Solve left(1+aright)^2left(60-35right)3500/35=2400/35-11left(60-35+11right)

$${ \left(1+a \right) }^{ 2 } \left( 60-35 \right) \frac{ 3500 }{ 35 } = \frac{ 2400 }{ 35-11 } \left( 60-35+11 \right)$$

Solve for a

$$a=\frac{1}{5}=0.2$$

$$a = -\frac{11}{5} = -2\frac{1}{5} = -2.2$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+a\right)^{2}.

$$\left(1+2a+a^{2}\right)\left(60-35\right)\times \frac{3500}{35}=\frac{2400}{35-11}\left(60-35+11\right)$$

Subtract 35 from 60 to get 25.

$$\left(1+2a+a^{2}\right)\times 25\times \frac{3500}{35}=\frac{2400}{35-11}\left(60-35+11\right)$$

Divide 3500 by 35 to get 100.

$$\left(1+2a+a^{2}\right)\times 25\times 100=\frac{2400}{35-11}\left(60-35+11\right)$$

Multiply 25 and 100 to get 2500.

$$\left(1+2a+a^{2}\right)\times 2500=\frac{2400}{35-11}\left(60-35+11\right)$$

Use the distributive property to multiply 1+2a+a^{2} by 2500.

$$2500+5000a+2500a^{2}=\frac{2400}{35-11}\left(60-35+11\right)$$

Subtract 11 from 35 to get 24.

$$2500+5000a+2500a^{2}=\frac{2400}{24}\left(60-35+11\right)$$

Divide 2400 by 24 to get 100.

$$2500+5000a+2500a^{2}=100\left(60-35+11\right)$$

Subtract 35 from 60 to get 25.

$$2500+5000a+2500a^{2}=100\left(25+11\right)$$

Add 25 and 11 to get 36.

$$2500+5000a+2500a^{2}=100\times 36$$

Multiply 100 and 36 to get 3600.

$$2500+5000a+2500a^{2}=3600$$

Subtract 3600 from both sides.

$$2500+5000a+2500a^{2}-3600=0$$

Subtract 3600 from 2500 to get -1100.

$$-1100+5000a+2500a^{2}=0$$

Divide both sides by 100.

$$-11+50a+25a^{2}=0$$

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

$$25a^{2}+50a-11=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 25a^{2}+aa+ba-11. To find a and b, set up a system to be solved.

$$a+b=50 ab=25\left(-11\right)=-275$$

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 -275.

$$-1,275 -5,55 -11,25$$

Calculate the sum for each pair.

$$-1+275=274 -5+55=50 -11+25=14$$

The solution is the pair that gives sum 50.

$$a=-5 b=55$$

Rewrite 25a^{2}+50a-11 as \left(25a^{2}-5a\right)+\left(55a-11\right).

$$\left(25a^{2}-5a\right)+\left(55a-11\right)$$

Factor out 5a in the first and 11 in the second group.

$$5a\left(5a-1\right)+11\left(5a-1\right)$$

Factor out common term 5a-1 by using distributive property.

$$\left(5a-1\right)\left(5a+11\right)$$

To find equation solutions, solve 5a-1=0 and 5a+11=0.

$$a=\frac{1}{5} a=-\frac{11}{5}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+a\right)^{2}.

$$\left(1+2a+a^{2}\right)\left(60-35\right)\times \frac{3500}{35}=\frac{2400}{35-11}\left(60-35+11\right)$$

Subtract 35 from 60 to get 25.

$$\left(1+2a+a^{2}\right)\times 25\times \frac{3500}{35}=\frac{2400}{35-11}\left(60-35+11\right)$$

Divide 3500 by 35 to get 100.

$$\left(1+2a+a^{2}\right)\times 25\times 100=\frac{2400}{35-11}\left(60-35+11\right)$$

Multiply 25 and 100 to get 2500.

$$\left(1+2a+a^{2}\right)\times 2500=\frac{2400}{35-11}\left(60-35+11\right)$$

Use the distributive property to multiply 1+2a+a^{2} by 2500.

$$2500+5000a+2500a^{2}=\frac{2400}{35-11}\left(60-35+11\right)$$

Subtract 11 from 35 to get 24.

$$2500+5000a+2500a^{2}=\frac{2400}{24}\left(60-35+11\right)$$

Divide 2400 by 24 to get 100.

$$2500+5000a+2500a^{2}=100\left(60-35+11\right)$$

Subtract 35 from 60 to get 25.

$$2500+5000a+2500a^{2}=100\left(25+11\right)$$

Add 25 and 11 to get 36.

$$2500+5000a+2500a^{2}=100\times 36$$

Multiply 100 and 36 to get 3600.

$$2500+5000a+2500a^{2}=3600$$

Subtract 3600 from both sides.

