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8b^{2}-22b+5=0
Divide both sides by 2.
a+b=-22 ab=8\times 5=40
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 8b^{2}+ab+bb+5. To find a and b, set up a system to be solved.
-1,-40 -2,-20 -4,-10 -5,-8
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 40.
-1-40=-41 -2-20=-22 -4-10=-14 -5-8=-13
Calculate the sum for each pair.
a=-20 b=-2
The solution is the pair that gives sum -22.
\left(8b^{2}-20b\right)+\left(-2b+5\right)
Rewrite 8b^{2}-22b+5 as \left(8b^{2}-20b\right)+\left(-2b+5\right).
4b\left(2b-5\right)-\left(2b-5\right)
Factor out 4b in the first and -1 in the second group.
\left(2b-5\right)\left(4b-1\right)
Factor out common term 2b-5 by using distributive property.
b=\frac{5}{2} b=\frac{1}{4}
To find equation solutions, solve 2b-5=0 and 4b-1=0.
16b^{2}-44b+10=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.
b=\frac{-\left(-44\right)±\sqrt{\left(-44\right)^{2}-4\times 16\times 10}}{2\times 16}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, -44 for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
b=\frac{-\left(-44\right)±\sqrt{1936-4\times 16\times 10}}{2\times 16}
Square -44.
b=\frac{-\left(-44\right)±\sqrt{1936-64\times 10}}{2\times 16}
Multiply -4 times 16.
b=\frac{-\left(-44\right)±\sqrt{1936-640}}{2\times 16}
Multiply -64 times 10.
b=\frac{-\left(-44\right)±\sqrt{1296}}{2\times 16}
Add 1936 to -640.
b=\frac{-\left(-44\right)±36}{2\times 16}
Take the square root of 1296.
b=\frac{44±36}{2\times 16}
The opposite of -44 is 44.
b=\frac{44±36}{32}
Multiply 2 times 16.
b=\frac{80}{32}
Now solve the equation b=\frac{44±36}{32} when ± is plus. Add 44 to 36.
b=\frac{5}{2}
Reduce the fraction \frac{80}{32} to lowest terms by extracting and canceling out 16.
b=\frac{8}{32}
Now solve the equation b=\frac{44±36}{32} when ± is minus. Subtract 36 from 44.
b=\frac{1}{4}
Reduce the fraction \frac{8}{32} to lowest terms by extracting and canceling out 8.
b=\frac{5}{2} b=\frac{1}{4}
The equation is now solved.
16b^{2}-44b+10=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.
16b^{2}-44b+10-10=-10
Subtract 10 from both sides of the equation.
16b^{2}-44b=-10
Subtracting 10 from itself leaves 0.
\frac{16b^{2}-44b}{16}=-\frac{10}{16}
Divide both sides by 16.
b^{2}+\left(-\frac{44}{16}\right)b=-\frac{10}{16}
Dividing by 16 undoes the multiplication by 16.
b^{2}-\frac{11}{4}b=-\frac{10}{16}
Reduce the fraction \frac{-44}{16} to lowest terms by extracting and canceling out 4.
b^{2}-\frac{11}{4}b=-\frac{5}{8}
Reduce the fraction \frac{-10}{16} to lowest terms by extracting and canceling out 2.
b^{2}-\frac{11}{4}b+\left(-\frac{11}{8}\right)^{2}=-\frac{5}{8}+\left(-\frac{11}{8}\right)^{2}
Divide -\frac{11}{4}, the coefficient of the x term, by 2 to get -\frac{11}{8}. Then add the square of -\frac{11}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
b^{2}-\frac{11}{4}b+\frac{121}{64}=-\frac{5}{8}+\frac{121}{64}
Square -\frac{11}{8} by squaring both the numerator and the denominator of the fraction.
b^{2}-\frac{11}{4}b+\frac{121}{64}=\frac{81}{64}
Add -\frac{5}{8} to \frac{121}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(b-\frac{11}{8}\right)^{2}=\frac{81}{64}
Factor b^{2}-\frac{11}{4}b+\frac{121}{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(b-\frac{11}{8}\right)^{2}}=\sqrt{\frac{81}{64}}
Take the square root of both sides of the equation.
b-\frac{11}{8}=\frac{9}{8} b-\frac{11}{8}=-\frac{9}{8}
Simplify.
b=\frac{5}{2} b=\frac{1}{4}
Add \frac{11}{8} to both sides of the equation.
x ^ 2 -\frac{11}{4}x +\frac{5}{8} = 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 16
r + s = \frac{11}{4} rs = \frac{5}{8}
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}{8} - u s = \frac{11}{8} + u
Two numbers r and s sum up to \frac{11}{4} exactly when the average of the two numbers is \frac{1}{2}*\frac{11}{4} = \frac{11}{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{11}{8} - u) (\frac{11}{8} + u) = \frac{5}{8}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{5}{8}
\frac{121}{64} - u^2 = \frac{5}{8}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{5}{8}-\frac{121}{64} = -\frac{81}{64}
Simplify the expression by subtracting \frac{121}{64} on both sides
u^2 = \frac{81}{64} u = \pm\sqrt{\frac{81}{64}} = \pm \frac{9}{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{11}{8} - \frac{9}{8} = 0.250 s = \frac{11}{8} + \frac{9}{8} = 2.500
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.