Solve a^2-19a^+48=0

$${ a }^{ 2 } -19 { a }^{ } +48 = 0$$

Solve for a

$$a=3$$

$$a=16$$

Steps Using Factoring

Calculate a to the power of 1 and get a.

$$a^{2}-19a+48=0$$

To solve the equation, factor a^{2}-19a+48 using formula a^{2}+\left(a+b\right)a+ab=\left(a+a\right)\left(a+b\right). To find a and b, set up a system to be solved.

$$a+b=-19 ab=48$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 48.

$$-1,-48 -2,-24 -3,-16 -4,-12 -6,-8$$

Calculate the sum for each pair.

$$-1-48=-49 -2-24=-26 -3-16=-19 -4-12=-16 -6-8=-14$$

The solution is the pair that gives sum -19.

$$a=-16 b=-3$$

Rewrite factored expression \left(a+a\right)\left(a+b\right) using the obtained values.

$$\left(a-16\right)\left(a-3\right)$$

To find equation solutions, solve a-16=0 and a-3=0.

$$a=16 a=3$$

Steps Using Factoring By Grouping

Calculate a to the power of 1 and get a.

$$a^{2}-19a+48=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as a^{2}+aa+ba+48. To find a and b, set up a system to be solved.

$$a+b=-19 ab=1\times 48=48$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 48.

$$-1,-48 -2,-24 -3,-16 -4,-12 -6,-8$$

Calculate the sum for each pair.

$$-1-48=-49 -2-24=-26 -3-16=-19 -4-12=-16 -6-8=-14$$

The solution is the pair that gives sum -19.

$$a=-16 b=-3$$

Rewrite a^{2}-19a+48 as \left(a^{2}-16a\right)+\left(-3a+48\right).

$$\left(a^{2}-16a\right)+\left(-3a+48\right)$$

Factor out a in the first and -3 in the second group.

$$a\left(a-16\right)-3\left(a-16\right)$$

Factor out common term a-16 by using distributive property.

$$\left(a-16\right)\left(a-3\right)$$

To find equation solutions, solve a-16=0 and a-3=0.

$$a=16 a=3$$

Steps Using the Quadratic Formula

Calculate a to the power of 1 and get a.

$$a^{2}-19a+48=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -19 for b, and 48 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-\left(-19\right)±\sqrt{\left(-19\right)^{2}-4\times 48}}{2}$$

Square -19.

$$a=\frac{-\left(-19\right)±\sqrt{361-4\times 48}}{2}$$

Multiply -4 times 48.

$$a=\frac{-\left(-19\right)±\sqrt{361-192}}{2}$$

Add 361 to -192.

$$a=\frac{-\left(-19\right)±\sqrt{169}}{2}$$

Take the square root of 169.

$$a=\frac{-\left(-19\right)±13}{2}$$

The opposite of -19 is 19.

$$a=\frac{19±13}{2}$$

Now solve the equation a=\frac{19±13}{2} when ± is plus. Add 19 to 13.

$$a=\frac{32}{2}$$

Divide 32 by 2.

$$a=16$$

Now solve the equation a=\frac{19±13}{2} when ± is minus. Subtract 13 from 19.

$$a=\frac{6}{2}$$

Divide 6 by 2.

$$a=3$$

The equation is now solved.

$$a=16 a=3$$

Steps for Completing the Square

Calculate a to the power of 1 and get a.

$$a^{2}-19a+48=0$$

Subtract 48 from both sides. Anything subtracted from zero gives its negation.

$$a^{2}-19a=-48$$

Divide -19, the coefficient of the x term, by 2 to get -\frac{19}{2}. Then add the square of -\frac{19}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}-19a+\left(-\frac{19}{2}\right)^{2}=-48+\left(-\frac{19}{2}\right)^{2}$$

Square -\frac{19}{2} by squaring both the numerator and the denominator of the fraction.

$$a^{2}-19a+\frac{361}{4}=-48+\frac{361}{4}$$

Add -48 to \frac{361}{4}.

$$a^{2}-19a+\frac{361}{4}=\frac{169}{4}$$

Factor a^{2}-19a+\frac{361}{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}.

$$\left(a-\frac{19}{2}\right)^{2}=\frac{169}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-\frac{19}{2}\right)^{2}}=\sqrt{\frac{169}{4}}$$

Simplify.

$$a-\frac{19}{2}=\frac{13}{2} a-\frac{19}{2}=-\frac{13}{2}$$

Add \frac{19}{2} to both sides of the equation.

$$a=16 a=3$$