Solve (4k+2)^2-20(k^2+1)=0

$$( 4 k + 2 ) ^ { 2 } - 20 ( k ^ { 2 } + 1 ) = 0$$

Solve for k

$$k=2$$

Steps Using Factoring By Grouping

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

$$16k^{2}+16k+4-20\left(k^{2}+1\right)=0$$

Use the distributive property to multiply -20 by k^{2}+1.

$$16k^{2}+16k+4-20k^{2}-20=0$$

Combine 16k^{2} and -20k^{2} to get -4k^{2}.

$$-4k^{2}+16k+4-20=0$$

Subtract 20 from 4 to get -16.

$$-4k^{2}+16k-16=0$$

Divide both sides by 4.

$$-k^{2}+4k-4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -k^{2}+ak+bk-4. To find a and b, set up a system to be solved.

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

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 4.

$$1,4 2,2$$

Calculate the sum for each pair.

$$1+4=5 2+2=4$$

The solution is the pair that gives sum 4.

$$a=2 b=2$$

Rewrite -k^{2}+4k-4 as \left(-k^{2}+2k\right)+\left(2k-4\right).

$$\left(-k^{2}+2k\right)+\left(2k-4\right)$$

Factor out -k in the first and 2 in the second group.

$$-k\left(k-2\right)+2\left(k-2\right)$$

Factor out common term k-2 by using distributive property.

$$\left(k-2\right)\left(-k+2\right)$$

To find equation solutions, solve k-2=0 and -k+2=0.

$$k=2 k=2$$

Steps Using the Quadratic Formula

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

$$16k^{2}+16k+4-20\left(k^{2}+1\right)=0$$

Use the distributive property to multiply -20 by k^{2}+1.

$$16k^{2}+16k+4-20k^{2}-20=0$$

Combine 16k^{2} and -20k^{2} to get -4k^{2}.

$$-4k^{2}+16k+4-20=0$$

Subtract 20 from 4 to get -16.

$$-4k^{2}+16k-16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 16 for b, and -16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-16±\sqrt{16^{2}-4\left(-4\right)\left(-16\right)}}{2\left(-4\right)}$$

Square 16.

$$k=\frac{-16±\sqrt{256-4\left(-4\right)\left(-16\right)}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$k=\frac{-16±\sqrt{256+16\left(-16\right)}}{2\left(-4\right)}$$

Multiply 16 times -16.

$$k=\frac{-16±\sqrt{256-256}}{2\left(-4\right)}$$

Add 256 to -256.

$$k=\frac{-16±\sqrt{0}}{2\left(-4\right)}$$

Take the square root of 0.

$$k=-\frac{16}{2\left(-4\right)}$$

Multiply 2 times -4.

$$k=-\frac{16}{-8}$$

Divide -16 by -8.

$$k=2$$

Steps for Completing the Square

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

$$16k^{2}+16k+4-20\left(k^{2}+1\right)=0$$

Use the distributive property to multiply -20 by k^{2}+1.

$$16k^{2}+16k+4-20k^{2}-20=0$$

Combine 16k^{2} and -20k^{2} to get -4k^{2}.

$$-4k^{2}+16k+4-20=0$$

Subtract 20 from 4 to get -16.

$$-4k^{2}+16k-16=0$$

Add 16 to both sides. Anything plus zero gives itself.

$$-4k^{2}+16k=16$$

Divide both sides by -4.

$$\frac{-4k^{2}+16k}{-4}=\frac{16}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$k^{2}+\frac{16}{-4}k=\frac{16}{-4}$$

Divide 16 by -4.

$$k^{2}-4k=\frac{16}{-4}$$

Divide 16 by -4.

$$k^{2}-4k=-4$$

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

$$k^{2}-4k+\left(-2\right)^{2}=-4+\left(-2\right)^{2}$$

Square -2.

$$k^{2}-4k+4=-4+4$$

Add -4 to 4.

$$k^{2}-4k+4=0$$

Factor k^{2}-4k+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}.

$$\left(k-2\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k-2\right)^{2}}=\sqrt{0}$$

Simplify.

$$k-2=0 k-2=0$$

Add 2 to both sides of the equation.

$$k=2 k=2$$

The equation is now solved. Solutions are the same.

$$k=2$$