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a+b=22 ab=48\left(-5\right)=-240
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 48x^{2}+ax+bx-5. To find a and b, set up a system to be solved.
-1,240 -2,120 -3,80 -4,60 -5,48 -6,40 -8,30 -10,24 -12,20 -15,16
Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -240.
-1+240=239 -2+120=118 -3+80=77 -4+60=56 -5+48=43 -6+40=34 -8+30=22 -10+24=14 -12+20=8 -15+16=1
Calculate the sum for each pair.
a=-8 b=30
The solution is the pair that gives sum 22.
\left(48x^{2}-8x\right)+\left(30x-5\right)
Rewrite 48x^{2}+22x-5 as \left(48x^{2}-8x\right)+\left(30x-5\right).
8x\left(6x-1\right)+5\left(6x-1\right)
Factor out 8x in the first and 5 in the second group.
\left(6x-1\right)\left(8x+5\right)
Factor out common term 6x-1 by using distributive property.
x=\frac{1}{6} x=-\frac{5}{8}
To find equation solutions, solve 6x-1=0 and 8x+5=0.
48x^{2}+22x-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{-22±\sqrt{22^{2}-4\times 48\left(-5\right)}}{2\times 48}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 48 for a, 22 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-22±\sqrt{484-4\times 48\left(-5\right)}}{2\times 48}
Square 22.
x=\frac{-22±\sqrt{484-192\left(-5\right)}}{2\times 48}
Multiply -4 times 48.
x=\frac{-22±\sqrt{484+960}}{2\times 48}
Multiply -192 times -5.
x=\frac{-22±\sqrt{1444}}{2\times 48}
Add 484 to 960.
x=\frac{-22±38}{2\times 48}
Take the square root of 1444.
x=\frac{-22±38}{96}
Multiply 2 times 48.
x=\frac{16}{96}
Now solve the equation x=\frac{-22±38}{96} when ± is plus. Add -22 to 38.
x=\frac{1}{6}
Reduce the fraction \frac{16}{96} to lowest terms by extracting and canceling out 16.
x=-\frac{60}{96}
Now solve the equation x=\frac{-22±38}{96} when ± is minus. Subtract 38 from -22.
x=-\frac{5}{8}
Reduce the fraction \frac{-60}{96} to lowest terms by extracting and canceling out 12.
x=\frac{1}{6} x=-\frac{5}{8}
The equation is now solved.
48x^{2}+22x-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.
48x^{2}+22x-5-\left(-5\right)=-\left(-5\right)
Add 5 to both sides of the equation.
48x^{2}+22x=-\left(-5\right)
Subtracting -5 from itself leaves 0.
48x^{2}+22x=5
Subtract -5 from 0.
\frac{48x^{2}+22x}{48}=\frac{5}{48}
Divide both sides by 48.
x^{2}+\frac{22}{48}x=\frac{5}{48}
Dividing by 48 undoes the multiplication by 48.
x^{2}+\frac{11}{24}x=\frac{5}{48}
Reduce the fraction \frac{22}{48} to lowest terms by extracting and canceling out 2.
x^{2}+\frac{11}{24}x+\left(\frac{11}{48}\right)^{2}=\frac{5}{48}+\left(\frac{11}{48}\right)^{2}
Divide \frac{11}{24}, the coefficient of the x term, by 2 to get \frac{11}{48}. Then add the square of \frac{11}{48} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{11}{24}x+\frac{121}{2304}=\frac{5}{48}+\frac{121}{2304}
Square \frac{11}{48} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{11}{24}x+\frac{121}{2304}=\frac{361}{2304}
Add \frac{5}{48} to \frac{121}{2304} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{11}{48}\right)^{2}=\frac{361}{2304}
Factor x^{2}+\frac{11}{24}x+\frac{121}{2304}. 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}{48}\right)^{2}}=\sqrt{\frac{361}{2304}}
Take the square root of both sides of the equation.
x+\frac{11}{48}=\frac{19}{48} x+\frac{11}{48}=-\frac{19}{48}
Simplify.
x=\frac{1}{6} x=-\frac{5}{8}
Subtract \frac{11}{48} from both sides of the equation.
x ^ 2 +\frac{11}{24}x -\frac{5}{48} = 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 48
r + s = -\frac{11}{24} rs = -\frac{5}{48}
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}{48} - u s = -\frac{11}{48} + u
Two numbers r and s sum up to -\frac{11}{24} exactly when the average of the two numbers is \frac{1}{2}*-\frac{11}{24} = -\frac{11}{48}. 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-gzdabgg4ehffg0hf.b01.azurefd.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>
(-\frac{11}{48} - u) (-\frac{11}{48} + u) = -\frac{5}{48}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{5}{48}
\frac{121}{2304} - u^2 = -\frac{5}{48}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{5}{48}-\frac{121}{2304} = -\frac{361}{2304}
Simplify the expression by subtracting \frac{121}{2304} on both sides
u^2 = \frac{361}{2304} u = \pm\sqrt{\frac{361}{2304}} = \pm \frac{19}{48}
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}{48} - \frac{19}{48} = -0.625 s = -\frac{11}{48} + \frac{19}{48} = 0.167
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.