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Differentiate w.r.t. x
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x ^ 2 +\frac{15}{2}x +18 = 0
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.This is achieved by dividing both sides of the equation by 2
r + s = -\frac{15}{2} rs = 18
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 = -\frac{15}{4} - u s = -\frac{15}{4} + u
Two numbers r and s sum up to -\frac{15}{2} exactly when the average of the two numbers is \frac{1}{2}*-\frac{15}{2} = -\frac{15}{4}. 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>
(-\frac{15}{4} - u) (-\frac{15}{4} + u) = 18
To solve for unknown quantity u, substitute these in the product equation rs = 18
\frac{225}{16} - u^2 = 18
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 18-\frac{225}{16} = \frac{63}{16}
Simplify the expression by subtracting \frac{225}{16} on both sides
u^2 = -\frac{63}{16} u = \pm\sqrt{-\frac{63}{16}} = \pm \frac{\sqrt{63}}{4}i
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{15}{4} - \frac{\sqrt{63}}{4}i = -3.750 - 1.984i s = -\frac{15}{4} + \frac{\sqrt{63}}{4}i = -3.750 + 1.984i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
2\times 2x^{2-1}+15x^{1-1}
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}.
4x^{2-1}+15x^{1-1}
Multiply 2 times 2.
4x^{1}+15x^{1-1}
Subtract 1 from 2.
4x^{1}+15x^{0}
Subtract 1 from 1.
4x+15x^{0}
For any term t, t^{1}=t.
4x+15\times 1
For any term t except 0, t^{0}=1.
4x+15
For any term t, t\times 1=t and 1t=t.