Solve 9/16=9-x^2/x^2+6x+9

$$\frac{ 9 }{ 16 } = \frac{ 9- { x }^{ 2 } }{ { x }^{ 2 } +6x+9 }$$

Solve for x

$$x=\frac{21}{25}=0.84$$

Steps Using Factoring By Grouping

Variable x cannot be equal to -3 since division by zero is not defined. Multiply both sides of the equation by 16\left(x+3\right)^{2}, the least common multiple of 16,x^{2}+6x+9.

$$9\left(x+3\right)^{2}=16\left(9-x^{2}\right)$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$9\left(x^{2}+6x+9\right)=16\left(9-x^{2}\right)$$

Use the distributive property to multiply 9 by x^{2}+6x+9.

$$9x^{2}+54x+81=16\left(9-x^{2}\right)$$

Use the distributive property to multiply 16 by 9-x^{2}.

$$9x^{2}+54x+81=144-16x^{2}$$

Subtract 144 from both sides.

$$9x^{2}+54x+81-144=-16x^{2}$$

Subtract 144 from 81 to get -63.

$$9x^{2}+54x-63=-16x^{2}$$

Add 16x^{2} to both sides.

$$9x^{2}+54x-63+16x^{2}=0$$

Combine 9x^{2} and 16x^{2} to get 25x^{2}.

$$25x^{2}+54x-63=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 25x^{2}+ax+bx-63. To find a and b, set up a system to be solved.

$$a+b=54 ab=25\left(-63\right)=-1575$$

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 -1575.

$$-1,1575 -3,525 -5,315 -7,225 -9,175 -15,105 -21,75 -25,63 -35,45$$

Calculate the sum for each pair.

$$-1+1575=1574 -3+525=522 -5+315=310 -7+225=218 -9+175=166 -15+105=90 -21+75=54 -25+63=38 -35+45=10$$

The solution is the pair that gives sum 54.

$$a=-21 b=75$$

Rewrite 25x^{2}+54x-63 as \left(25x^{2}-21x\right)+\left(75x-63\right).

$$\left(25x^{2}-21x\right)+\left(75x-63\right)$$

Factor out x in the first and 3 in the second group.

$$x\left(25x-21\right)+3\left(25x-21\right)$$

Factor out common term 25x-21 by using distributive property.

$$\left(25x-21\right)\left(x+3\right)$$

To find equation solutions, solve 25x-21=0 and x+3=0.

$$x=\frac{21}{25} x=-3$$

Variable x cannot be equal to -3.

$$x=\frac{21}{25}$$

Steps Using the Quadratic Formula

Variable x cannot be equal to -3 since division by zero is not defined. Multiply both sides of the equation by 16\left(x+3\right)^{2}, the least common multiple of 16,x^{2}+6x+9.

$$9\left(x+3\right)^{2}=16\left(9-x^{2}\right)$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$9\left(x^{2}+6x+9\right)=16\left(9-x^{2}\right)$$

Use the distributive property to multiply 9 by x^{2}+6x+9.

$$9x^{2}+54x+81=16\left(9-x^{2}\right)$$

Use the distributive property to multiply 16 by 9-x^{2}.

$$9x^{2}+54x+81=144-16x^{2}$$

Subtract 144 from both sides.

$$9x^{2}+54x+81-144=-16x^{2}$$

Subtract 144 from 81 to get -63.

$$9x^{2}+54x-63=-16x^{2}$$

Add 16x^{2} to both sides.

$$9x^{2}+54x-63+16x^{2}=0$$

Combine 9x^{2} and 16x^{2} to get 25x^{2}.

$$25x^{2}+54x-63=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 25 for a, 54 for b, and -63 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-54±\sqrt{54^{2}-4\times 25\left(-63\right)}}{2\times 25}$$

Square 54.

$$x=\frac{-54±\sqrt{2916-4\times 25\left(-63\right)}}{2\times 25}$$

Multiply -4 times 25.

$$x=\frac{-54±\sqrt{2916-100\left(-63\right)}}{2\times 25}$$

Multiply -100 times -63.

$$x=\frac{-54±\sqrt{2916+6300}}{2\times 25}$$

Add 2916 to 6300.

$$x=\frac{-54±\sqrt{9216}}{2\times 25}$$

Take the square root of 9216.

$$x=\frac{-54±96}{2\times 25}$$

Multiply 2 times 25.

$$x=\frac{-54±96}{50}$$

Now solve the equation x=\frac{-54±96}{50} when ± is plus. Add -54 to 96.

$$x=\frac{42}{50}$$

Reduce the fraction \frac{42}{50} to lowest terms by extracting and canceling out 2.

$$x=\frac{21}{25}$$

Now solve the equation x=\frac{-54±96}{50} when ± is minus. Subtract 96 from -54.

$$x=-\frac{150}{50}$$

Divide -150 by 50.

$$x=-3$$

The equation is now solved.

$$x=\frac{21}{25} x=-3$$

Variable x cannot be equal to -3.

$$x=\frac{21}{25}$$

Steps for Completing the Square

Variable x cannot be equal to -3 since division by zero is not defined. Multiply both sides of the equation by 16\left(x+3\right)^{2}, the least common multiple of 16,x^{2}+6x+9.

$$9\left(x+3\right)^{2}=16\left(9-x^{2}\right)$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$9\left(x^{2}+6x+9\right)=16\left(9-x^{2}\right)$$

Use the distributive property to multiply 9 by x^{2}+6x+9.

$$9x^{2}+54x+81=16\left(9-x^{2}\right)$$

Use the distributive property to multiply 16 by 9-x^{2}.

$$9x^{2}+54x+81=144-16x^{2}$$

Add 16x^{2} to both sides.

$$9x^{2}+54x+81+16x^{2}=144$$

Combine 9x^{2} and 16x^{2} to get 25x^{2}.

$$25x^{2}+54x+81=144$$

Subtract 81 from both sides.

$$25x^{2}+54x=144-81$$

Subtract 81 from 144 to get 63.

$$25x^{2}+54x=63$$

Divide both sides by 25.

$$\frac{25x^{2}+54x}{25}=\frac{63}{25}$$

Dividing by 25 undoes the multiplication by 25.

$$x^{2}+\frac{54}{25}x=\frac{63}{25}$$

Divide \frac{54}{25}, the coefficient of the x term, by 2 to get \frac{27}{25}. Then add the square of \frac{27}{25} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{54}{25}x+\left(\frac{27}{25}\right)^{2}=\frac{63}{25}+\left(\frac{27}{25}\right)^{2}$$

Square \frac{27}{25} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{54}{25}x+\frac{729}{625}=\frac{63}{25}+\frac{729}{625}$$

Add \frac{63}{25} to \frac{729}{625} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{54}{25}x+\frac{729}{625}=\frac{2304}{625}$$

Factor x^{2}+\frac{54}{25}x+\frac{729}{625}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{27}{25}\right)^{2}=\frac{2304}{625}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{27}{25}\right)^{2}}=\sqrt{\frac{2304}{625}}$$

Simplify.

$$x+\frac{27}{25}=\frac{48}{25} x+\frac{27}{25}=-\frac{48}{25}$$

Subtract \frac{27}{25} from both sides of the equation.

$$x=\frac{21}{25} x=-3$$

Variable x cannot be equal to -3.

$$x=\frac{21}{25}$$