Solve {l}{(7+D=11)}{(4+D)^2=3^4}

$$\left. \begin{array} { l } { ( 7 + D = 11 ) } \\ { ( 4 + D ) ^ { 2 } = 3 ^ { 4 } } \end{array} \right.$$

Solve for D

$$D=5$$

$$D=-13$$

Steps Using Factoring

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4+D\right)^{2}.

$$16+8D+D^{2}=3^{4}$$

Calculate 3 to the power of 4 and get 81.

$$16+8D+D^{2}=81$$

Subtract 81 from both sides.

$$16+8D+D^{2}-81=0$$

Subtract 81 from 16 to get -65.

$$-65+8D+D^{2}=0$$

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

$$D^{2}+8D-65=0$$

To solve the equation, factor D^{2}+8D-65 using formula D^{2}+\left(a+b\right)D+ab=\left(D+a\right)\left(D+b\right). To find a and b, set up a system to be solved.

$$a+b=8 ab=-65$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -65.

$$-1,65 -5,13$$

Calculate the sum for each pair.

$$-1+65=64 -5+13=8$$

The solution is the pair that gives sum 8.

$$a=-5 b=13$$

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

$$\left(D-5\right)\left(D+13\right)$$

To find equation solutions, solve D-5=0 and D+13=0.

$$D=5 D=-13$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4+D\right)^{2}.

$$16+8D+D^{2}=3^{4}$$

Calculate 3 to the power of 4 and get 81.

$$16+8D+D^{2}=81$$

Subtract 81 from both sides.

$$16+8D+D^{2}-81=0$$

Subtract 81 from 16 to get -65.

$$-65+8D+D^{2}=0$$

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

$$D^{2}+8D-65=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as D^{2}+aD+bD-65. To find a and b, set up a system to be solved.

$$a+b=8 ab=1\left(-65\right)=-65$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -65.

$$-1,65 -5,13$$

Calculate the sum for each pair.

$$-1+65=64 -5+13=8$$

The solution is the pair that gives sum 8.

$$a=-5 b=13$$

Rewrite D^{2}+8D-65 as \left(D^{2}-5D\right)+\left(13D-65\right).

$$\left(D^{2}-5D\right)+\left(13D-65\right)$$

Factor out D in the first and 13 in the second group.

$$D\left(D-5\right)+13\left(D-5\right)$$

Factor out common term D-5 by using distributive property.

$$\left(D-5\right)\left(D+13\right)$$

To find equation solutions, solve D-5=0 and D+13=0.

$$D=5 D=-13$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4+D\right)^{2}.

$$16+8D+D^{2}=3^{4}$$

Calculate 3 to the power of 4 and get 81.

$$16+8D+D^{2}=81$$

Subtract 81 from both sides.

$$16+8D+D^{2}-81=0$$

Subtract 81 from 16 to get -65.

$$-65+8D+D^{2}=0$$

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.

$$D^{2}+8D-65=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 8 for b, and -65 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$D=\frac{-8±\sqrt{8^{2}-4\left(-65\right)}}{2}$$

Square 8.

$$D=\frac{-8±\sqrt{64-4\left(-65\right)}}{2}$$

Multiply -4 times -65.

$$D=\frac{-8±\sqrt{64+260}}{2}$$

Add 64 to 260.

$$D=\frac{-8±\sqrt{324}}{2}$$

Take the square root of 324.

$$D=\frac{-8±18}{2}$$

Now solve the equation D=\frac{-8±18}{2} when ± is plus. Add -8 to 18.

$$D=\frac{10}{2}$$

Divide 10 by 2.

$$D=5$$

Now solve the equation D=\frac{-8±18}{2} when ± is minus. Subtract 18 from -8.

$$D=-\frac{26}{2}$$

Divide -26 by 2.

$$D=-13$$

The equation is now solved.

$$D=5 D=-13$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4+D\right)^{2}.

$$16+8D+D^{2}=3^{4}$$

Calculate 3 to the power of 4 and get 81.

$$16+8D+D^{2}=81$$

Subtract 16 from both sides.

$$8D+D^{2}=81-16$$

Subtract 16 from 81 to get 65.

$$8D+D^{2}=65$$

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.

$$D^{2}+8D=65$$

Divide 8, the coefficient of the x term, by 2 to get 4. Then add the square of 4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$D^{2}+8D+4^{2}=65+4^{2}$$

Square 4.

$$D^{2}+8D+16=65+16$$

Add 65 to 16.

$$D^{2}+8D+16=81$$

Factor D^{2}+8D+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}.

$$\left(D+4\right)^{2}=81$$

Take the square root of both sides of the equation.

$$\sqrt{\left(D+4\right)^{2}}=\sqrt{81}$$

Simplify.

$$D+4=9 D+4=-9$$

Subtract 4 from both sides of the equation.

$$D=5 D=-13$$