Solve 10v^2-29v+10

$$10 v ^ { 2 } - 29 v + 10$$

Factor

$$\left(2v-5\right)\left(5v-2\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 10v^{2}+av+bv+10. To find a and b, set up a system to be solved.

$$a+b=-29 ab=10\times 10=100$$

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 100.

$$-1,-100 -2,-50 -4,-25 -5,-20 -10,-10$$

Calculate the sum for each pair.

$$-1-100=-101 -2-50=-52 -4-25=-29 -5-20=-25 -10-10=-20$$

The solution is the pair that gives sum -29.

$$a=-25 b=-4$$

Rewrite 10v^{2}-29v+10 as \left(10v^{2}-25v\right)+\left(-4v+10\right).

$$\left(10v^{2}-25v\right)+\left(-4v+10\right)$$

Factor out 5v in the first and -2 in the second group.

$$5v\left(2v-5\right)-2\left(2v-5\right)$$

Factor out common term 2v-5 by using distributive property.

$$\left(2v-5\right)\left(5v-2\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$10v^{2}-29v+10=0$$

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.

$$v=\frac{-\left(-29\right)±\sqrt{\left(-29\right)^{2}-4\times 10\times 10}}{2\times 10}$$

Square -29.

$$v=\frac{-\left(-29\right)±\sqrt{841-4\times 10\times 10}}{2\times 10}$$

Multiply -4 times 10.

$$v=\frac{-\left(-29\right)±\sqrt{841-40\times 10}}{2\times 10}$$

Multiply -40 times 10.

$$v=\frac{-\left(-29\right)±\sqrt{841-400}}{2\times 10}$$

Add 841 to -400.

$$v=\frac{-\left(-29\right)±\sqrt{441}}{2\times 10}$$

Take the square root of 441.

$$v=\frac{-\left(-29\right)±21}{2\times 10}$$

The opposite of -29 is 29.

$$v=\frac{29±21}{2\times 10}$$

Multiply 2 times 10.

$$v=\frac{29±21}{20}$$

Now solve the equation v=\frac{29±21}{20} when ± is plus. Add 29 to 21.

$$v=\frac{50}{20}$$

Reduce the fraction \frac{50}{20} to lowest terms by extracting and canceling out 10.

$$v=\frac{5}{2}$$

Now solve the equation v=\frac{29±21}{20} when ± is minus. Subtract 21 from 29.

$$v=\frac{8}{20}$$

Reduce the fraction \frac{8}{20} to lowest terms by extracting and canceling out 4.

$$v=\frac{2}{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5}{2} for x_{1} and \frac{2}{5} for x_{2}.

$$10v^{2}-29v+10=10\left(v-\frac{5}{2}\right)\left(v-\frac{2}{5}\right)$$

Subtract \frac{5}{2} from v by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$10v^{2}-29v+10=10\times \frac{2v-5}{2}\left(v-\frac{2}{5}\right)$$

Subtract \frac{2}{5} from v by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$10v^{2}-29v+10=10\times \frac{2v-5}{2}\times \frac{5v-2}{5}$$

Multiply \frac{2v-5}{2} times \frac{5v-2}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$10v^{2}-29v+10=10\times \frac{\left(2v-5\right)\left(5v-2\right)}{2\times 5}$$

Multiply 2 times 5.

$$10v^{2}-29v+10=10\times \frac{\left(2v-5\right)\left(5v-2\right)}{10}$$

Cancel out 10, the greatest common factor in 10 and 10.

$$10v^{2}-29v+10=\left(2v-5\right)\left(5v-2\right)$$

Steps Using Direct Factoring Method

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 10

$$x ^ 2 -\frac{29}{10}x +1 = 0$$

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

$$r + s = \frac{29}{10} rs = 1$$

Two numbers r and s sum up to \frac{29}{10} exactly when the average of the two numbers is \frac{1}{2}*\frac{29}{10} = \frac{29}{20}. 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>

$$r = \frac{29}{20} - u s = \frac{29}{20} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 1

$$(\frac{29}{20} - u) (\frac{29}{20} + u) = 1$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{841}{400} - u^2 = 1$$

Simplify the expression by subtracting \frac{841}{400} on both sides

$$-u^2 = 1-\frac{841}{400} = -\frac{441}{400}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{441}{400} u = \pm\sqrt{\frac{441}{400}} = \pm \frac{21}{20}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =\frac{29}{20} - \frac{21}{20} = 0.400 s = \frac{29}{20} + \frac{21}{20} = 2.500$$

Evaluate

$$\left(2v-5\right)\left(5v-2\right)$$