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y^{2}-6y+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.
y=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 4}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -6 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
y=\frac{-\left(-6\right)±\sqrt{36-4\times 4}}{2}
Square -6.
y=\frac{-\left(-6\right)±\sqrt{36-16}}{2}
Multiply -4 times 4.
y=\frac{-\left(-6\right)±\sqrt{20}}{2}
Add 36 to -16.
y=\frac{-\left(-6\right)±2\sqrt{5}}{2}
Take the square root of 20.
y=\frac{6±2\sqrt{5}}{2}
The opposite of -6 is 6.
y=\frac{2\sqrt{5}+6}{2}
Now solve the equation y=\frac{6±2\sqrt{5}}{2} when ± is plus. Add 6 to 2\sqrt{5}.
y=\sqrt{5}+3
Divide 6+2\sqrt{5} by 2.
y=\frac{6-2\sqrt{5}}{2}
Now solve the equation y=\frac{6±2\sqrt{5}}{2} when ± is minus. Subtract 2\sqrt{5} from 6.
y=3-\sqrt{5}
Divide 6-2\sqrt{5} by 2.
y=\sqrt{5}+3 y=3-\sqrt{5}
The equation is now solved.
y^{2}-6y+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.
y^{2}-6y+4-4=-4
Subtract 4 from both sides of the equation.
y^{2}-6y=-4
Subtracting 4 from itself leaves 0.
y^{2}-6y+\left(-3\right)^{2}=-4+\left(-3\right)^{2}
Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
y^{2}-6y+9=-4+9
Square -3.
y^{2}-6y+9=5
Add -4 to 9.
\left(y-3\right)^{2}=5
Factor y^{2}-6y+9. 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(y-3\right)^{2}}=\sqrt{5}
Take the square root of both sides of the equation.
y-3=\sqrt{5} y-3=-\sqrt{5}
Simplify.
y=\sqrt{5}+3 y=3-\sqrt{5}
Add 3 to both sides of the equation.
x ^ 2 -6x +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 = 6 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 = 3 - u s = 3 + u
Two numbers r and s sum up to 6 exactly when the average of the two numbers is \frac{1}{2}*6 = 3. 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>
(3 - u) (3 + u) = 4
To solve for unknown quantity u, substitute these in the product equation rs = 4
9 - u^2 = 4
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 4-9 = -5
Simplify the expression by subtracting 9 on both sides
u^2 = 5 u = \pm\sqrt{5} = \pm \sqrt{5}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =3 - \sqrt{5} = 0.764 s = 3 + \sqrt{5} = 5.236
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.