Solve Q.No.6 Solve 25x ^ 2 - 1 = 0 . | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

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

Solve for x

$x=\frac{1}{5}=0.2$
$x=-\frac{1}{5}=-0.2$

Steps Using Difference of Squares

Steps by Finding Square Root

Steps Using the Quadratic Formula

Steps Using Difference of Squares

Consider $25x^{2}-1$. Rewrite $25x^{2}-1$ as $\left(5x\right)^{2}-1^{2}$. The difference of squares can be factored using the rule: $a^{2}-b^{2}=\left(a-b\right)\left(a+b\right)$.

$$\left(5x-1\right)\left(5x+1\right)=0$$

To find equation solutions, solve $5x-1=0$ and $5x+1=0$.

$$x=\frac{1}{5}$$ $$x=-\frac{1}{5}$$

Steps by Finding Square Root

Add $1$ to both sides. Anything plus zero gives itself.

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

Divide both sides by $25$.

$$x^{2}=\frac{1}{25}$$

Take the square root of both sides of the equation.

$$x=\frac{1}{5}$$ $$x=-\frac{1}{5}$$

Steps Using the Quadratic Formula

Quadratic equations like this one, with an $x^{2}$ term but no $x$ term, can still be solved using the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$, once they are put in standard form: $ax^{2}+bx+c=0$.

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

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $25$ for $a$, $0$ for $b$, and $-1$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{0±\sqrt{0^{2}-4\times 25\left(-1\right)}}{2\times 25}$$

Square $0$.

$$x=\frac{0±\sqrt{-4\times 25\left(-1\right)}}{2\times 25}$$

Multiply $-4$ times $25$.

$$x=\frac{0±\sqrt{-100\left(-1\right)}}{2\times 25}$$

Multiply $-100$ times $-1$.

$$x=\frac{0±\sqrt{100}}{2\times 25}$$

Take the square root of $100$.

$$x=\frac{0±10}{2\times 25}$$

Multiply $2$ times $25$.

$$x=\frac{0±10}{50}$$

Now solve the equation $x=\frac{0±10}{50}$ when $±$ is plus. Reduce the fraction $\frac{10}{50}$ to lowest terms by extracting and canceling out $10$.

$$x=\frac{1}{5}$$

Now solve the equation $x=\frac{0±10}{50}$ when $±$ is minus. Reduce the fraction $\frac{-10}{50}$ to lowest terms by extracting and canceling out $10$.

$$x=-\frac{1}{5}$$

The equation is now solved.

$$x=\frac{1}{5}$$ $$x=-\frac{1}{5}$$