Solve 5x^2=1+4x

$$5 x ^ { 2 } = 1 + 4 x$$

Solve for x

$$x=-\frac{1}{5}=-0.2$$

$$x=1$$

Steps Using Factoring By Grouping

Subtract 1 from both sides.

$$5x^{2}-1=4x$$

Subtract 4x from both sides.

$$5x^{2}-1-4x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$5x^{2}-4x-1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=-4 ab=5\left(-1\right)=-5$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=-5 b=1$$

Rewrite 5x^{2}-4x-1 as \left(5x^{2}-5x\right)+\left(x-1\right).

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

Factor out 5x in 5x^{2}-5x.

$$5x\left(x-1\right)+x-1$$

Factor out common term x-1 by using distributive property.

$$\left(x-1\right)\left(5x+1\right)$$

To find equation solutions, solve x-1=0 and 5x+1=0.

$$x=1 x=-\frac{1}{5}$$

Steps Using the Quadratic Formula

Subtract 1 from both sides.

$$5x^{2}-1=4x$$

Subtract 4x from both sides.

$$5x^{2}-1-4x=0$$

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.

$$5x^{2}-4x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -4 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 5\left(-1\right)}}{2\times 5}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 5\left(-1\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-4\right)±\sqrt{16-20\left(-1\right)}}{2\times 5}$$

Multiply -20 times -1.

$$x=\frac{-\left(-4\right)±\sqrt{16+20}}{2\times 5}$$

Add 16 to 20.

$$x=\frac{-\left(-4\right)±\sqrt{36}}{2\times 5}$$

Take the square root of 36.

$$x=\frac{-\left(-4\right)±6}{2\times 5}$$

The opposite of -4 is 4.

$$x=\frac{4±6}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{4±6}{10}$$

Now solve the equation x=\frac{4±6}{10} when ± is plus. Add 4 to 6.

$$x=\frac{10}{10}$$

Divide 10 by 10.

$$x=1$$

Now solve the equation x=\frac{4±6}{10} when ± is minus. Subtract 6 from 4.

$$x=-\frac{2}{10}$$

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

$$x=-\frac{1}{5}$$

The equation is now solved.

$$x=1 x=-\frac{1}{5}$$

Steps for Completing the Square

Subtract 4x from both sides.

$$5x^{2}-4x=1$$

Divide both sides by 5.

$$\frac{5x^{2}-4x}{5}=\frac{1}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}-\frac{4}{5}x=\frac{1}{5}$$

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

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

Square -\frac{2}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{4}{5}x+\frac{4}{25}=\frac{1}{5}+\frac{4}{25}$$

Add \frac{1}{5} to \frac{4}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{4}{5}x+\frac{4}{25}=\frac{9}{25}$$

Factor x^{2}-\frac{4}{5}x+\frac{4}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{2}{5}\right)^{2}=\frac{9}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{5}\right)^{2}}=\sqrt{\frac{9}{25}}$$

Simplify.

$$x-\frac{2}{5}=\frac{3}{5} x-\frac{2}{5}=-\frac{3}{5}$$

Add \frac{2}{5} to both sides of the equation.

$$x=1 x=-\frac{1}{5}$$