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4x^{2}-12x+9=0
Divide both sides by 3.
a+b=-12 ab=4\times 9=36
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx+9. To find a and b, set up a system to be solved.
-1,-36 -2,-18 -3,-12 -4,-9 -6,-6
Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 36.
-1-36=-37 -2-18=-20 -3-12=-15 -4-9=-13 -6-6=-12
Calculate the sum for each pair.
a=-6 b=-6
The solution is the pair that gives sum -12.
\left(4x^{2}-6x\right)+\left(-6x+9\right)
Rewrite 4x^{2}-12x+9 as \left(4x^{2}-6x\right)+\left(-6x+9\right).
2x\left(2x-3\right)-3\left(2x-3\right)
Factor out 2x in the first and -3 in the second group.
\left(2x-3\right)\left(2x-3\right)
Factor out common term 2x-3 by using distributive property.
\left(2x-3\right)^{2}
Rewrite as a binomial square.
x=\frac{3}{2}
To find equation solution, solve 2x-3=0.
12x^{2}-36x+27=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.
x=\frac{-\left(-36\right)±\sqrt{\left(-36\right)^{2}-4\times 12\times 27}}{2\times 12}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 12 for a, -36 for b, and 27 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-36\right)±\sqrt{1296-4\times 12\times 27}}{2\times 12}
Square -36.
x=\frac{-\left(-36\right)±\sqrt{1296-48\times 27}}{2\times 12}
Multiply -4 times 12.
x=\frac{-\left(-36\right)±\sqrt{1296-1296}}{2\times 12}
Multiply -48 times 27.
x=\frac{-\left(-36\right)±\sqrt{0}}{2\times 12}
Add 1296 to -1296.
x=-\frac{-36}{2\times 12}
Take the square root of 0.
x=\frac{36}{2\times 12}
The opposite of -36 is 36.
x=\frac{36}{24}
Multiply 2 times 12.
x=\frac{3}{2}
Reduce the fraction \frac{36}{24} to lowest terms by extracting and canceling out 12.
12x^{2}-36x+27=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.
12x^{2}-36x+27-27=-27
Subtract 27 from both sides of the equation.
12x^{2}-36x=-27
Subtracting 27 from itself leaves 0.
\frac{12x^{2}-36x}{12}=-\frac{27}{12}
Divide both sides by 12.
x^{2}+\left(-\frac{36}{12}\right)x=-\frac{27}{12}
Dividing by 12 undoes the multiplication by 12.
x^{2}-3x=-\frac{27}{12}
Divide -36 by 12.
x^{2}-3x=-\frac{9}{4}
Reduce the fraction \frac{-27}{12} to lowest terms by extracting and canceling out 3.
x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=-\frac{9}{4}+\left(-\frac{3}{2}\right)^{2}
Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-3x+\frac{9}{4}=\frac{-9+9}{4}
Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}-3x+\frac{9}{4}=0
Add -\frac{9}{4} to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{3}{2}\right)^{2}=0
Factor x^{2}-3x+\frac{9}{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(x-\frac{3}{2}\right)^{2}}=\sqrt{0}
Take the square root of both sides of the equation.
x-\frac{3}{2}=0 x-\frac{3}{2}=0
Simplify.
x=\frac{3}{2} x=\frac{3}{2}
Add \frac{3}{2} to both sides of the equation.
x=\frac{3}{2}
The equation is now solved. Solutions are the same.
x ^ 2 -3x +\frac{9}{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.This is achieved by dividing both sides of the equation by 12
r + s = 3 rs = \frac{9}{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 = \frac{3}{2} - u s = \frac{3}{2} + u
Two numbers r and s sum up to 3 exactly when the average of the two numbers is \frac{1}{2}*3 = \frac{3}{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>
(\frac{3}{2} - u) (\frac{3}{2} + u) = \frac{9}{4}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{9}{4}
\frac{9}{4} - u^2 = \frac{9}{4}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{9}{4}-\frac{9}{4} = 0
Simplify the expression by subtracting \frac{9}{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 = \frac{3}{2} = 1.500
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.