Solve 11k^2+4k-52=10k^2-7

$$11 k ^ { 2 } + 4 k - 52 = 10 k ^ { 2 } - 7$$

Solve for k

$$k=-9$$

$$k=5$$

Steps Using Factoring

Subtract 10k^{2} from both sides.

$$11k^{2}+4k-52-10k^{2}=-7$$

Combine 11k^{2} and -10k^{2} to get k^{2}.

$$k^{2}+4k-52=-7$$

Add 7 to both sides.

$$k^{2}+4k-52+7=0$$

Add -52 and 7 to get -45.

$$k^{2}+4k-45=0$$

To solve the equation, factor k^{2}+4k-45 using formula k^{2}+\left(a+b\right)k+ab=\left(k+a\right)\left(k+b\right). To find a and b, set up a system to be solved.

$$a+b=4 ab=-45$$

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 -45.

$$-1,45 -3,15 -5,9$$

Calculate the sum for each pair.

$$-1+45=44 -3+15=12 -5+9=4$$

The solution is the pair that gives sum 4.

$$a=-5 b=9$$

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

$$\left(k-5\right)\left(k+9\right)$$

To find equation solutions, solve k-5=0 and k+9=0.

$$k=5 k=-9$$

Steps Using Factoring By Grouping

Subtract 10k^{2} from both sides.

$$11k^{2}+4k-52-10k^{2}=-7$$

Combine 11k^{2} and -10k^{2} to get k^{2}.

$$k^{2}+4k-52=-7$$

Add 7 to both sides.

$$k^{2}+4k-52+7=0$$

Add -52 and 7 to get -45.

$$k^{2}+4k-45=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as k^{2}+ak+bk-45. To find a and b, set up a system to be solved.

$$a+b=4 ab=1\left(-45\right)=-45$$

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 -45.

$$-1,45 -3,15 -5,9$$

Calculate the sum for each pair.

$$-1+45=44 -3+15=12 -5+9=4$$

The solution is the pair that gives sum 4.

$$a=-5 b=9$$

Rewrite k^{2}+4k-45 as \left(k^{2}-5k\right)+\left(9k-45\right).

$$\left(k^{2}-5k\right)+\left(9k-45\right)$$

Factor out k in the first and 9 in the second group.

$$k\left(k-5\right)+9\left(k-5\right)$$

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

$$\left(k-5\right)\left(k+9\right)$$

To find equation solutions, solve k-5=0 and k+9=0.

$$k=5 k=-9$$

Steps Using the Quadratic Formula

Subtract 10k^{2} from both sides.

$$11k^{2}+4k-52-10k^{2}=-7$$

Combine 11k^{2} and -10k^{2} to get k^{2}.

$$k^{2}+4k-52=-7$$

Add 7 to both sides.

$$k^{2}+4k-52+7=0$$

Add -52 and 7 to get -45.

$$k^{2}+4k-45=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 4 for b, and -45 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-4±\sqrt{4^{2}-4\left(-45\right)}}{2}$$

Square 4.

$$k=\frac{-4±\sqrt{16-4\left(-45\right)}}{2}$$

Multiply -4 times -45.

$$k=\frac{-4±\sqrt{16+180}}{2}$$

Add 16 to 180.

$$k=\frac{-4±\sqrt{196}}{2}$$

Take the square root of 196.

$$k=\frac{-4±14}{2}$$

Now solve the equation k=\frac{-4±14}{2} when ± is plus. Add -4 to 14.

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

Divide 10 by 2.

$$k=5$$

Now solve the equation k=\frac{-4±14}{2} when ± is minus. Subtract 14 from -4.

$$k=-\frac{18}{2}$$

Divide -18 by 2.

$$k=-9$$

The equation is now solved.

$$k=5 k=-9$$

Steps for Completing the Square

Subtract 10k^{2} from both sides.

$$11k^{2}+4k-52-10k^{2}=-7$$

Combine 11k^{2} and -10k^{2} to get k^{2}.

$$k^{2}+4k-52=-7$$

Add 52 to both sides.

$$k^{2}+4k=-7+52$$

Add -7 and 52 to get 45.

$$k^{2}+4k=45$$

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.

$$k^{2}+4k+2^{2}=45+2^{2}$$

Square 2.

$$k^{2}+4k+4=45+4$$

Add 45 to 4.

$$k^{2}+4k+4=49$$

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}.

$$\left(k+2\right)^{2}=49$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k+2\right)^{2}}=\sqrt{49}$$

Simplify.

$$k+2=7 k+2=-7$$

Subtract 2 from both sides of the equation.

$$k=5 k=-9$$