Solve (2x-1)^2-(x-2)^2=24

$$( 2 x - 1 ) ^ { 2 } - ( x - 2 ) ^ { 2 } = 24$$

Solve for x

$$x=-3$$

$$x=3$$

Steps Using Difference of Squares

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

$$4x^{2}-4x+1-\left(x-2\right)^{2}=24$$

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

$$4x^{2}-4x+1-\left(x^{2}-4x+4\right)=24$$

To find the opposite of x^{2}-4x+4, find the opposite of each term.

$$4x^{2}-4x+1-x^{2}+4x-4=24$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}-4x+1+4x-4=24$$

Combine -4x and 4x to get 0.

$$3x^{2}+1-4=24$$

Subtract 4 from 1 to get -3.

$$3x^{2}-3=24$$

Subtract 24 from both sides.

$$3x^{2}-3-24=0$$

Subtract 24 from -3 to get -27.

$$3x^{2}-27=0$$

Divide both sides by 3.

$$x^{2}-9=0$$

Consider x^{2}-9. Rewrite x^{2}-9 as x^{2}-3^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-3\right)\left(x+3\right)=0$$

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

$$x=3 x=-3$$

Steps by Finding Square Root

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

$$4x^{2}-4x+1-\left(x-2\right)^{2}=24$$

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

$$4x^{2}-4x+1-\left(x^{2}-4x+4\right)=24$$

To find the opposite of x^{2}-4x+4, find the opposite of each term.

$$4x^{2}-4x+1-x^{2}+4x-4=24$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}-4x+1+4x-4=24$$

Combine -4x and 4x to get 0.

$$3x^{2}+1-4=24$$

Subtract 4 from 1 to get -3.

$$3x^{2}-3=24$$

Add 3 to both sides.

$$3x^{2}=24+3$$

Add 24 and 3 to get 27.

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

Divide both sides by 3.

$$x^{2}=\frac{27}{3}$$

Divide 27 by 3 to get 9.

$$x^{2}=9$$

Take the square root of both sides of the equation.

$$x=3 x=-3$$

Steps Using the Quadratic Formula

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

$$4x^{2}-4x+1-\left(x-2\right)^{2}=24$$

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

$$4x^{2}-4x+1-\left(x^{2}-4x+4\right)=24$$

To find the opposite of x^{2}-4x+4, find the opposite of each term.

$$4x^{2}-4x+1-x^{2}+4x-4=24$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}-4x+1+4x-4=24$$

Combine -4x and 4x to get 0.

$$3x^{2}+1-4=24$$

Subtract 4 from 1 to get -3.

$$3x^{2}-3=24$$

Subtract 24 from both sides.

$$3x^{2}-3-24=0$$

Subtract 24 from -3 to get -27.

$$3x^{2}-27=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 0 for b, and -27 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 3\left(-27\right)}}{2\times 3}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 3\left(-27\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{0±\sqrt{-12\left(-27\right)}}{2\times 3}$$

Multiply -12 times -27.

$$x=\frac{0±\sqrt{324}}{2\times 3}$$

Take the square root of 324.

$$x=\frac{0±18}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{0±18}{6}$$

Now solve the equation x=\frac{0±18}{6} when ± is plus. Divide 18 by 6.

$$x=3$$

Now solve the equation x=\frac{0±18}{6} when ± is minus. Divide -18 by 6.

$$x=-3$$

The equation is now solved.

$$x=3 x=-3$$