Solve (5-x)^2+3x^2=25

$$( 5 - x ) ^ { 2 } + 3 x ^ { 2 } = 25$$

Solve for x

$$x = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

$$x=0$$

Steps Using Factoring

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

$$25-10x+x^{2}+3x^{2}=25$$

Combine x^{2} and 3x^{2} to get 4x^{2}.

$$25-10x+4x^{2}=25$$

Subtract 25 from both sides.

$$25-10x+4x^{2}-25=0$$

Subtract 25 from 25 to get 0.

$$-10x+4x^{2}=0$$

Factor out x.

$$x\left(-10+4x\right)=0$$

To find equation solutions, solve x=0 and -10+4x=0.

$$x=0 x=\frac{5}{2}$$

Steps Using the Quadratic Formula

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

$$25-10x+x^{2}+3x^{2}=25$$

Combine x^{2} and 3x^{2} to get 4x^{2}.

$$25-10x+4x^{2}=25$$

Subtract 25 from both sides.

$$25-10x+4x^{2}-25=0$$

Subtract 25 from 25 to get 0.

$$-10x+4x^{2}=0$$

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.

$$4x^{2}-10x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -10 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}}}{2\times 4}$$

Take the square root of \left(-10\right)^{2}.

$$x=\frac{-\left(-10\right)±10}{2\times 4}$$

The opposite of -10 is 10.

$$x=\frac{10±10}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{10±10}{8}$$

Now solve the equation x=\frac{10±10}{8} when ± is plus. Add 10 to 10.

$$x=\frac{20}{8}$$

Reduce the fraction \frac{20}{8} to lowest terms by extracting and canceling out 4.

$$x=\frac{5}{2}$$

Now solve the equation x=\frac{10±10}{8} when ± is minus. Subtract 10 from 10.

$$x=\frac{0}{8}$$

Divide 0 by 8.

$$x=0$$

The equation is now solved.

$$x=\frac{5}{2} x=0$$

Steps for Completing the Square

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

$$25-10x+x^{2}+3x^{2}=25$$

Combine x^{2} and 3x^{2} to get 4x^{2}.

$$25-10x+4x^{2}=25$$

Subtract 25 from both sides.

$$-10x+4x^{2}=25-25$$

Subtract 25 from 25 to get 0.

$$-10x+4x^{2}=0$$

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.

$$4x^{2}-10x=0$$

Divide both sides by 4.

$$\frac{4x^{2}-10x}{4}=\frac{0}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{10}{4}\right)x=\frac{0}{4}$$

Reduce the fraction \frac{-10}{4} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{5}{2}x=\frac{0}{4}$$

Divide 0 by 4.

$$x^{2}-\frac{5}{2}x=0$$

Divide -\frac{5}{2}, the coefficient of the x term, by 2 to get -\frac{5}{4}. Then add the square of -\frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{5}{2}x+\left(-\frac{5}{4}\right)^{2}=\left(-\frac{5}{4}\right)^{2}$$

Square -\frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{25}{16}$$

Factor x^{2}-\frac{5}{2}x+\frac{25}{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}.

$$\left(x-\frac{5}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

$$x-\frac{5}{4}=\frac{5}{4} x-\frac{5}{4}=-\frac{5}{4}$$

Add \frac{5}{4} to both sides of the equation.

$$x=\frac{5}{2} x=0$$