Solve 4q^2-28q+40=-9

$$4 q ^ { 2 } - 28 q + 40 = - 9$$

Solve for q

$$q = \frac{7}{2} = 3\frac{1}{2} = 3.5$$

Steps Using Factoring By Grouping

Add 9 to both sides.

$$4q^{2}-28q+40+9=0$$

Add 40 and 9 to get 49.

$$4q^{2}-28q+49=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4q^{2}+aq+bq+49. To find a and b, set up a system to be solved.

$$a+b=-28 ab=4\times 49=196$$

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 196.

$$-1,-196 -2,-98 -4,-49 -7,-28 -14,-14$$

Calculate the sum for each pair.

$$-1-196=-197 -2-98=-100 -4-49=-53 -7-28=-35 -14-14=-28$$

The solution is the pair that gives sum -28.

$$a=-14 b=-14$$

Rewrite 4q^{2}-28q+49 as \left(4q^{2}-14q\right)+\left(-14q+49\right).

$$\left(4q^{2}-14q\right)+\left(-14q+49\right)$$

Factor out 2q in the first and -7 in the second group.

$$2q\left(2q-7\right)-7\left(2q-7\right)$$

Factor out common term 2q-7 by using distributive property.

$$\left(2q-7\right)\left(2q-7\right)$$

Rewrite as a binomial square.

$$\left(2q-7\right)^{2}$$

To find equation solution, solve 2q-7=0.

$$q=\frac{7}{2}$$

Steps Using the Quadratic Formula

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.

$$4q^{2}-28q+40=-9$$

Add 9 to both sides of the equation.

$$4q^{2}-28q+40-\left(-9\right)=-9-\left(-9\right)$$

Subtracting -9 from itself leaves 0.

$$4q^{2}-28q+40-\left(-9\right)=0$$

Subtract -9 from 40.

$$4q^{2}-28q+49=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -28 for b, and 49 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$q=\frac{-\left(-28\right)±\sqrt{\left(-28\right)^{2}-4\times 4\times 49}}{2\times 4}$$

Square -28.

$$q=\frac{-\left(-28\right)±\sqrt{784-4\times 4\times 49}}{2\times 4}$$

Multiply -4 times 4.

$$q=\frac{-\left(-28\right)±\sqrt{784-16\times 49}}{2\times 4}$$

Multiply -16 times 49.

$$q=\frac{-\left(-28\right)±\sqrt{784-784}}{2\times 4}$$

Add 784 to -784.

$$q=\frac{-\left(-28\right)±\sqrt{0}}{2\times 4}$$

Take the square root of 0.

$$q=-\frac{-28}{2\times 4}$$

The opposite of -28 is 28.

$$q=\frac{28}{2\times 4}$$

Multiply 2 times 4.

$$q=\frac{28}{8}$$

Reduce the fraction \frac{28}{8} to lowest terms by extracting and canceling out 4.

$$q=\frac{7}{2}$$

Steps for Completing the Square

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.

$$4q^{2}-28q+40=-9$$

Subtract 40 from both sides of the equation.

$$4q^{2}-28q+40-40=-9-40$$

Subtracting 40 from itself leaves 0.

$$4q^{2}-28q=-9-40$$

Subtract 40 from -9.

$$4q^{2}-28q=-49$$

Divide both sides by 4.

$$\frac{4q^{2}-28q}{4}=-\frac{49}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$q^{2}+\left(-\frac{28}{4}\right)q=-\frac{49}{4}$$

Divide -28 by 4.

$$q^{2}-7q=-\frac{49}{4}$$

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.

$$q^{2}-7q+\left(-\frac{7}{2}\right)^{2}=-\frac{49}{4}+\left(-\frac{7}{2}\right)^{2}$$

Square -\frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$q^{2}-7q+\frac{49}{4}=\frac{-49+49}{4}$$

Add -\frac{49}{4} to \frac{49}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$q^{2}-7q+\frac{49}{4}=0$$

Factor q^{2}-7q+\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}.

$$\left(q-\frac{7}{2}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(q-\frac{7}{2}\right)^{2}}=\sqrt{0}$$

Simplify.

$$q-\frac{7}{2}=0 q-\frac{7}{2}=0$$

Add \frac{7}{2} to both sides of the equation.

$$q=\frac{7}{2} q=\frac{7}{2}$$

The equation is now solved. Solutions are the same.

$$q=\frac{7}{2}$$