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t^{2}-4t+2=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.
t=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 2}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -4 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
t=\frac{-\left(-4\right)±\sqrt{16-4\times 2}}{2}
Square -4.
t=\frac{-\left(-4\right)±\sqrt{16-8}}{2}
Multiply -4 times 2.
t=\frac{-\left(-4\right)±\sqrt{8}}{2}
Add 16 to -8.
t=\frac{-\left(-4\right)±2\sqrt{2}}{2}
Take the square root of 8.
t=\frac{4±2\sqrt{2}}{2}
The opposite of -4 is 4.
t=\frac{2\sqrt{2}+4}{2}
Now solve the equation t=\frac{4±2\sqrt{2}}{2} when ± is plus. Add 4 to 2\sqrt{2}.
t=\sqrt{2}+2
Divide 4+2\sqrt{2} by 2.
t=\frac{4-2\sqrt{2}}{2}
Now solve the equation t=\frac{4±2\sqrt{2}}{2} when ± is minus. Subtract 2\sqrt{2} from 4.
t=2-\sqrt{2}
Divide 4-2\sqrt{2} by 2.
t=\sqrt{2}+2 t=2-\sqrt{2}
The equation is now solved.
t^{2}-4t+2=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.
t^{2}-4t+2-2=-2
Subtract 2 from both sides of the equation.
t^{2}-4t=-2
Subtracting 2 from itself leaves 0.
t^{2}-4t+\left(-2\right)^{2}=-2+\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.
t^{2}-4t+4=-2+4
Square -2.
t^{2}-4t+4=2
Add -2 to 4.
\left(t-2\right)^{2}=2
Factor t^{2}-4t+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(t-2\right)^{2}}=\sqrt{2}
Take the square root of both sides of the equation.
t-2=\sqrt{2} t-2=-\sqrt{2}
Simplify.
t=\sqrt{2}+2 t=2-\sqrt{2}
Add 2 to both sides of the equation.
x ^ 2 -4x +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.
r + s = 4 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 = 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) = 2
To solve for unknown quantity u, substitute these in the product equation rs = 2
4 - u^2 = 2
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 2-4 = -2
Simplify the expression by subtracting 4 on both sides
u^2 = 2 u = \pm\sqrt{2} = \pm \sqrt{2}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =2 - \sqrt{2} = 0.586 s = 2 + \sqrt{2} = 3.414
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.