Quadratic Equations and Inequalities

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Math › Quadratic Equations and Inequalities

Questions 1 - 10

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

Determine the discriminant:

$\textup{Cannot be determined.}$

Explanation

The discriminant is the term inside the square root of the quadratic equation.

The polynomial is provided in standard form .

Substitute the variables into the equation.

The answer is:

Expand:

$3x^{2}-25x-18$

$3x^{2}-29x-18$

$3x^{2}-18x-29$

$3x^{2}-18x-25$

None of the other answers

Explanation

To multiple these binomials, you can use the FOIL method to multiply each of the expressions individually.This will give you

or $3x^{2}-25x-18$ .

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

Evaluate the roots for:

$\frac{1}{2}\pm\frac{i\sqrt3}{2}$

$\frac{1}{4}\pm\frac{i\sqrt3}{4}$

$\frac{1}{4}\pm\frac{i\sqrt6}{4}$

$\frac{1}{3}\pm\frac{i\sqrt6}{3}$

$\frac{1}{4}\pm\frac{3i\sqrt3}{2}$

Explanation

Write the quadratic formula.

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

The coefficients of the variables can be determined by the given equation in standard form: .

Substitute the terms into the equation.

$x=\frac{-(-2)\pm\sqrt{(-2)^2-4(2)(2)}}{2(2)} = \frac{2\pm\sqrt{4-16}}{4}$

Simplify the terms.

$\frac{2\pm\sqrt{4-16}}{4} = \frac{2\pm\sqrt{-12}}{4} = \frac{2\pm 2i\sqrt{3}}{4}$

The answer is: $\frac{1}{2}\pm\frac{i\sqrt3}{2}$

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 by completing the square:

$\small x^2-8x+9=0$

$\small x=4\pm\sqrt7$

$\small x=9\pm2\sqrt2$

$\small x=-9\pm2\sqrt2$

$\small x=-4\pm\sqrt7$

Explanation

To complete the square, the equation must be in the form:

$\small \small ax^2+2ab+b^2=c$

$\small x^2-8x+9=0$

$\small a=1$

$\small 2ab=-8$

$\small b=-4$

$\small b^2=16$

$\small x^2-8x+9+7=0+7$

$\small x^2-8x+16=7$

$\small (x-4)^2=7$

$\small \small x-4=\pm\sqrt7$

$\small x=4\pm\sqrt7$

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.

In the form,

our , and we will complete the square to find the value.

Therefore we get:

$x^2 + 4x+\left(\frac{4}{2}\right)^2 = 5+\left(\frac{4}{2}\right)^2$

$x+2=\pm3$

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.

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