Solve z^2+36=13z

$$z ^ { 2 } + 36 = 13 z$$

Solve for z

$$z=4$$

$$z=9$$

Steps Using Factoring

Subtract 13z from both sides.

$$z^{2}+36-13z=0$$

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

$$z^{2}-13z+36=0$$

To solve the equation, factor z^{2}-13z+36 using formula z^{2}+\left(a+b\right)z+ab=\left(z+a\right)\left(z+b\right). To find a and b, set up a system to be solved.

$$a+b=-13 ab=36$$

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 36.

$$-1,-36 -2,-18 -3,-12 -4,-9 -6,-6$$

Calculate the sum for each pair.

$$-1-36=-37 -2-18=-20 -3-12=-15 -4-9=-13 -6-6=-12$$

The solution is the pair that gives sum -13.

$$a=-9 b=-4$$

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

$$\left(z-9\right)\left(z-4\right)$$

To find equation solutions, solve z-9=0 and z-4=0.

$$z=9 z=4$$

Steps Using Factoring By Grouping

Subtract 13z from both sides.

$$z^{2}+36-13z=0$$

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

$$z^{2}-13z+36=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as z^{2}+az+bz+36. To find a and b, set up a system to be solved.

$$a+b=-13 ab=1\times 36=36$$

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 36.

$$-1,-36 -2,-18 -3,-12 -4,-9 -6,-6$$

Calculate the sum for each pair.

$$-1-36=-37 -2-18=-20 -3-12=-15 -4-9=-13 -6-6=-12$$

The solution is the pair that gives sum -13.

$$a=-9 b=-4$$

Rewrite z^{2}-13z+36 as \left(z^{2}-9z\right)+\left(-4z+36\right).

$$\left(z^{2}-9z\right)+\left(-4z+36\right)$$

Factor out z in the first and -4 in the second group.

$$z\left(z-9\right)-4\left(z-9\right)$$

Factor out common term z-9 by using distributive property.

$$\left(z-9\right)\left(z-4\right)$$

To find equation solutions, solve z-9=0 and z-4=0.

$$z=9 z=4$$

Steps Using the Quadratic Formula

Subtract 13z from both sides.

$$z^{2}+36-13z=0$$

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.

$$z^{2}-13z+36=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -13 for b, and 36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$z=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 36}}{2}$$

Square -13.

$$z=\frac{-\left(-13\right)±\sqrt{169-4\times 36}}{2}$$

Multiply -4 times 36.

$$z=\frac{-\left(-13\right)±\sqrt{169-144}}{2}$$

Add 169 to -144.

$$z=\frac{-\left(-13\right)±\sqrt{25}}{2}$$

Take the square root of 25.

$$z=\frac{-\left(-13\right)±5}{2}$$

The opposite of -13 is 13.

$$z=\frac{13±5}{2}$$

Now solve the equation z=\frac{13±5}{2} when ± is plus. Add 13 to 5.

$$z=\frac{18}{2}$$

Divide 18 by 2.

$$z=9$$

Now solve the equation z=\frac{13±5}{2} when ± is minus. Subtract 5 from 13.

$$z=\frac{8}{2}$$

Divide 8 by 2.

$$z=4$$

The equation is now solved.

$$z=9 z=4$$

Steps for Completing the Square

Subtract 13z from both sides.

$$z^{2}+36-13z=0$$

Subtract 36 from both sides. Anything subtracted from zero gives its negation.

$$z^{2}-13z=-36$$

Divide -13, the coefficient of the x term, by 2 to get -\frac{13}{2}. Then add the square of -\frac{13}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$z^{2}-13z+\left(-\frac{13}{2}\right)^{2}=-36+\left(-\frac{13}{2}\right)^{2}$$

Square -\frac{13}{2} by squaring both the numerator and the denominator of the fraction.

$$z^{2}-13z+\frac{169}{4}=-36+\frac{169}{4}$$

Add -36 to \frac{169}{4}.

$$z^{2}-13z+\frac{169}{4}=\frac{25}{4}$$

Factor z^{2}-13z+\frac{169}{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}.

$$\left(z-\frac{13}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(z-\frac{13}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$z-\frac{13}{2}=\frac{5}{2} z-\frac{13}{2}=-\frac{5}{2}$$

Add \frac{13}{2} to both sides of the equation.

$$z=9 z=4$$