Skip to main content
Solve for x
Tick mark Image
Graph

Similar Problems from Web Search

Share

a+b=-5 ab=12\left(-3\right)=-36
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 12x^{2}+ax+bx-3. 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 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 -36.
1-36=-35 2-18=-16 3-12=-9 4-9=-5 6-6=0
Calculate the sum for each pair.
a=-9 b=4
The solution is the pair that gives sum -5.
\left(12x^{2}-9x\right)+\left(4x-3\right)
Rewrite 12x^{2}-5x-3 as \left(12x^{2}-9x\right)+\left(4x-3\right).
3x\left(4x-3\right)+4x-3
Factor out 3x in 12x^{2}-9x.
\left(4x-3\right)\left(3x+1\right)
Factor out common term 4x-3 by using distributive property.
x=\frac{3}{4} x=-\frac{1}{3}
To find equation solutions, solve 4x-3=0 and 3x+1=0.
12x^{2}-5x-3=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(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 12\left(-3\right)}}{2\times 12}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 12 for a, -5 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-5\right)±\sqrt{25-4\times 12\left(-3\right)}}{2\times 12}
Square -5.
x=\frac{-\left(-5\right)±\sqrt{25-48\left(-3\right)}}{2\times 12}
Multiply -4 times 12.
x=\frac{-\left(-5\right)±\sqrt{25+144}}{2\times 12}
Multiply -48 times -3.
x=\frac{-\left(-5\right)±\sqrt{169}}{2\times 12}
Add 25 to 144.
x=\frac{-\left(-5\right)±13}{2\times 12}
Take the square root of 169.
x=\frac{5±13}{2\times 12}
The opposite of -5 is 5.
x=\frac{5±13}{24}
Multiply 2 times 12.
x=\frac{18}{24}
Now solve the equation x=\frac{5±13}{24} when ± is plus. Add 5 to 13.
x=\frac{3}{4}
Reduce the fraction \frac{18}{24} to lowest terms by extracting and canceling out 6.
x=-\frac{8}{24}
Now solve the equation x=\frac{5±13}{24} when ± is minus. Subtract 13 from 5.
x=-\frac{1}{3}
Reduce the fraction \frac{-8}{24} to lowest terms by extracting and canceling out 8.
x=\frac{3}{4} x=-\frac{1}{3}
The equation is now solved.
12x^{2}-5x-3=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}-5x-3-\left(-3\right)=-\left(-3\right)
Add 3 to both sides of the equation.
12x^{2}-5x=-\left(-3\right)
Subtracting -3 from itself leaves 0.
12x^{2}-5x=3
Subtract -3 from 0.
\frac{12x^{2}-5x}{12}=\frac{3}{12}
Divide both sides by 12.
x^{2}-\frac{5}{12}x=\frac{3}{12}
Dividing by 12 undoes the multiplication by 12.
x^{2}-\frac{5}{12}x=\frac{1}{4}
Reduce the fraction \frac{3}{12} to lowest terms by extracting and canceling out 3.
x^{2}-\frac{5}{12}x+\left(-\frac{5}{24}\right)^{2}=\frac{1}{4}+\left(-\frac{5}{24}\right)^{2}
Divide -\frac{5}{12}, the coefficient of the x term, by 2 to get -\frac{5}{24}. Then add the square of -\frac{5}{24} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{5}{12}x+\frac{25}{576}=\frac{1}{4}+\frac{25}{576}
Square -\frac{5}{24} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{5}{12}x+\frac{25}{576}=\frac{169}{576}
Add \frac{1}{4} to \frac{25}{576} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{5}{24}\right)^{2}=\frac{169}{576}
Factor x^{2}-\frac{5}{12}x+\frac{25}{576}. 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{5}{24}\right)^{2}}=\sqrt{\frac{169}{576}}
Take the square root of both sides of the equation.
x-\frac{5}{24}=\frac{13}{24} x-\frac{5}{24}=-\frac{13}{24}
Simplify.
x=\frac{3}{4} x=-\frac{1}{3}
Add \frac{5}{24} to both sides of the equation.
x ^ 2 -\frac{5}{12}x -\frac{1}{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 = \frac{5}{12} rs = -\frac{1}{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{5}{24} - u s = \frac{5}{24} + u
Two numbers r and s sum up to \frac{5}{12} exactly when the average of the two numbers is \frac{1}{2}*\frac{5}{12} = \frac{5}{24}. 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{5}{24} - u) (\frac{5}{24} + u) = -\frac{1}{4}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{1}{4}
\frac{25}{576} - u^2 = -\frac{1}{4}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{1}{4}-\frac{25}{576} = -\frac{169}{576}
Simplify the expression by subtracting \frac{25}{576} on both sides
u^2 = \frac{169}{576} u = \pm\sqrt{\frac{169}{576}} = \pm \frac{13}{24}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{5}{24} - \frac{13}{24} = -0.333 s = \frac{5}{24} + \frac{13}{24} = 0.750
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.