Solve for u
u=2
Share
Copied to clipboard
a+b=4 ab=-\left(-4\right)=4
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -u^{2}+au+bu-4. To find a and b, set up a system to be solved.
1,4 2,2
Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 4.
1+4=5 2+2=4
Calculate the sum for each pair.
a=2 b=2
The solution is the pair that gives sum 4.
\left(-u^{2}+2u\right)+\left(2u-4\right)
Rewrite -u^{2}+4u-4 as \left(-u^{2}+2u\right)+\left(2u-4\right).
-u\left(u-2\right)+2\left(u-2\right)
Factor out -u in the first and 2 in the second group.
\left(u-2\right)\left(-u+2\right)
Factor out common term u-2 by using distributive property.
u=2 u=2
To find equation solutions, solve u-2=0 and -u+2=0.
-u^{2}+4u-4=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.
u=\frac{-4±\sqrt{4^{2}-4\left(-1\right)\left(-4\right)}}{2\left(-1\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 4 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
u=\frac{-4±\sqrt{16-4\left(-1\right)\left(-4\right)}}{2\left(-1\right)}
Square 4.
u=\frac{-4±\sqrt{16+4\left(-4\right)}}{2\left(-1\right)}
Multiply -4 times -1.
u=\frac{-4±\sqrt{16-16}}{2\left(-1\right)}
Multiply 4 times -4.
u=\frac{-4±\sqrt{0}}{2\left(-1\right)}
Add 16 to -16.
u=-\frac{4}{2\left(-1\right)}
Take the square root of 0.
u=-\frac{4}{-2}
Multiply 2 times -1.
u=2
Divide -4 by -2.
-u^{2}+4u-4=0
Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.
-u^{2}+4u-4-\left(-4\right)=-\left(-4\right)
Add 4 to both sides of the equation.
-u^{2}+4u=-\left(-4\right)
Subtracting -4 from itself leaves 0.
-u^{2}+4u=4
Subtract -4 from 0.
\frac{-u^{2}+4u}{-1}=\frac{4}{-1}
Divide both sides by -1.
u^{2}+\frac{4}{-1}u=\frac{4}{-1}
Dividing by -1 undoes the multiplication by -1.
u^{2}-4u=\frac{4}{-1}
Divide 4 by -1.
u^{2}-4u=-4
Divide 4 by -1.
u^{2}-4u+\left(-2\right)^{2}=-4+\left(-2\right)^{2}
Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
u^{2}-4u+4=-4+4
Square -2.
u^{2}-4u+4=0
Add -4 to 4.
\left(u-2\right)^{2}=0
Factor u^{2}-4u+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(u-2\right)^{2}}=\sqrt{0}
Take the square root of both sides of the equation.
u-2=0 u-2=0
Simplify.
u=2 u=2
Add 2 to both sides of the equation.
u=2
The equation is now solved. Solutions are the same.
x ^ 2 -4x +4 = 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.
r + s = 4 rs = 4
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 = 2 - u s = 2 + u
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>
(2 - u) (2 + u) = 4
To solve for unknown quantity u, substitute these in the product equation rs = 4
4 - u^2 = 4
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 4-4 = 0
Simplify the expression by subtracting 4 on both sides
u^2 = 0 u = 0
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r = s = 2
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
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}