Evaluate
2h^{2}-5h+4
Differentiate w.r.t. h
4h-5
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x ^ 2 -\frac{5}{2}x +2 = 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{5}{2} rs = 2
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{5}{4} - u s = \frac{5}{4} + u
Two numbers r and s sum up to \frac{5}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{5}{2} = \frac{5}{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{5}{4} - u) (\frac{5}{4} + u) = 2
To solve for unknown quantity u, substitute these in the product equation rs = 2
\frac{25}{16} - u^2 = 2
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 2-\frac{25}{16} = \frac{7}{16}
Simplify the expression by subtracting \frac{25}{16} on both sides
u^2 = -\frac{7}{16} u = \pm\sqrt{-\frac{7}{16}} = \pm \frac{\sqrt{7}}{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{5}{4} - \frac{\sqrt{7}}{4}i = 1.250 - 0.661i s = \frac{5}{4} + \frac{\sqrt{7}}{4}i = 1.250 + 0.661i
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 2h^{2-1}-5h^{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}.
4h^{2-1}-5h^{1-1}
Multiply 2 times 2.
4h^{1}-5h^{1-1}
Subtract 1 from 2.
4h^{1}-5h^{0}
Subtract 1 from 1.
4h-5h^{0}
For any term t, t^{1}=t.
4h-5
For any term t except 0, t^{0}=1.
Examples
Quadratic equation
{ x } ^ { 2 } - 4 x - 5 = 0
Trigonometry
4 \sin \theta \cos \theta = 2 \sin \theta
Linear equation
y = 3x + 4
Arithmetic
699 * 533
Matrix
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}