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Algebra II : Quadratic Formula

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #191 : Solving Quadratic Equations

Determine a possible root for: y=x^2+10x-1

Possible Answers:

$5+3\sqrt{2}$

$\sqrt{13}-5$

$\sqrt{26}-5$

$\textup{None of the answers are possible roots.}$

$5-2\sqrt6$

Correct answer:

$\sqrt{26}-5$

Explanation:

Write the quadratic formula.

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

Identify and substitute the terms.

a=1,b=10, c=-1

$x=\frac{-10\pm \sqrt{10^2-4(1)(-1)}}{2(1)}=\frac{-10\pm \sqrt{100+4}}{2}$

$x=\frac{-10\pm \sqrt{104}}{2}$

Factor the radical using factors of perfect squares.

$\sqrt{104} = \sqrt4 \cdot \sqrt{26} = 2\sqrt{26}$

$x=\frac{-10\pm2 \sqrt{26}}{2} = -5\pm \sqrt{26}$

These two answers are possible roots.

The answer is: $\sqrt{26}-5$

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Example Question #511 : Intermediate Single Variable Algebra

Find the roots of the quadratic function,

f(x)= -A^2x^2+2Ax+1

Where is any real number constant not equal to zero.

Possible Answers:

$x = \left \{\frac{1}{3}-\frac{\sqrt{3}}{A}i\: \: \: ,\: \: \: \frac{1}{3}+\frac{\sqrt{3}}{A}i\right \}$

$x = \left \{\frac{1}{A^2}-\frac{3\sqrt{3}}{A^2}\: \: \: ,\: \: \: \frac{1}{A^2}+\frac{3\sqrt{3}}{A^2}\right \}$

$x = \left \{\frac{1}{A^2}-\frac{2\sqrt{2}}{A}\: \: \: ,\: \: \: \frac{1}{A^2}+\frac{2\sqrt{2}}{A}\right \}$

$x = \left \{\frac{1}{A}-\frac{\sqrt{2}}{A}\: \: \: ,\: \: \: \frac{1}{A}+\frac{\sqrt{2}}{A}\right \}$

$x = \left \{\frac{1}{A}-\frac{\sqrt{2}}{A}i\: \: \: ,\: \: \: \frac{1}{A}+\frac{\sqrt{2}}{A}i\right \}$

Correct answer:

$x = \left \{\frac{1}{A}-\frac{\sqrt{2}}{A}\: \: \: ,\: \: \: \frac{1}{A}+\frac{\sqrt{2}}{A}\right \}$

Explanation:

f(x)= -A^2x^2+2Ax+1

To find the roots set the function to zero:

f(x)=0 ,

-A^2x^2 +2Ax+1=0 (1)

Apply the quadratic formula:

_________________________________________________________________

Reminder

Recall that for a quadratic ax^2 +bx+c=0 the general formula for the solution in terms of the constant coefficients is given by:

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

_____________________________________________________________

Use equation (2) to write a solution for equation (1).

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

$x = \frac{2A \pm \sqrt{8A^2}}{2A^2}$

If we simplify the right-hand term in the numerator we obtain:

$\sqrt{8A^2}=2\sqrt2A$

So now we have for ,

$x=\frac{2A}{2A^2} \pm \frac{2\sqrt{2}A}{2A^2}$

After all the cancellations in the expression above we obtain:

$x = \frac{1}{A} \pm \frac{\sqrt{2}}{A}$

Therefore, the solution set for this equation is:

$x = \left \{\frac{1}{A}-\frac{\sqrt{2}}{A}\: \: \: ,\: \: \: \frac{1}{A}+\frac{\sqrt{2}}{A}\right \}$

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Example Question #193 : Solving Quadratic Equations

Find the roots of the quadratic function,

f(x)=3x^2+7x+14

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

f(x)=3x^2+7x+14

The roots are the values of for which:

f(x)=0

3x^2+7x+14=0

______________________________________________________________

Reminder

Recall that for a quadratic ax^2 +bx+c=0 the general formula for the solution in terms of the constant coefficients is given by:

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

_____________________________________________________________

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:

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

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Example Question #191 : Solving Quadratic Equations

Find the roots using the quadratic formula.

x^2+9x+15

Possible Answers:

x= 2.21

x= -6.79,-2.21

x= 6.79, 2.21

x= -6.79

x= 7.89,5.65

Correct answer:

x= -6.79,-2.21

Explanation:

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

For this problem

a=1, coefficient of x^2 term

b=9, the coefficient of the x term

c=15, the constant term

$\frac{-9\pm \sqrt{9^2-4(1)(15)}}{2(1)}=\frac{-9\pm \sqrt{21}}{2}=\frac{-9}{2}\pm \frac{\sqrt{21}}{2}$

solving the expression shows the roots of -6.79 and -2.21

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Example Question #195 : Solving Quadratic Equations

Use the quadratic formula to solve for . Use a calculator to estimate the value to the closest hundredth.

3x^2 + 14x + 9 = 0

Possible Answers:

-0.77 and -3.90

4.42 and -23.33

No solution

13.38 and 3.21

9.12 and -4.53

Correct answer:

-0.77 and -3.90

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:

ax^2+bx+c=0

3x^2 + 14x + 9 = 0

a = 3

b = 14

c = 9

Substituting these values into the equation, you get:

$\frac{-14 \pm \sqrt{(14)^2 - 4(3)(9)}}{2(3)}$

$\frac{-14 \pm \sqrt{196 - 108}}{6}$

$\frac{-14 \pm \sqrt{88}}{6}$

Use a calculator to determine the final values.

$\frac{14+\sqrt{88}}{6}=3.90$

$\frac{14-\sqrt{88}}{6}=0.77$

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Algebra II : Quadratic Formula

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #191 : Solving Quadratic Equations

Example Question #511 : Intermediate Single Variable Algebra

Example Question #193 : Solving Quadratic Equations

Example Question #191 : Solving Quadratic Equations

Example Question #195 : Solving Quadratic Equations

All Algebra II Resources

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