How to solve 2y imes(y)^(1)=1

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve 2y\times(y)^(1)=1 | Equatix

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Question

$$2y \times { y }^{ 1 } =1$$

Solve for y

$y=\frac{\sqrt{2}}{2}\approx 0.707106781$
$y=-\frac{\sqrt{2}}{2}\approx -0.707106781$

Steps by Finding Square Root

Steps Using the Quadratic Formula

Steps by Finding Square Root

To multiply powers of the same base, add their exponents. Add $1$ and $1$ to get $2$.

$$2y^{2}=1$$

Divide both sides by $2$.

$$y^{2}=\frac{1}{2}$$

Take the square root of both sides of the equation.

$$y=\frac{\sqrt{2}}{2}$$ $$y=-\frac{\sqrt{2}}{2}$$

Steps Using the Quadratic Formula

To multiply powers of the same base, add their exponents. Add $1$ and $1$ to get $2$.

$$2y^{2}=1$$

Subtract $1$ from both sides.

$$2y^{2}-1=0$$

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

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

Square $0$.

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

Multiply $-4$ times $2$.

$$y=\frac{0±\sqrt{-8\left(-1\right)}}{2\times 2}$$

Multiply $-8$ times $-1$.

$$y=\frac{0±\sqrt{8}}{2\times 2}$$

Take the square root of $8$.

$$y=\frac{0±2\sqrt{2}}{2\times 2}$$

Multiply $2$ times $2$.

$$y=\frac{0±2\sqrt{2}}{4}$$

Now solve the equation $y=\frac{0±2\sqrt{2}}{4}$ when $±$ is plus.

$$y=\frac{\sqrt{2}}{2}$$

Now solve the equation $y=\frac{0±2\sqrt{2}}{4}$ when $±$ is minus.

$$y=-\frac{\sqrt{2}}{2}$$

The equation is now solved.

$$y=\frac{\sqrt{2}}{2}$$ $$y=-\frac{\sqrt{2}}{2}$$

Show Solution

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Example

Question

Solve for y

Steps by Finding Square Root

Steps Using the Quadratic Formula

Steps by Finding Square Root

Steps Using the Quadratic Formula