Algebra II - Quadratic Formula

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

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

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 roots of the quadratic function,

$x = \left { \frac{-7+\sqrt{119}i}{6}, \frac{-7-\sqrt{119}i}{6}\right },$

$x = \left { \frac{7+\sqrt{119}}{6}, \frac{7-\sqrt{119}}{6}\right },$

$x = \left { \frac{-3+\sqrt{217}}{2}, \frac{-3-\sqrt{217}}{2}\right },$

$x = \left { \frac{-3+\sqrt{217}i}{2}, \frac{-3-\sqrt{217}i}{2}\right },$

$x = \left { \frac{-7+\sqrt{117}i}{6}, \frac{-7-\sqrt{117}i}{6}\right },$

Explanation

The roots are the values of for which:

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Reminder

Recall that for a quadratic the general formula for the solution in terms of the constant coefficients is given by:

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

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Use the quadratic formula to find the roots.

$x = \frac{-7\pm\sqrt{7^2-4(14)(3)})}{2(3)}$

$x = \frac{7}{6}\pm \frac{\sqrt{-119}}{6}$

Notice that $\sqrt{-119}$ is not a real number, and therefore the roots will be complex numbers.

Using the definition of the imaginary unit $i = \sqrt{-1}$ we can rewrite $\sqrt{-119}$ as follows,

$\sqrt{-119}=\sqrt{119(-1)}=\sqrt{119}\sqrt{-1}=\sqrt{119}i$

Now we can write the solutions to this problem in the form:

$x = \left { \frac{7+\sqrt{119}}{6}i, \frac{7-\sqrt{119}}{6}i\right },$

Solve for . Use the quadratic formula to find your solution. Use a calculator to estimate the value to the closest hundredth.

and

No solution

and

Explanation

Recall that the quadratic formula is defined as:

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

For this question, the variables are as follows:

Substituting these values into the equation, you get:

$\frac{-(-7) \pm \sqrt{(-7)^2 - 4(1)(-5)}}{2(1)}$

$\frac{7 \pm \sqrt{49 - (-20)}}{2}$

$\frac{7 \pm \sqrt{49 +20}}{2}$

$\frac{7 \pm \sqrt{69}}{2}$

Use a calculator to determine the final values.

$\frac{7+\sqrt{69}}{2}=7.65$

$\frac{7-\sqrt{69}}{2}=-0.65$

The height of a kicked soccer ball can be modeled with the equation

where the height is given in meters and is the time in seconds. At what time(s) will the ball be 2 meters off the ground?

seconds

Explanation

Set up the equation to solve for the time when the height is at 2 meters:

Now put the equation into quadratic form so that we can solve it using the quadratic formula

The quadratic equation is

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

where , , and .

Solving for gives us two possible values,

seconds

seconds.

Use the quadratic formula to find the roots of the quadratic,

$x = \left{-\frac{3}{4} + \frac{\sqrt{65}}{4}, -\frac{3}{4} - \frac{\sqrt{65}}{4} \right}$

$x = \left{\frac{3}{4} + \frac{\sqrt{65}}{2}, \frac{3}{4} - \frac{\sqrt{65}}{2} \right}$

$x = \left{-\frac{3}{4} + \frac{3\sqrt{6}}{4} -\frac{3}{4} - \frac{3\sqrt{6}}{4} \right}$

$x = \left{-\frac{3}{4} + \frac{3\sqrt{6}}{4}i, -\frac{3}{4} - \frac{3\sqrt{6}}{4}i \right}$

$x = \left{-\frac{3}{4} + \frac{\sqrt{65}}{2}i, -\frac{3}{4} - \frac{\sqrt{65}}{2}i \right}$

Explanation

Recall the general form of a quadratic,

The solution set has the form,

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

For our particular case, , , and

$x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-7)}}{2(2)}$

$x = \frac{-3 \pm \sqrt{65}}{4}$

$x = \left{-\frac{3}{4} + \frac{\sqrt{65}}{4}, -\frac{3}{4} - \frac{\sqrt{65}}{4} \right}$

Use the quadratic formula to find the answer of the following quadratic equation.

$x=\frac{9+\sqrt{57}}{6},x=\frac{9-\sqrt{57}}{6}$

$x=\frac{9+\sqrt{59}}{6},x=\frac{9-\sqrt{59}}{6}$

$x=3+\sqrt{57},x=3-\sqrt{57}$

$x=\frac{3+\sqrt{57}}{2},x=\frac{3-\sqrt{57}}{2}$

$x=\frac{9+2\sqrt{57}}{6},x=\frac{9-2\sqrt{57}}{6}$

Explanation

The quadratic equation is:

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

$\a=3 \b=-9 \c=2$

Therefore:

$\frac{-(-9)\pm \sqrt{(-9)^2-4(3)(2)}}{2(3)}$

Which gives the answer:

$x=\frac{9\pm \sqrt{57}}{6}$

Solve for . Use the quadratic formula to find your solution. Use a calculator to estimate the value to the closest hundredth.

and

No solution

and

Explanation

Recall that the quadratic formula is defined as:

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

For this question, the variables are as follows:

Substituting these values into the equation, you get:

$\frac{-(-17) \pm \sqrt{(-17)^2 - 4(5)(12)}}{2(5)}$

$\frac{17 \pm \sqrt{289 - 240}}{10}$

$\frac{17 \pm \sqrt{49}}{10}$

$\frac{17 \pm 7}{10}$

Separate this expression into two fractions and simplify to determine the final values.

$\frac{17 + 7}{10} = \frac{24}{10} = 2.4$

$\frac{17 - 7}{10} = \frac{10}{10} = 1$

Find a root using the quadratic equation:

$- \frac{\sqrt{30}}{10}$

$- \frac{\sqrt{30}}{3}$

$- \frac{2\sqrt{15}}{3}$

$- \frac{\sqrt{30}}{2}$

$\frac{\sqrt{15}}{3}$

Explanation

Write the quadratic equation.

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

Rewrite the given equation in form.

We can determine the coefficients of the terms.

Substitute these values into the quadratic equation.

$x=\frac{\pm \sqrt{-4(-10)(3)}}{2(-10)} = \frac{\pm \sqrt{120}}{-20}$

Rewrite this fraction using common factors, and simplify each step.

$\frac{\pm \sqrt{120}}{-20} = \frac{\pm \sqrt{4\times 30}}{-20} = \frac{\pm \sqrt{4}\cdot\sqrt{30}}{-20}=\frac{\pm 2\cdot\sqrt{30}}{-20} =\pm \frac{\sqrt{30}}{10}$

One of the possible answers given is: $- \frac{\sqrt{30}}{10}$

Quadratic Formula

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