Solve y^2-2y+15

$$y ^ { 2 } - 2 y + 15$$

Evaluate

$$y^{2}-2y+15$$

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.

$$x ^ 2 -2x +15 = 0$$

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

$$r + s = 2 rs = 15$$

Two numbers r and s sum up to 2 exactly when the average of the two numbers is \frac{1}{2}*2 = 1. 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>

$$r = 1 - u s = 1 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 15

$$(1 - u) (1 + u) = 15$$

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

$$1 - u^2 = 15$$

Simplify the expression by subtracting 1 on both sides

$$-u^2 = 15-1 = 14$$

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

$$u^2 = -14 u = \pm\sqrt{-14} = \pm \sqrt{14}i$$

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

$$r =1 - \sqrt{14}i s = 1 + \sqrt{14}i$$

Differentiate w.r.t. y

$$2\left(y-1\right)$$

Steps Using Derivative Rule for Sum

The derivative of a polynomial is the sum of the derivatives of its terms. The derivative of a constant term is 0. The derivative of ax^{n} is nax^{n-1}.

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

Subtract 1 from 2.

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

Subtract 1 from 1.

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

For any term t, t^{1}=t.

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

For any term t except 0, t^{0}=1.

$$2y-2$$