$$2500+5000a+2500a^{2}-3600=0$$

Subtract 3600 from 2500 to get -1100.

$$-1100+5000a+2500a^{2}=0$$

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.

$$2500a^{2}+5000a-1100=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2500 for a, 5000 for b, and -1100 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-5000±\sqrt{5000^{2}-4\times 2500\left(-1100\right)}}{2\times 2500}$$

Square 5000.

$$a=\frac{-5000±\sqrt{25000000-4\times 2500\left(-1100\right)}}{2\times 2500}$$

Multiply -4 times 2500.

$$a=\frac{-5000±\sqrt{25000000-10000\left(-1100\right)}}{2\times 2500}$$

Multiply -10000 times -1100.

$$a=\frac{-5000±\sqrt{25000000+11000000}}{2\times 2500}$$

Add 25000000 to 11000000.

$$a=\frac{-5000±\sqrt{36000000}}{2\times 2500}$$

Take the square root of 36000000.

$$a=\frac{-5000±6000}{2\times 2500}$$

Multiply 2 times 2500.

$$a=\frac{-5000±6000}{5000}$$

Now solve the equation a=\frac{-5000±6000}{5000} when ± is plus. Add -5000 to 6000.

$$a=\frac{1000}{5000}$$

Reduce the fraction \frac{1000}{5000} to lowest terms by extracting and canceling out 1000.

$$a=\frac{1}{5}$$

Now solve the equation a=\frac{-5000±6000}{5000} when ± is minus. Subtract 6000 from -5000.

$$a=-\frac{11000}{5000}$$

Reduce the fraction \frac{-11000}{5000} to lowest terms by extracting and canceling out 1000.

$$a=-\frac{11}{5}$$

The equation is now solved.

$$a=\frac{1}{5} a=-\frac{11}{5}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+a\right)^{2}.

$$\left(1+2a+a^{2}\right)\left(60-35\right)\times \frac{3500}{35}=\frac{2400}{35-11}\left(60-35+11\right)$$

Subtract 35 from 60 to get 25.

$$\left(1+2a+a^{2}\right)\times 25\times \frac{3500}{35}=\frac{2400}{35-11}\left(60-35+11\right)$$

Divide 3500 by 35 to get 100.

$$\left(1+2a+a^{2}\right)\times 25\times 100=\frac{2400}{35-11}\left(60-35+11\right)$$

Multiply 25 and 100 to get 2500.

$$\left(1+2a+a^{2}\right)\times 2500=\frac{2400}{35-11}\left(60-35+11\right)$$

Use the distributive property to multiply 1+2a+a^{2} by 2500.

$$2500+5000a+2500a^{2}=\frac{2400}{35-11}\left(60-35+11\right)$$

Subtract 11 from 35 to get 24.

$$2500+5000a+2500a^{2}=\frac{2400}{24}\left(60-35+11\right)$$

Divide 2400 by 24 to get 100.

$$2500+5000a+2500a^{2}=100\left(60-35+11\right)$$

Subtract 35 from 60 to get 25.

$$2500+5000a+2500a^{2}=100\left(25+11\right)$$

Add 25 and 11 to get 36.

$$2500+5000a+2500a^{2}=100\times 36$$

Multiply 100 and 36 to get 3600.

$$2500+5000a+2500a^{2}=3600$$

Subtract 2500 from both sides.

$$5000a+2500a^{2}=3600-2500$$

Subtract 2500 from 3600 to get 1100.

$$5000a+2500a^{2}=1100$$

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.

$$2500a^{2}+5000a=1100$$

Divide both sides by 2500.

$$\frac{2500a^{2}+5000a}{2500}=\frac{1100}{2500}$$

Dividing by 2500 undoes the multiplication by 2500.

$$a^{2}+\frac{5000}{2500}a=\frac{1100}{2500}$$

Divide 5000 by 2500.

$$a^{2}+2a=\frac{1100}{2500}$$

Reduce the fraction \frac{1100}{2500} to lowest terms by extracting and canceling out 100.

$$a^{2}+2a=\frac{11}{25}$$

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}+2a+1^{2}=\frac{11}{25}+1^{2}$$

Square 1.

$$a^{2}+2a+1=\frac{11}{25}+1$$

Add \frac{11}{25} to 1.

$$a^{2}+2a+1=\frac{36}{25}$$

Factor a^{2}+2a+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(a+1\right)^{2}=\frac{36}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a+1\right)^{2}}=\sqrt{\frac{36}{25}}$$

Simplify.

$$a+1=\frac{6}{5} a+1=-\frac{6}{5}$$

Subtract 1 from both sides of the equation.

$$a=\frac{1}{5} a=-\frac{11}{5}$$