Skip to main content
Solve for x
Tick mark Image
Graph

Similar Problems from Web Search

Share

a+b=-1 ab=-4\times 3=-12
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -4x^{2}+ax+bx+3. To find a and b, set up a system to be solved.
1,-12 2,-6 3,-4
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 -12.
1-12=-11 2-6=-4 3-4=-1
Calculate the sum for each pair.
a=3 b=-4
The solution is the pair that gives sum -1.
\left(-4x^{2}+3x\right)+\left(-4x+3\right)
Rewrite -4x^{2}-x+3 as \left(-4x^{2}+3x\right)+\left(-4x+3\right).
-x\left(4x-3\right)-\left(4x-3\right)
Factor out -x in the first and -1 in the second group.
\left(4x-3\right)\left(-x-1\right)
Factor out common term 4x-3 by using distributive property.
x=\frac{3}{4} x=-1
To find equation solutions, solve 4x-3=0 and -x-1=0.
-4x^{2}-x+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(-1\right)±\sqrt{1-4\left(-4\right)\times 3}}{2\left(-4\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, -1 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-1\right)±\sqrt{1+16\times 3}}{2\left(-4\right)}
Multiply -4 times -4.
x=\frac{-\left(-1\right)±\sqrt{1+48}}{2\left(-4\right)}
Multiply 16 times 3.
x=\frac{-\left(-1\right)±\sqrt{49}}{2\left(-4\right)}
Add 1 to 48.
x=\frac{-\left(-1\right)±7}{2\left(-4\right)}
Take the square root of 49.
x=\frac{1±7}{2\left(-4\right)}
The opposite of -1 is 1.
x=\frac{1±7}{-8}
Multiply 2 times -4.
x=\frac{8}{-8}
Now solve the equation x=\frac{1±7}{-8} when ± is plus. Add 1 to 7.
x=-1
Divide 8 by -8.
x=-\frac{6}{-8}
Now solve the equation x=\frac{1±7}{-8} when ± is minus. Subtract 7 from 1.
x=\frac{3}{4}
Reduce the fraction \frac{-6}{-8} to lowest terms by extracting and canceling out 2.
x=-1 x=\frac{3}{4}
The equation is now solved.
-4x^{2}-x+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.
-4x^{2}-x+3-3=-3
Subtract 3 from both sides of the equation.
-4x^{2}-x=-3
Subtracting 3 from itself leaves 0.
\frac{-4x^{2}-x}{-4}=-\frac{3}{-4}
Divide both sides by -4.
x^{2}+\left(-\frac{1}{-4}\right)x=-\frac{3}{-4}
Dividing by -4 undoes the multiplication by -4.
x^{2}+\frac{1}{4}x=-\frac{3}{-4}
Divide -1 by -4.
x^{2}+\frac{1}{4}x=\frac{3}{4}
Divide -3 by -4.
x^{2}+\frac{1}{4}x+\left(\frac{1}{8}\right)^{2}=\frac{3}{4}+\left(\frac{1}{8}\right)^{2}
Divide \frac{1}{4}, the coefficient of the x term, by 2 to get \frac{1}{8}. Then add the square of \frac{1}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{1}{4}x+\frac{1}{64}=\frac{3}{4}+\frac{1}{64}
Square \frac{1}{8} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{1}{4}x+\frac{1}{64}=\frac{49}{64}
Add \frac{3}{4} to \frac{1}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{1}{8}\right)^{2}=\frac{49}{64}
Factor x^{2}+\frac{1}{4}x+\frac{1}{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{1}{8}\right)^{2}}=\sqrt{\frac{49}{64}}
Take the square root of both sides of the equation.
x+\frac{1}{8}=\frac{7}{8} x+\frac{1}{8}=-\frac{7}{8}
Simplify.
x=\frac{3}{4} x=-1
Subtract \frac{1}{8} from both sides of the equation.
x ^ 2 +\frac{1}{4}x -\frac{3}{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 = -\frac{1}{4} rs = -\frac{3}{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{1}{8} - u s = -\frac{1}{8} + u
Two numbers r and s sum up to -\frac{1}{4} exactly when the average of the two numbers is \frac{1}{2}*-\frac{1}{4} = -\frac{1}{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{1}{8} - u) (-\frac{1}{8} + u) = -\frac{3}{4}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{4}
\frac{1}{64} - u^2 = -\frac{3}{4}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{3}{4}-\frac{1}{64} = -\frac{49}{64}
Simplify the expression by subtracting \frac{1}{64} on both sides
u^2 = \frac{49}{64} u = \pm\sqrt{\frac{49}{64}} = \pm \frac{7}{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{1}{8} - \frac{7}{8} = -1 s = -\frac{1}{8} + \frac{7}{8} = 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.