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High School Math : Using the Quadratic Formula

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Example Questions

Example Question #341 : Algebra Ii

Solve using the quadratic formula:

2x^2+2x+3=0

Possible Answers:

$x= \frac{-1 \pm i\sqrt{5}}{3}$

$x= \frac{-2 \pm i\sqrt{5}}{3}$

$x= \frac{-1 \pm i\sqrt{5}}{2}$

$x= \frac{-3 \pm i\sqrt{5}}{2}$

$x= \frac{-2 \pm i\sqrt{5}}{2}$

Correct answer:

$x= \frac{-1 \pm i\sqrt{5}}{2}$

Explanation:

Use the quadratic formula to solve:

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

a = 2

b = 2

c = 3

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

$x = \frac{-2 \pm \sqrt{-20}}{4}$

$x = \frac{-2 \pm 2i\sqrt{5}}{4}$

$x = \frac{-1 \pm i\sqrt{5}}{2}$

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Example Question #91 : Intermediate Single Variable Algebra

Solve using the quadratic formula:

2a^2-5a+4=0

Possible Answers:

$a=\frac{4 \pm i\sqrt{7}}{4}$

$a=\frac{6 \pm i\sqrt{7}}{4}$

$a=\frac{3 \pm i\sqrt{7}}{4}$

$a=\frac{7 \pm i\sqrt{7}}{4}$

$a=\frac{5 \pm i\sqrt{7}}{4}$

Correct answer:

$a=\frac{5 \pm i\sqrt{7}}{4}$

Explanation:

Use the quadratic formula to solve:

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

a = 2

b = -5

c = 4

$a = \frac{-(-5) \pm \sqrt{(-5)^2-4(2)(4)}}{2(2)}$

$a = \frac{5 \pm \sqrt{-7}}{4}$

$a = \frac{5 \pm i\sqrt{7}}{4}$

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Example Question #92 : Intermediate Single Variable Algebra

Solve using the quadratic formula:

ax^2+bx+3b=0

Possible Answers:

$x=\frac{b \pm \sqrt{b^2-12ab}}{a}$

$x=\frac{b \pm \sqrt{b^2-12ab}}{2a}$

$x=\frac{-b \pm \sqrt{b^2-8ab}}{2a}$

$x=\frac{-b \pm \sqrt{b^2-12ab}}{a}$

$x=\frac{-b \pm \sqrt{b^2-12ab}}{2a}$

Correct answer:

$x=\frac{-b \pm \sqrt{b^2-12ab}}{2a}$

Explanation:

Use the quadratic formula to solve:

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

a = a

b = b

c = 3b

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

$x = \frac{-b \pm \sqrt{b^2-12ab}}{2a}$

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Example Question #93 : Intermediate Single Variable Algebra

Solve using the quadratric formula:

bx^2+acx+c=0

Possible Answers:

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

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

$x=\frac{-ac\sqrt{a^2c^2-2bc}}{2b}$

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

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

Correct answer:

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

Explanation:

Use the quadratic formula to solve:

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

a = b

b = ac

c = c

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

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

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Example Question #1 : Using The Quadratic Formula

A baseball that is thrown in the air follows a trajectory of h(t)= -4t^2 +12t+6 , where h(t) is the height of the ball in feet and is the time elapsed in seconds. How long does the ball stay in the air before it hits the ground?

Possible Answers:

Between 2 and 2.5 seconds

Between 4 and 4.5 seconds

Between 3 and 3.5 seconds

Between 3.5 and 4 seconds

Between 2.5 and 3 seconds

Correct answer:

Between 3 and 3.5 seconds

Explanation:

To solve this, we look at the equation h(t)=0 .

Setting the equation equal to 0 we get 0= -4t^2 +12t+6 .

Once in this form, we can use the Quadratic Formula to solve for .

The quadratic formula says that if 0= -ax^2 +bx+c , then

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

Plugging in our values:

$t= \frac{-12\pm\sqrt{(12^2-4(-4)(6)}}{2(-4)}=\frac{-12\pm \sqrt{240}}{-8}= \frac{12\pm15.5}{8}$

Therefore t=3.4375 or -.4375 and since we are looking only for positive values (because we can't have negative time), 3.4375 seconds is our answer.

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High School Math : Using the Quadratic Formula

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #341 : Algebra Ii

Example Question #91 : Intermediate Single Variable Algebra

Example Question #92 : Intermediate Single Variable Algebra

Example Question #93 : Intermediate Single Variable Algebra

Example Question #1 : Using The Quadratic Formula

All High School Math Resources

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