Quadratic Equations and Inequalities

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Algebra II › Quadratic Equations and Inequalities

Questions 1 - 10

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:

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

$x+2=\pm3$

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:

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

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

$\small a=1$

$\small 2ab=-8$

$\small b=-4$

$\small b^2=16$

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

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

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

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

$\small x=4\pm\sqrt7$

Multiply:

Explanation

To multiply these binomials, use FOIL. Remember to multiply the first terms:

$4x\cdot 2x=8x^2$

Then the outside terms:

$4x\cdot 3=12x$

Then the inside terms:

$-1\cdot 2x=-2x$

And the last terms:

$-1\cdot 3=-3$

Put those together to get:

Simplify to get your answer:

Solve the expression:

Explanation

Use the FOIL method to simplify this expression.

Multiply each term of the first binomial with the terms of the second binomial.

Simplify the expression.

Combine like terms.

The answer is:

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

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

Solve for the roots:

$\frac{1\pm\sqrt{5}}{4}$

$\frac{-1\pm\sqrt{5}}{4}$

$\frac{-1\pm i\sqrt{5}}{4}$

$\frac{-1\pm\sqrt{5}}{2}$

$\frac{1\pm\sqrt{5}}{16}$

Explanation

Write the quadratic formula.

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

The given equation is in standard form of:

The coefficients correspond to the values that go inside the quadratic equation.

Substitute the values into the equation.

$x= \frac{-(-4)\pm\sqrt{(-4)^2-4(8)(-2)}}{2(8)}$

Simplify this equation.

$x= \frac{4\pm\sqrt{16+64}}{16} = \frac{4\pm\sqrt{80}}{16}$

The radical can be rewritten as:

$\sqrt{80} = \sqrt{16}\cdot \sqrt 5 = 4\sqrt5$

Substitute and simplify the fraction.

$x=\frac{4\pm4\sqrt{5}}{16}=\frac{1\pm\sqrt{5}}{4}$

The answer is: $\frac{1\pm\sqrt{5}}{4}$

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

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

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