Solve y^2+4y-0

$$y ^ { 2 } + 4 y - 0$$

Evaluate

$$y\left(y+4\right)$$

Solution Steps

Multiply -1 and 0 to get 0.

$$y^{2}+4y+0$$

Anything plus zero gives itself.

$$y^{2}+4y$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.

$$x ^ 2 +4x +0 = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = -4 rs = 0$$

Two numbers r and s sum up to -4 exactly when the average of the two numbers is \frac{1}{2}*-4 = -2. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = -2 - u s = -2 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 0

$$(-2 - u) (-2 + u) = 0$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$4 - u^2 = 0$$

Simplify the expression by subtracting 4 on both sides

$$-u^2 = 0-4 = -4$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = 4 u = \pm\sqrt{4} = \pm 2$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =-2 - 2 = -4 s = -2 + 2 = 0$$

Factor

$$y\left(y+4\right)$$

Solution Steps

Factor out y.

$$y\left(y+4\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$y^{2}+4y=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$y=\frac{-4±\sqrt{4^{2}}}{2}$$

Take the square root of 4^{2}.

$$y=\frac{-4±4}{2}$$

Now solve the equation y=\frac{-4±4}{2} when ± is plus. Add -4 to 4.

$$y=\frac{0}{2}$$

Divide 0 by 2.

$$y=0$$

Now solve the equation y=\frac{-4±4}{2} when ± is minus. Subtract 4 from -4.

$$y=-\frac{8}{2}$$

Divide -8 by 2.

$$y=-4$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 0 for x_{1} and -4 for x_{2}.

$$y^{2}+4y=y\left(y-\left(-4\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$y^{2}+4y=y\left(y+4\right)$$