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High School Math : Solving Quadratic Equations

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Example Questions

Example Question #1 : Completing The Square

$f(x)=x^{2}-4x+7$

Find the vertex of the parabola by completing the square.

Possible Answers:

$\small (7,0)$

$\small (2,3)$

$\small (1,4)$

$\small (3,2)$

$\small (0,7)$

Correct answer:

$\small (2,3)$

Explanation:

To find the vertex of a parabola, we must put the equation into the vertex form:

$\small f(x)=a[b(x-h)]^{2}+k$

The vertex can then be found with the coordinates (h, k).

To put the parabola's equation into vertex form, you have to complete the square. Completing the square just means adding the same number to both sides of the equation -- which, remember, doesn't change the value of the equation -- in order to create a perfect square.

Start with the original equation:

$f(x)=x^{2}-4x+7$

Put all of the terms on one side:

$f(x)-7=x^{2}-4x$

Now we know that we have to add something to both sides in order to create a perfect square:

$f(x)+7+\square =x^{2}-4x+\square$

In this case, we need to add 4 on both sides so that the right-hand side of the equation factors neatly.

$f(x)-7+4=x^{2}-4x+4$

Now we factor:

$f(x)-3=(x-2)^{2}$

Once we isolate f(x) , we have the equation in vertex form:

$f(x)=(x-2)^{2}+3$

Thus, the parabola's vertex can be found at $\small (2,3)$ .

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Example Question #2 : Completing The Square

Complete the square:

2x^2+3x+2=0

Possible Answers:

$x=\frac{-2 \pm i\sqrt{7}}{4}$

$x=\frac{-2 \pm i\sqrt{7}}{3}$

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

$x=\frac{-1 \pm i\sqrt{7}}{4}$

$x=\frac{-3 \pm i\sqrt{7}}{3}$

Correct answer:

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

Explanation:

Begin by dividing the equation by and subtracting from each side:

2x^2+3x+2=0

$x^2+\frac{3}{2}x = -1$

Square the value in front of the and add to each side:

$x^2+\frac{3}{2}x +\frac{9}{16} = -1+\frac{9}{16}$

Factor the left side of the equation:

$(x+\frac{3}{4})^2 = \frac{-7}{16}$

Take the square root of both sides and simplify:

$x+\frac{3}{4} = \frac{\pm \sqrt{-7}}{4}$

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

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Example Question #3 : Completing The Square

Use factoring to solve the quadratic equation:

6x^2+7x-3=0

Possible Answers:

$x=-\frac{3}{2},\frac{1}{3}$

$x=-\frac{3}{2},\frac{1}{4}$

$x=-\frac{3}{4},\frac{1}{3}$

$x=-\frac{3}{4},\frac{1}{4}$

$x=-\frac{3}{2},\frac{2}{3}$

Correct answer:

$x=-\frac{3}{2},\frac{1}{3}$

Explanation:

Factor and solve:

6x^2+7x-3=0

6x^2 +9x-2x-3=0

Factor like terms:

3x(2x+3)-1(2x+3)=0

Combine like terms:

(2x+3)(3x-1)=0

$x=-\frac{3}{2},\frac{1}{3}$

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Example Question #4 : Completing The Square

Complete the square:

2n^2+n-21=0

Possible Answers:

$n=-\frac{5}{2},3$

$n=-\frac{7}{2},3$

$n=-\frac{7}{2},4$

$n=-\frac{9}{2},3$

$n=-\frac{7}{2},2$

Correct answer:

$n=-\frac{7}{2},3$

Explanation:

Begin by dividing the equation by and adding 10.5 to each side:

2n^2+n-21=0

$n^2+\frac{1}{2}n = \frac{21}{2}$

Square the value in front of the and add to each side:

$n^2+\frac{1}{2}n + \frac{1}{16} = \frac{21}{2} + \frac{1}{16}$

Factor the left side of the equation:

$(n+\frac{1}{4})^2 = \frac{169}{16}$

Take the square root of both sides and simplify:

$n+\frac{1}{4} = \frac{\pm 13}{4}$

$n = \frac{-1 \pm 13}{4}$

$n=-\frac{7}{2},3$

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Example Question #5 : Completing The Square

Complete the square:

3x^2+4x+2=0

Possible Answers:

$x=\frac{-4 \pm i\sqrt{2}}{3}$

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

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

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

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

Correct answer:

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

Explanation:

Begin by dividing the equation by and subtracting $\frac{2}{3}$ from each side:

3x^2+4x+2=0

$x^2+\frac{4}{3}x = -\frac{2}{3}$

Square the value in front of the and add to each side:

$x^2+\frac{4}{3}x +\frac{4}{9} = -\frac{2}{3}+\frac{4}{9}$

Factor the left side of the equation:

$(x+\frac{2}{3})^2 = \frac{-2}{9}$

Take the square root of both sides and simplify:

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

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

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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 #94 : Intermediate Single Variable Algebra

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 2.5 and 3 seconds

Between 2 and 2.5 seconds

Between 3.5 and 4 seconds

Between 3 and 3.5 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 .

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 : Solving Quadratic Equations

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #1 : Completing The Square

Example Question #2 : Completing The Square

Example Question #3 : Completing The Square

Example Question #4 : Completing The Square

Example Question #5 : Completing The Square

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 #94 : Intermediate Single Variable Algebra

All High School Math Resources

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