Solve for x
x=-\frac{7}{9}\approx -0.777777778
x = \frac{5}{4} = 1\frac{1}{4} = 1.25
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a+b=-17 ab=36\left(-35\right)=-1260
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 36x^{2}+ax+bx-35. To find a and b, set up a system to be solved.
1,-1260 2,-630 3,-420 4,-315 5,-252 6,-210 7,-180 9,-140 10,-126 12,-105 14,-90 15,-84 18,-70 20,-63 21,-60 28,-45 30,-42 35,-36
Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -1260.
1-1260=-1259 2-630=-628 3-420=-417 4-315=-311 5-252=-247 6-210=-204 7-180=-173 9-140=-131 10-126=-116 12-105=-93 14-90=-76 15-84=-69 18-70=-52 20-63=-43 21-60=-39 28-45=-17 30-42=-12 35-36=-1
Calculate the sum for each pair.
a=-45 b=28
The solution is the pair that gives sum -17.
\left(36x^{2}-45x\right)+\left(28x-35\right)
Rewrite 36x^{2}-17x-35 as \left(36x^{2}-45x\right)+\left(28x-35\right).
9x\left(4x-5\right)+7\left(4x-5\right)
Factor out 9x in the first and 7 in the second group.
\left(4x-5\right)\left(9x+7\right)
Factor out common term 4x-5 by using distributive property.
x=\frac{5}{4} x=-\frac{7}{9}
To find equation solutions, solve 4x-5=0 and 9x+7=0.
36x^{2}-17x-35=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(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 36\left(-35\right)}}{2\times 36}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 36 for a, -17 for b, and -35 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-17\right)±\sqrt{289-4\times 36\left(-35\right)}}{2\times 36}
Square -17.
x=\frac{-\left(-17\right)±\sqrt{289-144\left(-35\right)}}{2\times 36}
Multiply -4 times 36.
x=\frac{-\left(-17\right)±\sqrt{289+5040}}{2\times 36}
Multiply -144 times -35.
x=\frac{-\left(-17\right)±\sqrt{5329}}{2\times 36}
Add 289 to 5040.
x=\frac{-\left(-17\right)±73}{2\times 36}
Take the square root of 5329.
x=\frac{17±73}{2\times 36}
The opposite of -17 is 17.
x=\frac{17±73}{72}
Multiply 2 times 36.
x=\frac{90}{72}
Now solve the equation x=\frac{17±73}{72} when ± is plus. Add 17 to 73.
x=\frac{5}{4}
Reduce the fraction \frac{90}{72} to lowest terms by extracting and canceling out 18.
x=-\frac{56}{72}
Now solve the equation x=\frac{17±73}{72} when ± is minus. Subtract 73 from 17.
x=-\frac{7}{9}
Reduce the fraction \frac{-56}{72} to lowest terms by extracting and canceling out 8.
x=\frac{5}{4} x=-\frac{7}{9}
The equation is now solved.
36x^{2}-17x-35=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.
36x^{2}-17x-35-\left(-35\right)=-\left(-35\right)
Add 35 to both sides of the equation.
36x^{2}-17x=-\left(-35\right)
Subtracting -35 from itself leaves 0.
36x^{2}-17x=35
Subtract -35 from 0.
\frac{36x^{2}-17x}{36}=\frac{35}{36}
Divide both sides by 36.
x^{2}-\frac{17}{36}x=\frac{35}{36}
Dividing by 36 undoes the multiplication by 36.
x^{2}-\frac{17}{36}x+\left(-\frac{17}{72}\right)^{2}=\frac{35}{36}+\left(-\frac{17}{72}\right)^{2}
Divide -\frac{17}{36}, the coefficient of the x term, by 2 to get -\frac{17}{72}. Then add the square of -\frac{17}{72} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{17}{36}x+\frac{289}{5184}=\frac{35}{36}+\frac{289}{5184}
Square -\frac{17}{72} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{17}{36}x+\frac{289}{5184}=\frac{5329}{5184}
Add \frac{35}{36} to \frac{289}{5184} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{17}{72}\right)^{2}=\frac{5329}{5184}
Factor x^{2}-\frac{17}{36}x+\frac{289}{5184}. 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{17}{72}\right)^{2}}=\sqrt{\frac{5329}{5184}}
Take the square root of both sides of the equation.
x-\frac{17}{72}=\frac{73}{72} x-\frac{17}{72}=-\frac{73}{72}
Simplify.
x=\frac{5}{4} x=-\frac{7}{9}
Add \frac{17}{72} to both sides of the equation.
x ^ 2 -\frac{17}{36}x -\frac{35}{36} = 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 36
r + s = \frac{17}{36} rs = -\frac{35}{36}
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{17}{72} - u s = \frac{17}{72} + u
Two numbers r and s sum up to \frac{17}{36} exactly when the average of the two numbers is \frac{1}{2}*\frac{17}{36} = \frac{17}{72}. 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{17}{72} - u) (\frac{17}{72} + u) = -\frac{35}{36}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{35}{36}
\frac{289}{5184} - u^2 = -\frac{35}{36}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{35}{36}-\frac{289}{5184} = -\frac{5329}{5184}
Simplify the expression by subtracting \frac{289}{5184} on both sides
u^2 = \frac{5329}{5184} u = \pm\sqrt{\frac{5329}{5184}} = \pm \frac{73}{72}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{17}{72} - \frac{73}{72} = -0.778 s = \frac{17}{72} + \frac{73}{72} = 1.250
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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