Skip to main content
Solve for x
Tick mark Image
Graph

Similar Problems from Web Search

Share

a+b=-11 ab=-2\left(-5\right)=10
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -2x^{2}+ax+bx-5. To find a and b, set up a system to be solved.
-1,-10 -2,-5
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 10.
-1-10=-11 -2-5=-7
Calculate the sum for each pair.
a=-1 b=-10
The solution is the pair that gives sum -11.
\left(-2x^{2}-x\right)+\left(-10x-5\right)
Rewrite -2x^{2}-11x-5 as \left(-2x^{2}-x\right)+\left(-10x-5\right).
-x\left(2x+1\right)-5\left(2x+1\right)
Factor out -x in the first and -5 in the second group.
\left(2x+1\right)\left(-x-5\right)
Factor out common term 2x+1 by using distributive property.
x=-\frac{1}{2} x=-5
To find equation solutions, solve 2x+1=0 and -x-5=0.
-2x^{2}-11x-5=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(-11\right)±\sqrt{\left(-11\right)^{2}-4\left(-2\right)\left(-5\right)}}{2\left(-2\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -11 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-11\right)±\sqrt{121-4\left(-2\right)\left(-5\right)}}{2\left(-2\right)}
Square -11.
x=\frac{-\left(-11\right)±\sqrt{121+8\left(-5\right)}}{2\left(-2\right)}
Multiply -4 times -2.
x=\frac{-\left(-11\right)±\sqrt{121-40}}{2\left(-2\right)}
Multiply 8 times -5.
x=\frac{-\left(-11\right)±\sqrt{81}}{2\left(-2\right)}
Add 121 to -40.
x=\frac{-\left(-11\right)±9}{2\left(-2\right)}
Take the square root of 81.
x=\frac{11±9}{2\left(-2\right)}
The opposite of -11 is 11.
x=\frac{11±9}{-4}
Multiply 2 times -2.
x=\frac{20}{-4}
Now solve the equation x=\frac{11±9}{-4} when ± is plus. Add 11 to 9.
x=-5
Divide 20 by -4.
x=\frac{2}{-4}
Now solve the equation x=\frac{11±9}{-4} when ± is minus. Subtract 9 from 11.
x=-\frac{1}{2}
Reduce the fraction \frac{2}{-4} to lowest terms by extracting and canceling out 2.
x=-5 x=-\frac{1}{2}
The equation is now solved.
-2x^{2}-11x-5=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.
-2x^{2}-11x-5-\left(-5\right)=-\left(-5\right)
Add 5 to both sides of the equation.
-2x^{2}-11x=-\left(-5\right)
Subtracting -5 from itself leaves 0.
-2x^{2}-11x=5
Subtract -5 from 0.
\frac{-2x^{2}-11x}{-2}=\frac{5}{-2}
Divide both sides by -2.
x^{2}+\left(-\frac{11}{-2}\right)x=\frac{5}{-2}
Dividing by -2 undoes the multiplication by -2.
x^{2}+\frac{11}{2}x=\frac{5}{-2}
Divide -11 by -2.
x^{2}+\frac{11}{2}x=-\frac{5}{2}
Divide 5 by -2.
x^{2}+\frac{11}{2}x+\left(\frac{11}{4}\right)^{2}=-\frac{5}{2}+\left(\frac{11}{4}\right)^{2}
Divide \frac{11}{2}, the coefficient of the x term, by 2 to get \frac{11}{4}. Then add the square of \frac{11}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{11}{2}x+\frac{121}{16}=-\frac{5}{2}+\frac{121}{16}
Square \frac{11}{4} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{11}{2}x+\frac{121}{16}=\frac{81}{16}
Add -\frac{5}{2} to \frac{121}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{11}{4}\right)^{2}=\frac{81}{16}
Factor x^{2}+\frac{11}{2}x+\frac{121}{16}. 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{11}{4}\right)^{2}}=\sqrt{\frac{81}{16}}
Take the square root of both sides of the equation.
x+\frac{11}{4}=\frac{9}{4} x+\frac{11}{4}=-\frac{9}{4}
Simplify.
x=-\frac{1}{2} x=-5
Subtract \frac{11}{4} from both sides of the equation.
x ^ 2 +\frac{11}{2}x +\frac{5}{2} = 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{11}{2} rs = \frac{5}{2}
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{11}{4} - u s = -\frac{11}{4} + u
Two numbers r and s sum up to -\frac{11}{2} exactly when the average of the two numbers is \frac{1}{2}*-\frac{11}{2} = -\frac{11}{4}. 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{11}{4} - u) (-\frac{11}{4} + u) = \frac{5}{2}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{5}{2}
\frac{121}{16} - u^2 = \frac{5}{2}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{5}{2}-\frac{121}{16} = -\frac{81}{16}
Simplify the expression by subtracting \frac{121}{16} on both sides
u^2 = \frac{81}{16} u = \pm\sqrt{\frac{81}{16}} = \pm \frac{9}{4}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{11}{4} - \frac{9}{4} = -5 s = -\frac{11}{4} + \frac{9}{4} = -0.500
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.