Solve 17x^2-(3x-1)^2=x+2

$$17 x ^ { 2 } - ( 3 x - 1 ) ^ { 2 } = x + 2$$

Solve for x

$$x=-1$$

$$x=\frac{3}{8}=0.375$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-1\right)^{2}.

$$17x^{2}-\left(9x^{2}-6x+1\right)=x+2$$

To find the opposite of 9x^{2}-6x+1, find the opposite of each term.

$$17x^{2}-9x^{2}+6x-1=x+2$$

Combine 17x^{2} and -9x^{2} to get 8x^{2}.

$$8x^{2}+6x-1=x+2$$

Subtract x from both sides.

$$8x^{2}+6x-1-x=2$$

Combine 6x and -x to get 5x.

$$8x^{2}+5x-1=2$$

Subtract 2 from both sides.

$$8x^{2}+5x-1-2=0$$

Subtract 2 from -1 to get -3.

$$8x^{2}+5x-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 8x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=5 ab=8\left(-3\right)=-24$$

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 -24.

$$-1,24 -2,12 -3,8 -4,6$$

Calculate the sum for each pair.

$$-1+24=23 -2+12=10 -3+8=5 -4+6=2$$

The solution is the pair that gives sum 5.

$$a=-3 b=8$$

Rewrite 8x^{2}+5x-3 as \left(8x^{2}-3x\right)+\left(8x-3\right).

$$\left(8x^{2}-3x\right)+\left(8x-3\right)$$

Factor out x in 8x^{2}-3x.

$$x\left(8x-3\right)+8x-3$$

Factor out common term 8x-3 by using distributive property.

$$\left(8x-3\right)\left(x+1\right)$$

To find equation solutions, solve 8x-3=0 and x+1=0.

$$x=\frac{3}{8} x=-1$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-1\right)^{2}.

$$17x^{2}-\left(9x^{2}-6x+1\right)=x+2$$

To find the opposite of 9x^{2}-6x+1, find the opposite of each term.

$$17x^{2}-9x^{2}+6x-1=x+2$$

Combine 17x^{2} and -9x^{2} to get 8x^{2}.

$$8x^{2}+6x-1=x+2$$

Subtract x from both sides.

$$8x^{2}+6x-1-x=2$$

Combine 6x and -x to get 5x.

$$8x^{2}+5x-1=2$$

Subtract 2 from both sides.

$$8x^{2}+5x-1-2=0$$

Subtract 2 from -1 to get -3.

$$8x^{2}+5x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, 5 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\times 8\left(-3\right)}}{2\times 8}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\times 8\left(-3\right)}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-5±\sqrt{25-32\left(-3\right)}}{2\times 8}$$

Multiply -32 times -3.

$$x=\frac{-5±\sqrt{25+96}}{2\times 8}$$

Add 25 to 96.

$$x=\frac{-5±\sqrt{121}}{2\times 8}$$

Take the square root of 121.

$$x=\frac{-5±11}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{-5±11}{16}$$

Now solve the equation x=\frac{-5±11}{16} when ± is plus. Add -5 to 11.

$$x=\frac{6}{16}$$

Reduce the fraction \frac{6}{16} to lowest terms by extracting and canceling out 2.

$$x=\frac{3}{8}$$

Now solve the equation x=\frac{-5±11}{16} when ± is minus. Subtract 11 from -5.

$$x=-\frac{16}{16}$$

Divide -16 by 16.

$$x=-1$$

The equation is now solved.

$$x=\frac{3}{8} x=-1$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-1\right)^{2}.

$$17x^{2}-\left(9x^{2}-6x+1\right)=x+2$$

To find the opposite of 9x^{2}-6x+1, find the opposite of each term.

$$17x^{2}-9x^{2}+6x-1=x+2$$

Combine 17x^{2} and -9x^{2} to get 8x^{2}.

$$8x^{2}+6x-1=x+2$$

Subtract x from both sides.

$$8x^{2}+6x-1-x=2$$

Combine 6x and -x to get 5x.

$$8x^{2}+5x-1=2$$

Add 1 to both sides.

$$8x^{2}+5x=2+1$$

Add 2 and 1 to get 3.

$$8x^{2}+5x=3$$

Divide both sides by 8.

$$\frac{8x^{2}+5x}{8}=\frac{3}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$x^{2}+\frac{5}{8}x=\frac{3}{8}$$

Divide \frac{5}{8}, the coefficient of the x term, by 2 to get \frac{5}{16}. Then add the square of \frac{5}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{5}{8}x+\left(\frac{5}{16}\right)^{2}=\frac{3}{8}+\left(\frac{5}{16}\right)^{2}$$

Square \frac{5}{16} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{8}x+\frac{25}{256}=\frac{3}{8}+\frac{25}{256}$$

Add \frac{3}{8} to \frac{25}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{5}{8}x+\frac{25}{256}=\frac{121}{256}$$

Factor x^{2}+\frac{5}{8}x+\frac{25}{256}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{16}\right)^{2}=\frac{121}{256}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{16}\right)^{2}}=\sqrt{\frac{121}{256}}$$

Simplify.

$$x+\frac{5}{16}=\frac{11}{16} x+\frac{5}{16}=-\frac{11}{16}$$

Subtract \frac{5}{16} from both sides of the equation.

$$x=\frac{3}{8} x=-1$$