Solve for v
v=\frac{\sqrt{33}}{6}+\frac{7}{2}\approx 4.457427108
v=-\frac{\sqrt{33}}{6}+\frac{7}{2}\approx 2.542572892
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3v^{2}-21v+34=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(-21\right)±\sqrt{\left(-21\right)^{2}-4\times 3\times 34}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -21 for b, and 34 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
v=\frac{-\left(-21\right)±\sqrt{441-4\times 3\times 34}}{2\times 3}
Square -21.
v=\frac{-\left(-21\right)±\sqrt{441-12\times 34}}{2\times 3}
Multiply -4 times 3.
v=\frac{-\left(-21\right)±\sqrt{441-408}}{2\times 3}
Multiply -12 times 34.
v=\frac{-\left(-21\right)±\sqrt{33}}{2\times 3}
Add 441 to -408.
v=\frac{21±\sqrt{33}}{2\times 3}
The opposite of -21 is 21.
v=\frac{21±\sqrt{33}}{6}
Multiply 2 times 3.
v=\frac{\sqrt{33}+21}{6}
Now solve the equation v=\frac{21±\sqrt{33}}{6} when ± is plus. Add 21 to \sqrt{33}.
v=\frac{\sqrt{33}}{6}+\frac{7}{2}
Divide 21+\sqrt{33} by 6.
v=\frac{21-\sqrt{33}}{6}
Now solve the equation v=\frac{21±\sqrt{33}}{6} when ± is minus. Subtract \sqrt{33} from 21.
v=-\frac{\sqrt{33}}{6}+\frac{7}{2}
Divide 21-\sqrt{33} by 6.
v=\frac{\sqrt{33}}{6}+\frac{7}{2} v=-\frac{\sqrt{33}}{6}+\frac{7}{2}
The equation is now solved.
3v^{2}-21v+34=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.
3v^{2}-21v+34-34=-34
Subtract 34 from both sides of the equation.
3v^{2}-21v=-34
Subtracting 34 from itself leaves 0.
\frac{3v^{2}-21v}{3}=-\frac{34}{3}
Divide both sides by 3.
v^{2}+\left(-\frac{21}{3}\right)v=-\frac{34}{3}
Dividing by 3 undoes the multiplication by 3.
v^{2}-7v=-\frac{34}{3}
Divide -21 by 3.
v^{2}-7v+\left(-\frac{7}{2}\right)^{2}=-\frac{34}{3}+\left(-\frac{7}{2}\right)^{2}
Divide -7, the coefficient of the x term, by 2 to get -\frac{7}{2}. Then add the square of -\frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
v^{2}-7v+\frac{49}{4}=-\frac{34}{3}+\frac{49}{4}
Square -\frac{7}{2} by squaring both the numerator and the denominator of the fraction.
v^{2}-7v+\frac{49}{4}=\frac{11}{12}
Add -\frac{34}{3} to \frac{49}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(v-\frac{7}{2}\right)^{2}=\frac{11}{12}
Factor v^{2}-7v+\frac{49}{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{7}{2}\right)^{2}}=\sqrt{\frac{11}{12}}
Take the square root of both sides of the equation.
v-\frac{7}{2}=\frac{\sqrt{33}}{6} v-\frac{7}{2}=-\frac{\sqrt{33}}{6}
Simplify.
v=\frac{\sqrt{33}}{6}+\frac{7}{2} v=-\frac{\sqrt{33}}{6}+\frac{7}{2}
Add \frac{7}{2} to both sides of the equation.
x ^ 2 -7x +\frac{34}{3} = 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 3
r + s = 7 rs = \frac{34}{3}
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{7}{2} - u s = \frac{7}{2} + u
Two numbers r and s sum up to 7 exactly when the average of the two numbers is \frac{1}{2}*7 = \frac{7}{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{7}{2} - u) (\frac{7}{2} + u) = \frac{34}{3}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{34}{3}
\frac{49}{4} - u^2 = \frac{34}{3}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{34}{3}-\frac{49}{4} = -\frac{11}{12}
Simplify the expression by subtracting \frac{49}{4} on both sides
u^2 = \frac{11}{12} u = \pm\sqrt{\frac{11}{12}} = \pm \frac{\sqrt{11}}{\sqrt{12}}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{7}{2} - \frac{\sqrt{11}}{\sqrt{12}} = 2.543 s = \frac{7}{2} + \frac{\sqrt{11}}{\sqrt{12}} = 4.457
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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