High School Math : Using the Quadratic Formula

Study concepts, example questions & explanations for High School Math

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Example Questions

Example Question #341 : Algebra Ii

Solve using the quadratic formula:

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

Use the quadratic formula to solve:

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

\displaystyle a = 2

\displaystyle b = 2

\displaystyle c = 3

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

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

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

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

Example Question #91 : Intermediate Single Variable Algebra

Solve using the quadratic formula:

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

Use the quadratic formula to solve:

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

\displaystyle a = 2

\displaystyle b = -5

\displaystyle c = 4

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

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

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

Example Question #341 : Algebra Ii

Solve using the quadratic formula:

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

Use the quadratic formula to solve:

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

\displaystyle a = a

\displaystyle b = b

\displaystyle c = 3b

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

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

Example Question #61 : Quadratic Equations And Inequalities

Solve using the quadratric formula:

\displaystyle bx^2+acx+c=0

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

Use the quadratic formula to solve:

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

\displaystyle a = b

\displaystyle b = ac

\displaystyle c = c

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

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

Example Question #1601 : Algebra Ii

A baseball that is thrown in the air follows a trajectory of \displaystyle h(t)= -4t^2 +12t+6, where \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 2.5 and 3 seconds

 Between 4 and 4.5 seconds 

Between 2 and 2.5 seconds

Between 3.5 and 4 seconds

Between 3 and 3.5 seconds

Correct answer:

Between 3 and 3.5 seconds

Explanation:

To solve this, we look at the equation \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 

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

Plugging in our values:

 \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 \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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