Solve (5x+1)(5x-1)=[1]

$$( 5 x + 1 ) ( 5 x - 1 ) = [ 1 ]$$

Solve for x

$$x=\frac{\sqrt{2}}{5}\approx 0.282842712$$

$$x=-\frac{\sqrt{2}}{5}\approx -0.282842712$$

Steps by Finding Square Root

Consider \left(5x+1\right)\left(5x-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$\left(5x\right)^{2}-1=1$$

Expand \left(5x\right)^{2}.

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

Calculate 5 to the power of 2 and get 25.

$$25x^{2}-1=1$$

Add 1 to both sides.

$$25x^{2}=1+1$$

Add 1 and 1 to get 2.

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

Divide both sides by 25.

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

Take the square root of both sides of the equation.

$$x=\frac{\sqrt{2}}{5} x=-\frac{\sqrt{2}}{5}$$

Steps Using the Quadratic Formula

Consider \left(5x+1\right)\left(5x-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$\left(5x\right)^{2}-1=1$$

Expand \left(5x\right)^{2}.

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

Calculate 5 to the power of 2 and get 25.

$$25x^{2}-1=1$$

Subtract 1 from both sides.

$$25x^{2}-1-1=0$$

Subtract 1 from -1 to get -2.

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

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

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

Square 0.

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

Multiply -4 times 25.

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

Multiply -100 times -2.

$$x=\frac{0±\sqrt{200}}{2\times 25}$$

Take the square root of 200.

$$x=\frac{0±10\sqrt{2}}{2\times 25}$$

Multiply 2 times 25.

$$x=\frac{0±10\sqrt{2}}{50}$$

Now solve the equation x=\frac{0±10\sqrt{2}}{50} when ± is plus.

$$x=\frac{\sqrt{2}}{5}$$

Now solve the equation x=\frac{0±10\sqrt{2}}{50} when ± is minus.

$$x=-\frac{\sqrt{2}}{5}$$

The equation is now solved.

$$x=\frac{\sqrt{2}}{5} x=-\frac{\sqrt{2}}{5}$$