Solve for x
x=\frac{3}{4}=0.75
x=12
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a+b=-51 ab=4\times 36=144
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx+36. To find a and b, set up a system to be solved.
-1,-144 -2,-72 -3,-48 -4,-36 -6,-24 -8,-18 -9,-16 -12,-12
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 144.
-1-144=-145 -2-72=-74 -3-48=-51 -4-36=-40 -6-24=-30 -8-18=-26 -9-16=-25 -12-12=-24
Calculate the sum for each pair.
a=-48 b=-3
The solution is the pair that gives sum -51.
\left(4x^{2}-48x\right)+\left(-3x+36\right)
Rewrite 4x^{2}-51x+36 as \left(4x^{2}-48x\right)+\left(-3x+36\right).
4x\left(x-12\right)-3\left(x-12\right)
Factor out 4x in the first and -3 in the second group.
\left(x-12\right)\left(4x-3\right)
Factor out common term x-12 by using distributive property.
x=12 x=\frac{3}{4}
To find equation solutions, solve x-12=0 and 4x-3=0.
4x^{2}-51x+36=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(-51\right)±\sqrt{\left(-51\right)^{2}-4\times 4\times 36}}{2\times 4}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -51 for b, and 36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-51\right)±\sqrt{2601-4\times 4\times 36}}{2\times 4}
Square -51.
x=\frac{-\left(-51\right)±\sqrt{2601-16\times 36}}{2\times 4}
Multiply -4 times 4.
x=\frac{-\left(-51\right)±\sqrt{2601-576}}{2\times 4}
Multiply -16 times 36.
x=\frac{-\left(-51\right)±\sqrt{2025}}{2\times 4}
Add 2601 to -576.
x=\frac{-\left(-51\right)±45}{2\times 4}
Take the square root of 2025.
x=\frac{51±45}{2\times 4}
The opposite of -51 is 51.
x=\frac{51±45}{8}
Multiply 2 times 4.
x=\frac{96}{8}
Now solve the equation x=\frac{51±45}{8} when ± is plus. Add 51 to 45.
x=12
Divide 96 by 8.
x=\frac{6}{8}
Now solve the equation x=\frac{51±45}{8} when ± is minus. Subtract 45 from 51.
x=\frac{3}{4}
Reduce the fraction \frac{6}{8} to lowest terms by extracting and canceling out 2.
x=12 x=\frac{3}{4}
The equation is now solved.
4x^{2}-51x+36=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.
4x^{2}-51x+36-36=-36
Subtract 36 from both sides of the equation.
4x^{2}-51x=-36
Subtracting 36 from itself leaves 0.
\frac{4x^{2}-51x}{4}=-\frac{36}{4}
Divide both sides by 4.
x^{2}-\frac{51}{4}x=-\frac{36}{4}
Dividing by 4 undoes the multiplication by 4.
x^{2}-\frac{51}{4}x=-9
Divide -36 by 4.
x^{2}-\frac{51}{4}x+\left(-\frac{51}{8}\right)^{2}=-9+\left(-\frac{51}{8}\right)^{2}
Divide -\frac{51}{4}, the coefficient of the x term, by 2 to get -\frac{51}{8}. Then add the square of -\frac{51}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{51}{4}x+\frac{2601}{64}=-9+\frac{2601}{64}
Square -\frac{51}{8} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{51}{4}x+\frac{2601}{64}=\frac{2025}{64}
Add -9 to \frac{2601}{64}.
\left(x-\frac{51}{8}\right)^{2}=\frac{2025}{64}
Factor x^{2}-\frac{51}{4}x+\frac{2601}{64}. 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{51}{8}\right)^{2}}=\sqrt{\frac{2025}{64}}
Take the square root of both sides of the equation.
x-\frac{51}{8}=\frac{45}{8} x-\frac{51}{8}=-\frac{45}{8}
Simplify.
x=12 x=\frac{3}{4}
Add \frac{51}{8} to both sides of the equation.
x ^ 2 -\frac{51}{4}x +9 = 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 4
r + s = \frac{51}{4} rs = 9
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{51}{8} - u s = \frac{51}{8} + u
Two numbers r and s sum up to \frac{51}{4} exactly when the average of the two numbers is \frac{1}{2}*\frac{51}{4} = \frac{51}{8}. 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{51}{8} - u) (\frac{51}{8} + u) = 9
To solve for unknown quantity u, substitute these in the product equation rs = 9
\frac{2601}{64} - u^2 = 9
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 9-\frac{2601}{64} = -\frac{2025}{64}
Simplify the expression by subtracting \frac{2601}{64} on both sides
u^2 = \frac{2025}{64} u = \pm\sqrt{\frac{2025}{64}} = \pm \frac{45}{8}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{51}{8} - \frac{45}{8} = 0.750 s = \frac{51}{8} + \frac{45}{8} = 12
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
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