Solve 1300(1+x)left(1+xright)^1=1573

$$1300(1+x) { \left(1+x \right) }^{ 1 } =1573$$

Solve for x

$$x=\frac{1}{10}=0.1$$

$$x = -\frac{21}{10} = -2\frac{1}{10} = -2.1$$

Steps Using Factoring By Grouping

To multiply powers of the same base, add their exponents. Add 1 and 1 to get 2.

$$1300\left(1+x\right)^{2}=1573$$

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

$$1300\left(1+2x+x^{2}\right)=1573$$

Use the distributive property to multiply 1300 by 1+2x+x^{2}.

$$1300+2600x+1300x^{2}=1573$$

Subtract 1573 from both sides.

$$1300+2600x+1300x^{2}-1573=0$$

Subtract 1573 from 1300 to get -273.

$$-273+2600x+1300x^{2}=0$$

Divide both sides by 13.

$$-21+200x+100x^{2}=0$$

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

$$100x^{2}+200x-21=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 100x^{2}+ax+bx-21. To find a and b, set up a system to be solved.

$$a+b=200 ab=100\left(-21\right)=-2100$$

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 -2100.

$$-1,2100 -2,1050 -3,700 -4,525 -5,420 -6,350 -7,300 -10,210 -12,175 -14,150 -15,140 -20,105 -21,100 -25,84 -28,75 -30,70 -35,60 -42,50$$

Calculate the sum for each pair.

$$-1+2100=2099 -2+1050=1048 -3+700=697 -4+525=521 -5+420=415 -6+350=344 -7+300=293 -10+210=200 -12+175=163 -14+150=136 -15+140=125 -20+105=85 -21+100=79 -25+84=59 -28+75=47 -30+70=40 -35+60=25 -42+50=8$$

The solution is the pair that gives sum 200.

$$a=-10 b=210$$

Rewrite 100x^{2}+200x-21 as \left(100x^{2}-10x\right)+\left(210x-21\right).

$$\left(100x^{2}-10x\right)+\left(210x-21\right)$$

Factor out 10x in the first and 21 in the second group.

$$10x\left(10x-1\right)+21\left(10x-1\right)$$

Factor out common term 10x-1 by using distributive property.

$$\left(10x-1\right)\left(10x+21\right)$$

To find equation solutions, solve 10x-1=0 and 10x+21=0.

$$x=\frac{1}{10} x=-\frac{21}{10}$$

Steps Using the Quadratic Formula

To multiply powers of the same base, add their exponents. Add 1 and 1 to get 2.

$$1300\left(1+x\right)^{2}=1573$$

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

$$1300\left(1+2x+x^{2}\right)=1573$$

Use the distributive property to multiply 1300 by 1+2x+x^{2}.

$$1300+2600x+1300x^{2}=1573$$

Subtract 1573 from both sides.

$$1300+2600x+1300x^{2}-1573=0$$

Subtract 1573 from 1300 to get -273.

$$-273+2600x+1300x^{2}=0$$

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.

$$1300x^{2}+2600x-273=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1300 for a, 2600 for b, and -273 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2600±\sqrt{2600^{2}-4\times 1300\left(-273\right)}}{2\times 1300}$$

Square 2600.

$$x=\frac{-2600±\sqrt{6760000-4\times 1300\left(-273\right)}}{2\times 1300}$$

Multiply -4 times 1300.

$$x=\frac{-2600±\sqrt{6760000-5200\left(-273\right)}}{2\times 1300}$$

Multiply -5200 times -273.

$$x=\frac{-2600±\sqrt{6760000+1419600}}{2\times 1300}$$

Add 6760000 to 1419600.

$$x=\frac{-2600±\sqrt{8179600}}{2\times 1300}$$

Take the square root of 8179600.

$$x=\frac{-2600±2860}{2\times 1300}$$

Multiply 2 times 1300.

$$x=\frac{-2600±2860}{2600}$$

Now solve the equation x=\frac{-2600±2860}{2600} when ± is plus. Add -2600 to 2860.

$$x=\frac{260}{2600}$$

Reduce the fraction \frac{260}{2600} to lowest terms by extracting and canceling out 260.

$$x=\frac{1}{10}$$

Now solve the equation x=\frac{-2600±2860}{2600} when ± is minus. Subtract 2860 from -2600.

$$x=-\frac{5460}{2600}$$

Reduce the fraction \frac{-5460}{2600} to lowest terms by extracting and canceling out 260.

$$x=-\frac{21}{10}$$

The equation is now solved.

$$x=\frac{1}{10} x=-\frac{21}{10}$$

Steps for Completing the Square

To multiply powers of the same base, add their exponents. Add 1 and 1 to get 2.

$$1300\left(1+x\right)^{2}=1573$$

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

$$1300\left(1+2x+x^{2}\right)=1573$$

Use the distributive property to multiply 1300 by 1+2x+x^{2}.

$$1300+2600x+1300x^{2}=1573$$

Subtract 1300 from both sides.

$$2600x+1300x^{2}=1573-1300$$

Subtract 1300 from 1573 to get 273.

$$2600x+1300x^{2}=273$$

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.

$$1300x^{2}+2600x=273$$

Divide both sides by 1300.

$$\frac{1300x^{2}+2600x}{1300}=\frac{273}{1300}$$

Dividing by 1300 undoes the multiplication by 1300.

$$x^{2}+\frac{2600}{1300}x=\frac{273}{1300}$$

Divide 2600 by 1300.

$$x^{2}+2x=\frac{273}{1300}$$

Reduce the fraction \frac{273}{1300} to lowest terms by extracting and canceling out 13.

$$x^{2}+2x=\frac{21}{100}$$

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.

$$x^{2}+2x+1^{2}=\frac{21}{100}+1^{2}$$

Square 1.

$$x^{2}+2x+1=\frac{21}{100}+1$$

Add \frac{21}{100} to 1.

$$x^{2}+2x+1=\frac{121}{100}$$

Factor x^{2}+2x+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}.

$$\left(x+1\right)^{2}=\frac{121}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{\frac{121}{100}}$$

Simplify.

$$x+1=\frac{11}{10} x+1=-\frac{11}{10}$$

Subtract 1 from both sides of the equation.

$$x=\frac{1}{10} x=-\frac{21}{10}$$