Solve (4x)^2+(3x)^2=(25)^2

$$( 4 x ) ^ { 2 } + ( 3 x ) ^ { 2 } = ( 25 ) ^ { 2 }$$

Solve for x

$$x=-5$$

$$x=5$$

Steps Using Difference of Squares

Expand \left(4x\right)^{2}.

$$4^{2}x^{2}+\left(3x\right)^{2}=25^{2}$$

Calculate 4 to the power of 2 and get 16.

$$16x^{2}+\left(3x\right)^{2}=25^{2}$$

Expand \left(3x\right)^{2}.

$$16x^{2}+3^{2}x^{2}=25^{2}$$

Calculate 3 to the power of 2 and get 9.

$$16x^{2}+9x^{2}=25^{2}$$

Combine 16x^{2} and 9x^{2} to get 25x^{2}.

$$25x^{2}=25^{2}$$

Calculate 25 to the power of 2 and get 625.

$$25x^{2}=625$$

Subtract 625 from both sides.

$$25x^{2}-625=0$$

Divide both sides by 25.

$$x^{2}-25=0$$

Consider x^{2}-25. Rewrite x^{2}-25 as x^{2}-5^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-5\right)\left(x+5\right)=0$$

To find equation solutions, solve x-5=0 and x+5=0.

$$x=5 x=-5$$

Steps by Finding Square Root

Expand \left(4x\right)^{2}.

$$4^{2}x^{2}+\left(3x\right)^{2}=25^{2}$$

Calculate 4 to the power of 2 and get 16.

$$16x^{2}+\left(3x\right)^{2}=25^{2}$$

Expand \left(3x\right)^{2}.

$$16x^{2}+3^{2}x^{2}=25^{2}$$

Calculate 3 to the power of 2 and get 9.

$$16x^{2}+9x^{2}=25^{2}$$

Combine 16x^{2} and 9x^{2} to get 25x^{2}.

$$25x^{2}=25^{2}$$

Calculate 25 to the power of 2 and get 625.

$$25x^{2}=625$$

Divide both sides by 25.

$$x^{2}=\frac{625}{25}$$

Divide 625 by 25 to get 25.

$$x^{2}=25$$

Take the square root of both sides of the equation.

$$x=5 x=-5$$

Steps Using the Quadratic Formula

Expand \left(4x\right)^{2}.

$$4^{2}x^{2}+\left(3x\right)^{2}=25^{2}$$

Calculate 4 to the power of 2 and get 16.

$$16x^{2}+\left(3x\right)^{2}=25^{2}$$

Expand \left(3x\right)^{2}.

$$16x^{2}+3^{2}x^{2}=25^{2}$$

Calculate 3 to the power of 2 and get 9.

$$16x^{2}+9x^{2}=25^{2}$$

Combine 16x^{2} and 9x^{2} to get 25x^{2}.

$$25x^{2}=25^{2}$$

Calculate 25 to the power of 2 and get 625.

$$25x^{2}=625$$

Subtract 625 from both sides.

$$25x^{2}-625=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 25 for a, 0 for b, and -625 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 25\left(-625\right)}}{2\times 25}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 25\left(-625\right)}}{2\times 25}$$

Multiply -4 times 25.

$$x=\frac{0±\sqrt{-100\left(-625\right)}}{2\times 25}$$

Multiply -100 times -625.

$$x=\frac{0±\sqrt{62500}}{2\times 25}$$

Take the square root of 62500.

$$x=\frac{0±250}{2\times 25}$$

Multiply 2 times 25.

$$x=\frac{0±250}{50}$$

Now solve the equation x=\frac{0±250}{50} when ± is plus. Divide 250 by 50.

$$x=5$$

Now solve the equation x=\frac{0±250}{50} when ± is minus. Divide -250 by 50.

$$x=-5$$

The equation is now solved.

$$x=5 x=-5$$