Solve {left(5/2+3/2xright)}^2-x^2=40

$${ \left( \frac{ 5 }{ 2 } + \frac{ 3 }{ 2 } x \right) }^{ 2 } - { x }^{ 2 } =40$$

Solve for x

$$x=-9$$

$$x=3$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\frac{5}{2}+\frac{3}{2}x\right)^{2}.

$$\frac{25}{4}+\frac{15}{2}x+\frac{9}{4}x^{2}-x^{2}=40$$

Combine \frac{9}{4}x^{2} and -x^{2} to get \frac{5}{4}x^{2}.

$$\frac{25}{4}+\frac{15}{2}x+\frac{5}{4}x^{2}=40$$

Subtract 40 from both sides.

$$\frac{25}{4}+\frac{15}{2}x+\frac{5}{4}x^{2}-40=0$$

Subtract 40 from \frac{25}{4} to get -\frac{135}{4}.

$$-\frac{135}{4}+\frac{15}{2}x+\frac{5}{4}x^{2}=0$$

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.

$$\frac{5}{4}x^{2}+\frac{15}{2}x-\frac{135}{4}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{5}{4} for a, \frac{15}{2} for b, and -\frac{135}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{15}{2}±\sqrt{\left(\frac{15}{2}\right)^{2}-4\times \frac{5}{4}\left(-\frac{135}{4}\right)}}{2\times \frac{5}{4}}$$

Square \frac{15}{2} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{15}{2}±\sqrt{\frac{225}{4}-4\times \frac{5}{4}\left(-\frac{135}{4}\right)}}{2\times \frac{5}{4}}$$

Multiply -4 times \frac{5}{4}.

$$x=\frac{-\frac{15}{2}±\sqrt{\frac{225}{4}-5\left(-\frac{135}{4}\right)}}{2\times \frac{5}{4}}$$

Multiply -5 times -\frac{135}{4}.

$$x=\frac{-\frac{15}{2}±\sqrt{\frac{225+675}{4}}}{2\times \frac{5}{4}}$$

Add \frac{225}{4} to \frac{675}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{15}{2}±\sqrt{225}}{2\times \frac{5}{4}}$$

Take the square root of 225.

$$x=\frac{-\frac{15}{2}±15}{2\times \frac{5}{4}}$$

Multiply 2 times \frac{5}{4}.

$$x=\frac{-\frac{15}{2}±15}{\frac{5}{2}}$$

Now solve the equation x=\frac{-\frac{15}{2}±15}{\frac{5}{2}} when ± is plus. Add -\frac{15}{2} to 15.

$$x=\frac{\frac{15}{2}}{\frac{5}{2}}$$

Divide \frac{15}{2} by \frac{5}{2} by multiplying \frac{15}{2} by the reciprocal of \frac{5}{2}.

$$x=3$$

Now solve the equation x=\frac{-\frac{15}{2}±15}{\frac{5}{2}} when ± is minus. Subtract 15 from -\frac{15}{2}.

$$x=-\frac{\frac{45}{2}}{\frac{5}{2}}$$

Divide -\frac{45}{2} by \frac{5}{2} by multiplying -\frac{45}{2} by the reciprocal of \frac{5}{2}.

$$x=-9$$

The equation is now solved.

$$x=3 x=-9$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\frac{5}{2}+\frac{3}{2}x\right)^{2}.

$$\frac{25}{4}+\frac{15}{2}x+\frac{9}{4}x^{2}-x^{2}=40$$

Combine \frac{9}{4}x^{2} and -x^{2} to get \frac{5}{4}x^{2}.

$$\frac{25}{4}+\frac{15}{2}x+\frac{5}{4}x^{2}=40$$

Subtract \frac{25}{4} from both sides.

$$\frac{15}{2}x+\frac{5}{4}x^{2}=40-\frac{25}{4}$$

Subtract \frac{25}{4} from 40 to get \frac{135}{4}.

$$\frac{15}{2}x+\frac{5}{4}x^{2}=\frac{135}{4}$$

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.

$$\frac{5}{4}x^{2}+\frac{15}{2}x=\frac{135}{4}$$

Divide both sides of the equation by \frac{5}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{5}{4}x^{2}+\frac{15}{2}x}{\frac{5}{4}}=\frac{\frac{135}{4}}{\frac{5}{4}}$$

Dividing by \frac{5}{4} undoes the multiplication by \frac{5}{4}.

$$x^{2}+\frac{\frac{15}{2}}{\frac{5}{4}}x=\frac{\frac{135}{4}}{\frac{5}{4}}$$

Divide \frac{15}{2} by \frac{5}{4} by multiplying \frac{15}{2} by the reciprocal of \frac{5}{4}.

$$x^{2}+6x=\frac{\frac{135}{4}}{\frac{5}{4}}$$

Divide \frac{135}{4} by \frac{5}{4} by multiplying \frac{135}{4} by the reciprocal of \frac{5}{4}.

$$x^{2}+6x=27$$

Divide 6, the coefficient of the x term, by 2 to get 3. Then add the square of 3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+6x+3^{2}=27+3^{2}$$

Square 3.

$$x^{2}+6x+9=27+9$$

Add 27 to 9.

$$x^{2}+6x+9=36$$

Factor x^{2}+6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+3\right)^{2}=36$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+3\right)^{2}}=\sqrt{36}$$

Simplify.

$$x+3=6 x+3=-6$$

Subtract 3 from both sides of the equation.

$$x=3 x=-9$$