Solving Quadratic Equations

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Algebra II › Solving Quadratic Equations

Questions 1 - 10

Solve:

$(-\infty, -4)$

$(8, \infty)$

$(10, \infty)$

Explanation

Start by setting the inequality to zero and by solving for .

Now, plot these two points on to a number line.

Notice that these two numbers effectively divide up the number line into three regions:

$(-\infty, -4)$ , , and $(8, \infty)$

Now, choose a number in each of these regions and put it back in the factored inequality to see which cases are true.

For $(-\infty, -4)$ , let

Since this is not less than , the solution to this inequality cannot lie in this region.

For , let .

Since this will make the inequality true, the solution can lie in this region.

Finally, for $(8, \infty)$ , let

Since this number is not less than zero, the solution cannot lie in this region.

Thus, the solution to this inequality is

Josephine wanted to solve the quadratic equation below by completing the square. Her first two steps are shown below:

Equation:

Step 1:

Step 2:

Which of the following equations would best represent the next step in solving the equation?

Explanation

To solve an equation by completing the square, you must factor the perfect square. The factored form of is . Once the left side of the equation is factored, you may take the square root of both sides.

Use the quadratic formula to determine a root:

$\frac{1-i\sqrt{35}}{6}$

$\frac{1-i\sqrt{35}}{2}$

$\frac{1-i\sqrt{37}}{6}$

$\frac{1-i\sqrt{37}}{2}$

$\frac{1-i\sqrt{37}}{3}$

Explanation

Write the quadratic formula for polynomials in the form of

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Substitute the known values.

$x=\frac{-(-1)\pm\sqrt{(-1)^2-4(3)(3)}}{2(3)}$

Simplify the equation.

$x=\frac{1\pm\sqrt{1-36}}{6} = \frac{1\pm\sqrt{-35}}{6}= \frac{1\pm i\sqrt{35}}{6}$

The roots to this parabola is: $x= \frac{1\pm i\sqrt{35}}{6}$

The answer is $\frac{1-i\sqrt{35}}{6}$ .

Find the zeros of ?

$x=-1, -\frac{1}{2}$

$x=1, -\frac{1}{2}$

$x=1, -\frac{3}{4}$

$x=-1, -\frac{3}{4}$

$x=1, \frac{3}{4}$

Explanation

This specific function cannot be factored, so use the quadratic equation:

$\frac{-b\pm \sqrt{b^2 -4ac}}{2a}$

Our function is in the form where,

Therefore the quadratic equation becomes,

$\frac{-3\pm \sqrt{3^2 -4(2)(1)}}{2(2)}$

$\frac{-3\pm \sqrt{9 -8}}{4}$

$\frac{-3\pm \sqrt{1}}{4}$

$\frac{-3+1}{4}$ OR $\frac{-3-1}{4}$

$\frac{-2}{4}$ OR $\frac{-4}{4}$

$x=-\frac{1}{2}$ OR

Which of the following is the same after completing the square?

$(x+\frac{1}{6})^2 =\frac{73}{36}$

$(x-\frac{1}{6})^2 =\frac{73}{36}$

$(x+\frac{1}{6})^2 =\frac{1}{2}$

$(x+\frac{1}{3})^2 =\frac{7}{2}$

$(x-\frac{1}{3})^2 =\frac{7}{2}$

Explanation

Divide by three on both sides.

$\frac{3x^2+x-6 }{3}=\frac{0}{3}$

$x^2+\frac{1}{3}x-2 = 0$

Add two on both sides.

$x^2+\frac{1}{3}x-2 +2= 0+2$

$x^2+\frac{1}{3}x = 2$

To complete the square, we will need to divide the one-third coefficient by two, which is similar to multiplying by one half, square the quantity, and add the two values on both sides.

$x^2+\frac{1}{3}x+ (\frac{1}{3}\cdot \frac{1}{2})^2 = 2+ (\frac{1}{3}\cdot \frac{1}{2})^2$

Simplify both sides.

$x^2+\frac{1}{3}x+ \frac{1}{36} = 2+ \frac{1}{36}$

Factor the left side, and combine the terms on the right.

$(x+\frac{1}{6})^2 = \frac{72}{36}+ \frac{1}{36}$

The answer is: $(x+\frac{1}{6})^2 =\frac{73}{36}$

Solve for by completing the square.

$x=0.24\text{ and }x=-8.24$

$x=0.39\text{ and }x=-4.39$

$x=1.24\text{ and }x=-10.24$

$x=0.97\text{ and }x=-8.97$

Explanation

Start by adding to both sides so that the terms with the are together on the left side of the equation.

Now, look at the coefficient of the -term. To complete the square, divide this coefficient by , then square the result. Add this term to both sides of the equation.

Rewrite the left side of the equation in the squared form.

Take the square root of both sides.

$x+4=\pm \sqrt{18}$

Now solve for .

$x=-4+\sqrt{18}=0.24$

$x=-4-\sqrt{18}=-8.24$

Round to two places after the decimal.

Which of the following is the same after completing the square?

$(x+\frac{1}{6})^2 =\frac{73}{36}$

$(x-\frac{1}{6})^2 =\frac{73}{36}$

$(x+\frac{1}{6})^2 =\frac{1}{2}$

$(x+\frac{1}{3})^2 =\frac{7}{2}$

$(x-\frac{1}{3})^2 =\frac{7}{2}$

Explanation

Divide by three on both sides.

$\frac{3x^2+x-6 }{3}=\frac{0}{3}$

$x^2+\frac{1}{3}x-2 = 0$

Add two on both sides.

$x^2+\frac{1}{3}x-2 +2= 0+2$

$x^2+\frac{1}{3}x = 2$

To complete the square, we will need to divide the one-third coefficient by two, which is similar to multiplying by one half, square the quantity, and add the two values on both sides.

$x^2+\frac{1}{3}x+ (\frac{1}{3}\cdot \frac{1}{2})^2 = 2+ (\frac{1}{3}\cdot \frac{1}{2})^2$

Simplify both sides.

$x^2+\frac{1}{3}x+ \frac{1}{36} = 2+ \frac{1}{36}$

Factor the left side, and combine the terms on the right.

$(x+\frac{1}{6})^2 = \frac{72}{36}+ \frac{1}{36}$

The answer is: $(x+\frac{1}{6})^2 =\frac{73}{36}$

Solve for by completing the square.

$x=0.24\text{ and }x=-8.24$

$x=0.39\text{ and }x=-4.39$

$x=1.24\text{ and }x=-10.24$

$x=0.97\text{ and }x=-8.97$

Explanation

Start by adding to both sides so that the terms with the are together on the left side of the equation.

Now, look at the coefficient of the -term. To complete the square, divide this coefficient by , then square the result. Add this term to both sides of the equation.

Rewrite the left side of the equation in the squared form.

Take the square root of both sides.

$x+4=\pm \sqrt{18}$

Now solve for .

$x=-4+\sqrt{18}=0.24$

$x=-4-\sqrt{18}=-8.24$

Round to two places after the decimal.

Josephine wanted to solve the quadratic equation below by completing the square. Her first two steps are shown below:

Equation:

Step 1:

Step 2:

Which of the following equations would best represent the next step in solving the equation?

Explanation

What is/are the solution(s) to the quadratic equation

Hint: Complete the square

Explanation

When using the complete the square method we will divide the coefficient by two and then square it. This will become our term which we will add to both sides.