Solve 6(1+x)^2=1734

$$6 ( 1 + x ) ^ { 2 } = 1734$$

Solve for x

$$x=16$$

$$x=-18$$

Steps Using Factoring

Divide both sides by 6.

$$\left(1+x\right)^{2}=\frac{1734}{6}$$

Divide 1734 by 6 to get 289.

$$\left(1+x\right)^{2}=289$$

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

$$1+2x+x^{2}=289$$

Subtract 289 from both sides.

$$1+2x+x^{2}-289=0$$

Subtract 289 from 1 to get -288.

$$-288+2x+x^{2}=0$$

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

$$x^{2}+2x-288=0$$

To solve the equation, factor x^{2}+2x-288 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=2 ab=-288$$

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 -288.

$$-1,288 -2,144 -3,96 -4,72 -6,48 -8,36 -9,32 -12,24 -16,18$$

Calculate the sum for each pair.

$$-1+288=287 -2+144=142 -3+96=93 -4+72=68 -6+48=42 -8+36=28 -9+32=23 -12+24=12 -16+18=2$$

The solution is the pair that gives sum 2.

$$a=-16 b=18$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-16\right)\left(x+18\right)$$

To find equation solutions, solve x-16=0 and x+18=0.

$$x=16 x=-18$$

Steps Using Factoring By Grouping

Divide both sides by 6.

$$\left(1+x\right)^{2}=\frac{1734}{6}$$

Divide 1734 by 6 to get 289.

$$\left(1+x\right)^{2}=289$$

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

$$1+2x+x^{2}=289$$

Subtract 289 from both sides.

$$1+2x+x^{2}-289=0$$

Subtract 289 from 1 to get -288.

$$-288+2x+x^{2}=0$$

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

$$x^{2}+2x-288=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-288. To find a and b, set up a system to be solved.

$$a+b=2 ab=1\left(-288\right)=-288$$

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 -288.

$$-1,288 -2,144 -3,96 -4,72 -6,48 -8,36 -9,32 -12,24 -16,18$$

Calculate the sum for each pair.

$$-1+288=287 -2+144=142 -3+96=93 -4+72=68 -6+48=42 -8+36=28 -9+32=23 -12+24=12 -16+18=2$$

The solution is the pair that gives sum 2.

$$a=-16 b=18$$

Rewrite x^{2}+2x-288 as \left(x^{2}-16x\right)+\left(18x-288\right).

$$\left(x^{2}-16x\right)+\left(18x-288\right)$$

Factor out x in the first and 18 in the second group.

$$x\left(x-16\right)+18\left(x-16\right)$$

Factor out common term x-16 by using distributive property.

$$\left(x-16\right)\left(x+18\right)$$

To find equation solutions, solve x-16=0 and x+18=0.

$$x=16 x=-18$$

Steps Using the Quadratic Formula

Divide both sides by 6.

$$\left(1+x\right)^{2}=\frac{1734}{6}$$

Divide 1734 by 6 to get 289.

$$\left(1+x\right)^{2}=289$$

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

$$1+2x+x^{2}=289$$

Subtract 289 from both sides.

$$1+2x+x^{2}-289=0$$

Subtract 289 from 1 to get -288.

$$-288+2x+x^{2}=0$$

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.

$$x^{2}+2x-288=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 2 for b, and -288 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-288\right)}}{2}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\left(-288\right)}}{2}$$

Multiply -4 times -288.

$$x=\frac{-2±\sqrt{4+1152}}{2}$$

Add 4 to 1152.

$$x=\frac{-2±\sqrt{1156}}{2}$$

Take the square root of 1156.

$$x=\frac{-2±34}{2}$$

Now solve the equation x=\frac{-2±34}{2} when ± is plus. Add -2 to 34.

$$x=\frac{32}{2}$$

Divide 32 by 2.

$$x=16$$

Now solve the equation x=\frac{-2±34}{2} when ± is minus. Subtract 34 from -2.

$$x=-\frac{36}{2}$$

Divide -36 by 2.

$$x=-18$$

The equation is now solved.

$$x=16 x=-18$$

Steps for Completing the Square

Divide both sides by 6.

$$\left(1+x\right)^{2}=\frac{1734}{6}$$

Divide 1734 by 6 to get 289.

$$\left(1+x\right)^{2}=289$$

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

$$1+2x+x^{2}=289$$

Subtract 1 from both sides.

$$2x+x^{2}=289-1$$

Subtract 1 from 289 to get 288.

$$2x+x^{2}=288$$

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.

$$x^{2}+2x=288$$

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.

$$x^{2}+2x+1^{2}=288+1^{2}$$

Square 1.

$$x^{2}+2x+1=288+1$$

Add 288 to 1.

$$x^{2}+2x+1=289$$

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}.

$$\left(x+1\right)^{2}=289$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{289}$$

Simplify.

$$x+1=17 x+1=-17$$

Subtract 1 from both sides of the equation.

$$x=16 x=-18$$