Solve for v
v=1
v=\frac{5}{8}=0.625
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a+b=-13 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 8v^{2}+av+bv+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=-8 b=-5
The solution is the pair that gives sum -13.
\left(8v^{2}-8v\right)+\left(-5v+5\right)
Rewrite 8v^{2}-13v+5 as \left(8v^{2}-8v\right)+\left(-5v+5\right).
8v\left(v-1\right)-5\left(v-1\right)
Factor out 8v in the first and -5 in the second group.
\left(v-1\right)\left(8v-5\right)
Factor out common term v-1 by using distributive property.
v=1 v=\frac{5}{8}
To find equation solutions, solve v-1=0 and 8v-5=0.
8v^{2}-13v+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.
v=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 8\times 5}}{2\times 8}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, -13 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
v=\frac{-\left(-13\right)±\sqrt{169-4\times 8\times 5}}{2\times 8}
Square -13.
v=\frac{-\left(-13\right)±\sqrt{169-32\times 5}}{2\times 8}
Multiply -4 times 8.
v=\frac{-\left(-13\right)±\sqrt{169-160}}{2\times 8}
Multiply -32 times 5.
v=\frac{-\left(-13\right)±\sqrt{9}}{2\times 8}
Add 169 to -160.
v=\frac{-\left(-13\right)±3}{2\times 8}
Take the square root of 9.
v=\frac{13±3}{2\times 8}
The opposite of -13 is 13.
v=\frac{13±3}{16}
Multiply 2 times 8.
v=\frac{16}{16}
Now solve the equation v=\frac{13±3}{16} when ± is plus. Add 13 to 3.
v=1
Divide 16 by 16.
v=\frac{10}{16}
Now solve the equation v=\frac{13±3}{16} when ± is minus. Subtract 3 from 13.
v=\frac{5}{8}
Reduce the fraction \frac{10}{16} to lowest terms by extracting and canceling out 2.
v=1 v=\frac{5}{8}
The equation is now solved.
8v^{2}-13v+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.
8v^{2}-13v+5-5=-5
Subtract 5 from both sides of the equation.
8v^{2}-13v=-5
Subtracting 5 from itself leaves 0.
\frac{8v^{2}-13v}{8}=-\frac{5}{8}
Divide both sides by 8.
v^{2}-\frac{13}{8}v=-\frac{5}{8}
Dividing by 8 undoes the multiplication by 8.
v^{2}-\frac{13}{8}v+\left(-\frac{13}{16}\right)^{2}=-\frac{5}{8}+\left(-\frac{13}{16}\right)^{2}
Divide -\frac{13}{8}, the coefficient of the x term, by 2 to get -\frac{13}{16}. Then add the square of -\frac{13}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
v^{2}-\frac{13}{8}v+\frac{169}{256}=-\frac{5}{8}+\frac{169}{256}
Square -\frac{13}{16} by squaring both the numerator and the denominator of the fraction.
v^{2}-\frac{13}{8}v+\frac{169}{256}=\frac{9}{256}
Add -\frac{5}{8} to \frac{169}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(v-\frac{13}{16}\right)^{2}=\frac{9}{256}
Factor v^{2}-\frac{13}{8}v+\frac{169}{256}. 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(v-\frac{13}{16}\right)^{2}}=\sqrt{\frac{9}{256}}
Take the square root of both sides of the equation.
v-\frac{13}{16}=\frac{3}{16} v-\frac{13}{16}=-\frac{3}{16}
Simplify.
v=1 v=\frac{5}{8}
Add \frac{13}{16} to both sides of the equation.
x ^ 2 -\frac{13}{8}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 8
r + s = \frac{13}{8} 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{13}{16} - u s = \frac{13}{16} + u
Two numbers r and s sum up to \frac{13}{8} exactly when the average of the two numbers is \frac{1}{2}*\frac{13}{8} = \frac{13}{16}. 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{13}{16} - u) (\frac{13}{16} + u) = \frac{5}{8}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{5}{8}
\frac{169}{256} - u^2 = \frac{5}{8}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{5}{8}-\frac{169}{256} = -\frac{9}{256}
Simplify the expression by subtracting \frac{169}{256} on both sides
u^2 = \frac{9}{256} u = \pm\sqrt{\frac{9}{256}} = \pm \frac{3}{16}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{13}{16} - \frac{3}{16} = 0.625 s = \frac{13}{16} + \frac{3}{16} = 1
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
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