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a+b=-2 ab=15\left(-8\right)=-120
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 15v^{2}+av+bv-8. To find a and b, set up a system to be solved.
1,-120 2,-60 3,-40 4,-30 5,-24 6,-20 8,-15 10,-12
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 -120.
1-120=-119 2-60=-58 3-40=-37 4-30=-26 5-24=-19 6-20=-14 8-15=-7 10-12=-2
Calculate the sum for each pair.
a=-12 b=10
The solution is the pair that gives sum -2.
\left(15v^{2}-12v\right)+\left(10v-8\right)
Rewrite 15v^{2}-2v-8 as \left(15v^{2}-12v\right)+\left(10v-8\right).
3v\left(5v-4\right)+2\left(5v-4\right)
Factor out 3v in the first and 2 in the second group.
\left(5v-4\right)\left(3v+2\right)
Factor out common term 5v-4 by using distributive property.
v=\frac{4}{5} v=-\frac{2}{3}
To find equation solutions, solve 5v-4=0 and 3v+2=0.
15v^{2}-2v-8=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(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 15\left(-8\right)}}{2\times 15}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 15 for a, -2 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
v=\frac{-\left(-2\right)±\sqrt{4-4\times 15\left(-8\right)}}{2\times 15}
Square -2.
v=\frac{-\left(-2\right)±\sqrt{4-60\left(-8\right)}}{2\times 15}
Multiply -4 times 15.
v=\frac{-\left(-2\right)±\sqrt{4+480}}{2\times 15}
Multiply -60 times -8.
v=\frac{-\left(-2\right)±\sqrt{484}}{2\times 15}
Add 4 to 480.
v=\frac{-\left(-2\right)±22}{2\times 15}
Take the square root of 484.
v=\frac{2±22}{2\times 15}
The opposite of -2 is 2.
v=\frac{2±22}{30}
Multiply 2 times 15.
v=\frac{24}{30}
Now solve the equation v=\frac{2±22}{30} when ± is plus. Add 2 to 22.
v=\frac{4}{5}
Reduce the fraction \frac{24}{30} to lowest terms by extracting and canceling out 6.
v=-\frac{20}{30}
Now solve the equation v=\frac{2±22}{30} when ± is minus. Subtract 22 from 2.
v=-\frac{2}{3}
Reduce the fraction \frac{-20}{30} to lowest terms by extracting and canceling out 10.
v=\frac{4}{5} v=-\frac{2}{3}
The equation is now solved.
15v^{2}-2v-8=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.
15v^{2}-2v-8-\left(-8\right)=-\left(-8\right)
Add 8 to both sides of the equation.
15v^{2}-2v=-\left(-8\right)
Subtracting -8 from itself leaves 0.
15v^{2}-2v=8
Subtract -8 from 0.
\frac{15v^{2}-2v}{15}=\frac{8}{15}
Divide both sides by 15.
v^{2}-\frac{2}{15}v=\frac{8}{15}
Dividing by 15 undoes the multiplication by 15.
v^{2}-\frac{2}{15}v+\left(-\frac{1}{15}\right)^{2}=\frac{8}{15}+\left(-\frac{1}{15}\right)^{2}
Divide -\frac{2}{15}, the coefficient of the x term, by 2 to get -\frac{1}{15}. Then add the square of -\frac{1}{15} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
v^{2}-\frac{2}{15}v+\frac{1}{225}=\frac{8}{15}+\frac{1}{225}
Square -\frac{1}{15} by squaring both the numerator and the denominator of the fraction.
v^{2}-\frac{2}{15}v+\frac{1}{225}=\frac{121}{225}
Add \frac{8}{15} to \frac{1}{225} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(v-\frac{1}{15}\right)^{2}=\frac{121}{225}
Factor v^{2}-\frac{2}{15}v+\frac{1}{225}. 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{1}{15}\right)^{2}}=\sqrt{\frac{121}{225}}
Take the square root of both sides of the equation.
v-\frac{1}{15}=\frac{11}{15} v-\frac{1}{15}=-\frac{11}{15}
Simplify.
v=\frac{4}{5} v=-\frac{2}{3}
Add \frac{1}{15} to both sides of the equation.