Solve x^2-49/4=9-{left(x-7/2right)}^2

$${ x }^{ 2 } - \frac{ 49 }{ 4 } =9- { \left(x- \frac{ 7 }{ 2 } \right) }^{ 2 }$$

Solve for x

$$x=-1$$

$$x = \frac{9}{2} = 4\frac{1}{2} = 4.5$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-\frac{7}{2}\right)^{2}.

$$x^{2}-\frac{49}{4}=9-\left(x^{2}-7x+\frac{49}{4}\right)$$

To find the opposite of x^{2}-7x+\frac{49}{4}, find the opposite of each term.

$$x^{2}-\frac{49}{4}=9-x^{2}+7x-\frac{49}{4}$$

Subtract \frac{49}{4} from 9 to get -\frac{13}{4}.

$$x^{2}-\frac{49}{4}=-\frac{13}{4}-x^{2}+7x$$

Subtract -\frac{13}{4} from both sides.

$$x^{2}-\frac{49}{4}-\left(-\frac{13}{4}\right)=-x^{2}+7x$$

The opposite of -\frac{13}{4} is \frac{13}{4}.

$$x^{2}-\frac{49}{4}+\frac{13}{4}=-x^{2}+7x$$

Add x^{2} to both sides.

$$x^{2}-\frac{49}{4}+\frac{13}{4}+x^{2}=7x$$

Add -\frac{49}{4} and \frac{13}{4} to get -9.

$$x^{2}-9+x^{2}=7x$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}-9=7x$$

Subtract 7x from both sides.

$$2x^{2}-9-7x=0$$

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

$$2x^{2}-7x-9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-9. To find a and b, set up a system to be solved.

$$a+b=-7 ab=2\left(-9\right)=-18$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -18.

$$1,-18 2,-9 3,-6$$

Calculate the sum for each pair.

$$1-18=-17 2-9=-7 3-6=-3$$

The solution is the pair that gives sum -7.

$$a=-9 b=2$$

Rewrite 2x^{2}-7x-9 as \left(2x^{2}-9x\right)+\left(2x-9\right).

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

Factor out x in 2x^{2}-9x.

$$x\left(2x-9\right)+2x-9$$

Factor out common term 2x-9 by using distributive property.

$$\left(2x-9\right)\left(x+1\right)$$

To find equation solutions, solve 2x-9=0 and x+1=0.

$$x=\frac{9}{2} x=-1$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-\frac{7}{2}\right)^{2}.

$$x^{2}-\frac{49}{4}=9-\left(x^{2}-7x+\frac{49}{4}\right)$$

To find the opposite of x^{2}-7x+\frac{49}{4}, find the opposite of each term.

$$x^{2}-\frac{49}{4}=9-x^{2}+7x-\frac{49}{4}$$

Subtract \frac{49}{4} from 9 to get -\frac{13}{4}.

$$x^{2}-\frac{49}{4}=-\frac{13}{4}-x^{2}+7x$$

Subtract -\frac{13}{4} from both sides.

$$x^{2}-\frac{49}{4}-\left(-\frac{13}{4}\right)=-x^{2}+7x$$

The opposite of -\frac{13}{4} is \frac{13}{4}.

$$x^{2}-\frac{49}{4}+\frac{13}{4}=-x^{2}+7x$$

Add x^{2} to both sides.

$$x^{2}-\frac{49}{4}+\frac{13}{4}+x^{2}=7x$$

Add -\frac{49}{4} and \frac{13}{4} to get -9.

$$x^{2}-9+x^{2}=7x$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}-9=7x$$

Subtract 7x from both sides.

$$2x^{2}-9-7x=0$$

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.

$$2x^{2}-7x-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -7 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 2\left(-9\right)}}{2\times 2}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 2\left(-9\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-7\right)±\sqrt{49-8\left(-9\right)}}{2\times 2}$$

Multiply -8 times -9.

$$x=\frac{-\left(-7\right)±\sqrt{49+72}}{2\times 2}$$

Add 49 to 72.

$$x=\frac{-\left(-7\right)±\sqrt{121}}{2\times 2}$$

Take the square root of 121.

$$x=\frac{-\left(-7\right)±11}{2\times 2}$$

The opposite of -7 is 7.

$$x=\frac{7±11}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{7±11}{4}$$

Now solve the equation x=\frac{7±11}{4} when ± is plus. Add 7 to 11.

$$x=\frac{18}{4}$$

Reduce the fraction \frac{18}{4} to lowest terms by extracting and canceling out 2.

$$x=\frac{9}{2}$$

Now solve the equation x=\frac{7±11}{4} when ± is minus. Subtract 11 from 7.

$$x=-\frac{4}{4}$$

Divide -4 by 4.

$$x=-1$$

The equation is now solved.

$$x=\frac{9}{2} x=-1$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-\frac{7}{2}\right)^{2}.

$$x^{2}-\frac{49}{4}=9-\left(x^{2}-7x+\frac{49}{4}\right)$$

To find the opposite of x^{2}-7x+\frac{49}{4}, find the opposite of each term.

$$x^{2}-\frac{49}{4}=9-x^{2}+7x-\frac{49}{4}$$

Subtract \frac{49}{4} from 9 to get -\frac{13}{4}.

$$x^{2}-\frac{49}{4}=-\frac{13}{4}-x^{2}+7x$$

Add x^{2} to both sides.

$$x^{2}-\frac{49}{4}+x^{2}=-\frac{13}{4}+7x$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}-\frac{49}{4}=-\frac{13}{4}+7x$$

Subtract 7x from both sides.

$$2x^{2}-\frac{49}{4}-7x=-\frac{13}{4}$$

Add \frac{49}{4} to both sides.

$$2x^{2}-7x=-\frac{13}{4}+\frac{49}{4}$$

Add -\frac{13}{4} and \frac{49}{4} to get 9.

$$2x^{2}-7x=9$$

Divide both sides by 2.

$$\frac{2x^{2}-7x}{2}=\frac{9}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{7}{2}x=\frac{9}{2}$$

Divide -\frac{7}{2}, the coefficient of the x term, by 2 to get -\frac{7}{4}. Then add the square of -\frac{7}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{2}x+\left(-\frac{7}{4}\right)^{2}=\frac{9}{2}+\left(-\frac{7}{4}\right)^{2}$$

Square -\frac{7}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=\frac{9}{2}+\frac{49}{16}$$

Add \frac{9}{2} to \frac{49}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=\frac{121}{16}$$

Factor x^{2}-\frac{7}{2}x+\frac{49}{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}.

$$\left(x-\frac{7}{4}\right)^{2}=\frac{121}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{4}\right)^{2}}=\sqrt{\frac{121}{16}}$$

Simplify.

$$x-\frac{7}{4}=\frac{11}{4} x-\frac{7}{4}=-\frac{11}{4}$$

Add \frac{7}{4} to both sides of the equation.

$$x=\frac{9}{2} x=-1$$