x=0$$

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.

$$-25x^{2}+200x-25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -25 for a, 200 for b, and -25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-200±\sqrt{200^{2}-4\left(-25\right)\left(-25\right)}}{2\left(-25\right)}$$

Square 200.

$$x=\frac{-200±\sqrt{40000-4\left(-25\right)\left(-25\right)}}{2\left(-25\right)}$$

Multiply -4 times -25.

$$x=\frac{-200±\sqrt{40000+100\left(-25\right)}}{2\left(-25\right)}$$

Multiply 100 times -25.

$$x=\frac{-200±\sqrt{40000-2500}}{2\left(-25\right)}$$

Add 40000 to -2500.

$$x=\frac{-200±\sqrt{37500}}{2\left(-25\right)}$$

Take the square root of 37500.

$$x=\frac{-200±50\sqrt{15}}{2\left(-25\right)}$$

Multiply 2 times -25.

$$x=\frac{-200±50\sqrt{15}}{-50}$$

Now solve the equation x=\frac{-200±50\sqrt{15}}{-50} when ± is plus. Add -200 to 50\sqrt{15}.

$$x=\frac{50\sqrt{15}-200}{-50}$$

Divide -200+50\sqrt{15} by -50.

$$x=4-\sqrt{15}$$

Now solve the equation x=\frac{-200±50\sqrt{15}}{-50} when ± is minus. Subtract 50\sqrt{15} from -200.

$$x=\frac{-50\sqrt{15}-200}{-50}$$

Divide -200-50\sqrt{15} by -50.

$$x=\sqrt{15}+4$$

The equation is now solved.

$$x=4-\sqrt{15} x=\sqrt{15}+4$$

Substitute 4-\sqrt{15} for x in the equation 5-5x=\frac{5\sqrt{3\left(x^{2}+1\right)}}{2}.

$$5-5\left(4-\sqrt{15}\right)=\frac{5\sqrt{3\left(\left(4-\sqrt{15}\right)^{2}+1\right)}}{2}$$

Simplify. The value x=4-\sqrt{15} satisfies the equation.

$$-15+5\times 15^{\frac{1}{2}}=5\times 15^{\frac{1}{2}}-15$$

Substitute \sqrt{15}+4 for x in the equation 5-5x=\frac{5\sqrt{3\left(x^{2}+1\right)}}{2}.

$$5-5\left(\sqrt{15}+4\right)=\frac{5\sqrt{3\left(\left(\sqrt{15}+4\right)^{2}+1\right)}}{2}$$

Simplify. The value x=\sqrt{15}+4 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-15-5\times 15^{\frac{1}{2}}=5\times 15^{\frac{1}{2}}+15$$

Equation -5\sqrt{3x^{2}+3}=10x-10 has a unique solution.

$$x=4-\sqrt{15}$$