Solve 2u^2-34u+60

$$2 u ^ { 2 } - 34 u + 60$$

Factor

$$2\left(u-15\right)\left(u-2\right)$$

Solution Steps

Factor out 2.

$$2\left(u^{2}-17u+30\right)$$

Consider u^{2}-17u+30. Factor the expression by grouping. First, the expression needs to be rewritten as u^{2}+au+bu+30. To find a and b, set up a system to be solved.

$$a+b=-17 ab=1\times 30=30$$

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 30.

$$-1,-30 -2,-15 -3,-10 -5,-6$$

Calculate the sum for each pair.

$$-1-30=-31 -2-15=-17 -3-10=-13 -5-6=-11$$

The solution is the pair that gives sum -17.

$$a=-15 b=-2$$

Rewrite u^{2}-17u+30 as \left(u^{2}-15u\right)+\left(-2u+30\right).

$$\left(u^{2}-15u\right)+\left(-2u+30\right)$$

Factor out u in the first and -2 in the second group.

$$u\left(u-15\right)-2\left(u-15\right)$$

Factor out common term u-15 by using distributive property.

$$\left(u-15\right)\left(u-2\right)$$

Rewrite the complete factored expression.

$$2\left(u-15\right)\left(u-2\right)$$

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.

$$2u^{2}-34u+60=0$$

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.

$$u=\frac{-\left(-34\right)±\sqrt{\left(-34\right)^{2}-4\times 2\times 60}}{2\times 2}$$

Square -34.

$$u=\frac{-\left(-34\right)±\sqrt{1156-4\times 2\times 60}}{2\times 2}$$

Multiply -4 times 2.

$$u=\frac{-\left(-34\right)±\sqrt{1156-8\times 60}}{2\times 2}$$

Multiply -8 times 60.

$$u=\frac{-\left(-34\right)±\sqrt{1156-480}}{2\times 2}$$

Add 1156 to -480.

$$u=\frac{-\left(-34\right)±\sqrt{676}}{2\times 2}$$

Take the square root of 676.

$$u=\frac{-\left(-34\right)±26}{2\times 2}$$

The opposite of -34 is 34.

$$u=\frac{34±26}{2\times 2}$$

Multiply 2 times 2.

$$u=\frac{34±26}{4}$$

Now solve the equation u=\frac{34±26}{4} when ± is plus. Add 34 to 26.

$$u=\frac{60}{4}$$

Divide 60 by 4.

$$u=15$$

Now solve the equation u=\frac{34±26}{4} when ± is minus. Subtract 26 from 34.

$$u=\frac{8}{4}$$

Divide 8 by 4.

$$u=2$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 15 for x_{1} and 2 for x_{2}.

$$2u^{2}-34u+60=2\left(u-15\right)\left(u-2\right)$$

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 2

$$x ^ 2 -17x +30 = 0$$

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

$$r + s = 17 rs = 30$$

Two numbers r and s sum up to 17 exactly when the average of the two numbers is \frac{1}{2}*17 = \frac{17}{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>

$$r = \frac{17}{2} - u s = \frac{17}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 30

$$(\frac{17}{2} - u) (\frac{17}{2} + u) = 30$$

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

$$\frac{289}{4} - u^2 = 30$$

Simplify the expression by subtracting \frac{289}{4} on both sides

$$-u^2 = 30-\frac{289}{4} = -\frac{169}{4}$$

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

$$u^2 = \frac{169}{4} u = \pm\sqrt{\frac{169}{4}} = \pm \frac{13}{2}$$

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

$$r =\frac{17}{2} - \frac{13}{2} = 2 s = \frac{17}{2} + \frac{13}{2} = 15$$

Evaluate

$$2\left(u-15\right)\left(u-2\right)$$