Solve 10q^2-43q+45

$$10 q ^ { 2 } - 43 q + 45$$

Factor

$$\left(2q-5\right)\left(5q-9\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 10q^{2}+aq+bq+45. To find a and b, set up a system to be solved.

$$a+b=-43 ab=10\times 45=450$$

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 450.

$$-1,-450 -2,-225 -3,-150 -5,-90 -6,-75 -9,-50 -10,-45 -15,-30 -18,-25$$

Calculate the sum for each pair.

$$-1-450=-451 -2-225=-227 -3-150=-153 -5-90=-95 -6-75=-81 -9-50=-59 -10-45=-55 -15-30=-45 -18-25=-43$$

The solution is the pair that gives sum -43.

$$a=-25 b=-18$$

Rewrite 10q^{2}-43q+45 as \left(10q^{2}-25q\right)+\left(-18q+45\right).

$$\left(10q^{2}-25q\right)+\left(-18q+45\right)$$

Factor out 5q in the first and -9 in the second group.

$$5q\left(2q-5\right)-9\left(2q-5\right)$$

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

$$\left(2q-5\right)\left(5q-9\right)$$

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.

$$10q^{2}-43q+45=0$$

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.

$$q=\frac{-\left(-43\right)±\sqrt{\left(-43\right)^{2}-4\times 10\times 45}}{2\times 10}$$

Square -43.

$$q=\frac{-\left(-43\right)±\sqrt{1849-4\times 10\times 45}}{2\times 10}$$

Multiply -4 times 10.

$$q=\frac{-\left(-43\right)±\sqrt{1849-40\times 45}}{2\times 10}$$

Multiply -40 times 45.

$$q=\frac{-\left(-43\right)±\sqrt{1849-1800}}{2\times 10}$$

Add 1849 to -1800.

$$q=\frac{-\left(-43\right)±\sqrt{49}}{2\times 10}$$

Take the square root of 49.

$$q=\frac{-\left(-43\right)±7}{2\times 10}$$

The opposite of -43 is 43.

$$q=\frac{43±7}{2\times 10}$$

Multiply 2 times 10.

$$q=\frac{43±7}{20}$$

Now solve the equation q=\frac{43±7}{20} when ± is plus. Add 43 to 7.

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

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

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

Now solve the equation q=\frac{43±7}{20} when ± is minus. Subtract 7 from 43.

$$q=\frac{36}{20}$$

Reduce the fraction \frac{36}{20} to lowest terms by extracting and canceling out 4.

$$q=\frac{9}{5}$$

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{9}{5} for x_{2}.

$$10q^{2}-43q+45=10\left(q-\frac{5}{2}\right)\left(q-\frac{9}{5}\right)$$

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

$$10q^{2}-43q+45=10\times \frac{2q-5}{2}\left(q-\frac{9}{5}\right)$$

Subtract \frac{9}{5} from q by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$10q^{2}-43q+45=10\times \frac{2q-5}{2}\times \frac{5q-9}{5}$$

Multiply \frac{2q-5}{2} times \frac{5q-9}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$10q^{2}-43q+45=10\times \frac{\left(2q-5\right)\left(5q-9\right)}{2\times 5}$$

Multiply 2 times 5.

$$10q^{2}-43q+45=10\times \frac{\left(2q-5\right)\left(5q-9\right)}{10}$$

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

$$10q^{2}-43q+45=\left(2q-5\right)\left(5q-9\right)$$

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

$$x ^ 2 -\frac{43}{10}x +\frac{9}{2} = 0$$

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

$$r + s = \frac{43}{10} rs = \frac{9}{2}$$

Two numbers r and s sum up to \frac{43}{10} exactly when the average of the two numbers is \frac{1}{2}*\frac{43}{10} = \frac{43}{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>

$$r = \frac{43}{20} - u s = \frac{43}{20} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{9}{2}

$$(\frac{43}{20} - u) (\frac{43}{20} + u) = \frac{9}{2}$$

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

$$\frac{1849}{400} - u^2 = \frac{9}{2}$$

Simplify the expression by subtracting \frac{1849}{400} on both sides

$$-u^2 = \frac{9}{2}-\frac{1849}{400} = -\frac{49}{400}$$

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

$$u^2 = \frac{49}{400} u = \pm\sqrt{\frac{49}{400}} = \pm \frac{7}{20}$$

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

$$r =\frac{43}{20} - \frac{7}{20} = 1.800 s = \frac{43}{20} + \frac{7}{20} = 2.500$$

Evaluate

$$\left(2q-5\right)\left(5q-9\right)$$