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a+b=-11 ab=28
To solve the equation, factor v^{2}-11v+28 using formula v^{2}+\left(a+b\right)v+ab=\left(v+a\right)\left(v+b\right). To find a and b, set up a system to be solved.
-1,-28 -2,-14 -4,-7
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 28.
-1-28=-29 -2-14=-16 -4-7=-11
Calculate the sum for each pair.
a=-7 b=-4
The solution is the pair that gives sum -11.
\left(v-7\right)\left(v-4\right)
Rewrite factored expression \left(v+a\right)\left(v+b\right) using the obtained values.
v=7 v=4
To find equation solutions, solve v-7=0 and v-4=0.
a+b=-11 ab=1\times 28=28
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as v^{2}+av+bv+28. To find a and b, set up a system to be solved.
-1,-28 -2,-14 -4,-7
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 28.
-1-28=-29 -2-14=-16 -4-7=-11
Calculate the sum for each pair.
a=-7 b=-4
The solution is the pair that gives sum -11.
\left(v^{2}-7v\right)+\left(-4v+28\right)
Rewrite v^{2}-11v+28 as \left(v^{2}-7v\right)+\left(-4v+28\right).
v\left(v-7\right)-4\left(v-7\right)
Factor out v in the first and -4 in the second group.
\left(v-7\right)\left(v-4\right)
Factor out common term v-7 by using distributive property.
v=7 v=4
To find equation solutions, solve v-7=0 and v-4=0.
v^{2}-11v+28=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(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 28}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -11 for b, and 28 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
v=\frac{-\left(-11\right)±\sqrt{121-4\times 28}}{2}
Square -11.
v=\frac{-\left(-11\right)±\sqrt{121-112}}{2}
Multiply -4 times 28.
v=\frac{-\left(-11\right)±\sqrt{9}}{2}
Add 121 to -112.
v=\frac{-\left(-11\right)±3}{2}
Take the square root of 9.
v=\frac{11±3}{2}
The opposite of -11 is 11.
v=\frac{14}{2}
Now solve the equation v=\frac{11±3}{2} when ± is plus. Add 11 to 3.
v=7
Divide 14 by 2.
v=\frac{8}{2}
Now solve the equation v=\frac{11±3}{2} when ± is minus. Subtract 3 from 11.
v=4
Divide 8 by 2.
v=7 v=4
The equation is now solved.
v^{2}-11v+28=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.
v^{2}-11v+28-28=-28
Subtract 28 from both sides of the equation.
v^{2}-11v=-28
Subtracting 28 from itself leaves 0.
v^{2}-11v+\left(-\frac{11}{2}\right)^{2}=-28+\left(-\frac{11}{2}\right)^{2}
Divide -11, the coefficient of the x term, by 2 to get -\frac{11}{2}. Then add the square of -\frac{11}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
v^{2}-11v+\frac{121}{4}=-28+\frac{121}{4}
Square -\frac{11}{2} by squaring both the numerator and the denominator of the fraction.
v^{2}-11v+\frac{121}{4}=\frac{9}{4}
Add -28 to \frac{121}{4}.
\left(v-\frac{11}{2}\right)^{2}=\frac{9}{4}
Factor v^{2}-11v+\frac{121}{4}. 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{11}{2}\right)^{2}}=\sqrt{\frac{9}{4}}
Take the square root of both sides of the equation.
v-\frac{11}{2}=\frac{3}{2} v-\frac{11}{2}=-\frac{3}{2}
Simplify.
v=7 v=4
Add \frac{11}{2} to both sides of the equation.
x ^ 2 -11x +28 = 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 = 11 rs = 28
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}{2} - u s = \frac{11}{2} + u
Two numbers r and s sum up to 11 exactly when the average of the two numbers is \frac{1}{2}*11 = \frac{11}{2}. 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}{2} - u) (\frac{11}{2} + u) = 28
To solve for unknown quantity u, substitute these in the product equation rs = 28
\frac{121}{4} - u^2 = 28
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 28-\frac{121}{4} = -\frac{9}{4}
Simplify the expression by subtracting \frac{121}{4} on both sides
u^2 = \frac{9}{4} u = \pm\sqrt{\frac{9}{4}} = \pm \frac{3}{2}
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}{2} - \frac{3}{2} = 4 s = \frac{11}{2} + \frac{3}{2} = 7
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.