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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:

$\displaystyle 2x^2+2x+3=0$

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

Use the quadratic formula to solve:

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

a = 2 $\displaystyle a = 2$

b = 2 $\displaystyle b = 2$

c = 3 $\displaystyle c = 3$

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

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

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

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

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Example Question #91 : Intermediate Single Variable Algebra

Solve using the quadratic formula:

$\displaystyle 2a^2-5a+4=0$

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

Use the quadratic formula to solve:

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

a = 2 $\displaystyle a = 2$

b = -5 $\displaystyle b = -5$

c = 4 $\displaystyle c = 4$

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

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

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

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Example Question #1 : Using The Quadratic Formula

Solve using the quadratic formula:

$\displaystyle ax^2+bx+3b=0$

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

Use the quadratic formula to solve:

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

a = a $\displaystyle a = a$

b = b $\displaystyle b = b$

c = 3b $\displaystyle c = 3b$

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

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

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Example Question #1 : Using The Quadratic Formula

Solve using the quadratric formula:

$\displaystyle bx^2+acx+c=0$

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

Use the quadratic formula to solve:

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

a = b $\displaystyle a = b$

b = ac $\displaystyle b = ac$

c = c $\displaystyle c = c$

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

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

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Example Question #1 : Using The Quadratic Formula

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

Possible Answers:

Between 3 and 3.5 seconds

Between 2 and 2.5 seconds

Between 3.5 and 4 seconds

Between 2.5 and 3 seconds

Between 4 and 4.5 seconds

Correct answer:

Between 3 and 3.5 seconds

Explanation:

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

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

Once in this form, we can use the Quadratic Formula to solve for $\displaystyle t$.

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

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

Plugging in our values:

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

Therefore $\displaystyle t=3.4375$ or -.4375 $\displaystyle -.4375$ 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 #1 : Using The Quadratic Formula

Example Question #1 : Using The Quadratic Formula

Example Question #1 : Using The Quadratic Formula

All High School Math Resources

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