Solve 5a^2+a=a(5a+a)

$$5 a ^ { 2 } + a = a ( 5 a + a )$$

Solve for a

$$a=1$$

$$a=0$$

Steps Using Factoring

Combine 5a and a to get 6a.

$$5a^{2}+a=a\times 6a$$

Multiply a and a to get a^{2}.

$$5a^{2}+a=a^{2}\times 6$$

Subtract a^{2}\times 6 from both sides.

$$5a^{2}+a-a^{2}\times 6=0$$

Combine 5a^{2} and -a^{2}\times 6 to get -a^{2}.

$$-a^{2}+a=0$$

Factor out a.

$$a\left(-a+1\right)=0$$

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

$$a=0 a=1$$

Steps Using the Quadratic Formula

Combine 5a and a to get 6a.

$$5a^{2}+a=a\times 6a$$

Multiply a and a to get a^{2}.

$$5a^{2}+a=a^{2}\times 6$$

Subtract a^{2}\times 6 from both sides.

$$5a^{2}+a-a^{2}\times 6=0$$

Combine 5a^{2} and -a^{2}\times 6 to get -a^{2}.

$$-a^{2}+a=0$$

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

$$a=\frac{-1±\sqrt{1^{2}}}{2\left(-1\right)}$$

Take the square root of 1^{2}.

$$a=\frac{-1±1}{2\left(-1\right)}$$

Multiply 2 times -1.

$$a=\frac{-1±1}{-2}$$

Now solve the equation a=\frac{-1±1}{-2} when ± is plus. Add -1 to 1.

$$a=\frac{0}{-2}$$

Divide 0 by -2.

$$a=0$$

Now solve the equation a=\frac{-1±1}{-2} when ± is minus. Subtract 1 from -1.

$$a=-\frac{2}{-2}$$

Divide -2 by -2.

$$a=1$$

The equation is now solved.

$$a=0 a=1$$

Steps for Completing the Square

Combine 5a and a to get 6a.

$$5a^{2}+a=a\times 6a$$

Multiply a and a to get a^{2}.

$$5a^{2}+a=a^{2}\times 6$$

Subtract a^{2}\times 6 from both sides.

$$5a^{2}+a-a^{2}\times 6=0$$

Combine 5a^{2} and -a^{2}\times 6 to get -a^{2}.

$$-a^{2}+a=0$$

Divide both sides by -1.

$$\frac{-a^{2}+a}{-1}=\frac{0}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$a^{2}+\frac{1}{-1}a=\frac{0}{-1}$$

Divide 1 by -1.

$$a^{2}-a=\frac{0}{-1}$$

Divide 0 by -1.

$$a^{2}-a=0$$

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}-a+\left(-\frac{1}{2}\right)^{2}=\left(-\frac{1}{2}\right)^{2}$$

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$a^{2}-a+\frac{1}{4}=\frac{1}{4}$$

Factor a^{2}-a+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(a-\frac{1}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-\frac{1}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$a-\frac{1}{2}=\frac{1}{2} a-\frac{1}{2}=-\frac{1}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$a=1 a=0$$