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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #483 : Intermediate Single Variable Algebra

Solve this quadratic equation by using the quadratic formula: x^2+4x+2

Possible Answers:

$x={-2\pm \sqrt{2}$

$x=\frac{4\pm \sqrt{8}}{2}$

$x=\frac{-4\pm \sqrt{8}}{8}$

$x=\frac{-4\pm \sqrt{8}}{1}$

$x=\frac{-4\pm \sqrt{8}}{4}$

Correct answer:

$x={-2\pm \sqrt{2}$

Explanation:

You must know the quadratic equation $x=\frac{-b\pm \sqrt{b^2-4ac}}{2}$ .

To plug in the right terms, recognize that polynomials in standard form are symbolized as ax^2+bx+c .

Plug in the values from your equation

$x=\frac{-4\pm \sqrt{4^2-4*1*2}}{2}$

simplify within the radical:

$x=\frac{-4\pm \sqrt{8}}{2}$

Simplify the radical:

$x=\frac{-4\pm2 \sqrt{2}}{2}$

Reduce:

$x={-2\pm \sqrt{2}$

Note that this represents two values since there is a $\pm$ in the equation. One is solved with an addition sign and the other is solved with a subtraction sign to yield two answers or roots where this equation crosses the x axis.

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Example Question #161 : Solving Quadratic Equations

Solve for .

$x^{2}-4x-4=0$

Possible Answers:

x=-2

$x=2\ \textup{or}\ x=-2$

x=4

$x=2\pm 2\sqrt{2}$

x=2

Correct answer:

$x=2\pm 2\sqrt{2}$

Explanation:

1) Begin the problem by factoring the final term. Include the negative when factoring.

–2 + 2 = 0

–4 + 1 = –3

–1 + 4 = 3

All options are exhausted, therefore the problem cannot be solved by factoring, which means that the roots either do not exist or are not rational numbers. We must use the quadratic formula.

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

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

$x=\frac{4\pm \sqrt{16+16}}{2}$

$x=\frac{4\pm \sqrt{32}}{2}$

$x=\frac{4\pm \sqrt{16\cdot 2}}{2}=\frac{4\pm \sqrt{16}\cdot \sqrt{2}}{2}=\frac{4\pm4\sqrt{2}}{2}$

$x=2\pm2\sqrt{2}$

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Example Question #484 : Intermediate Single Variable Algebra

Solve the quadratic equation:

$2x^{2}+12x+15 = 0$

Possible Answers:

x=-3,x=5

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

$x=-3\pm \sqrt{6}$

$x=-3\pm \frac{\sqrt{6}}2{}$

$x=\pm \frac{5}{2}$

Correct answer:

$x=-3\pm \frac{\sqrt{6}}2{}$

Explanation:

The standard form of a quadratic equation is $ax^{2}+bx+c=0$ , where a,b, and c are constants. Plug these constants into the quadratic formula to solve for x.

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

$x=\frac{-12\pm \sqrt{(12)^{2}-4(2)(15)}}{2(2)}$

$x=\frac{-12\pm \sqrt{144-120}}{4}=\frac{-12\pm \sqrt{24}}{4}$

$x=\frac{-12\pm 2\sqrt{6}}{4}$ $x=-3\pm \frac{\sqrt{6}}2{}$

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Example Question #1621 : Algebra Ii

Solve for by using the quadratic formula:

$y=x^{2}+14x+2$

Possible Answers:

$\frac{{-14\pm\sqrt{47}}{}}{2}$

${-7\pm\sqrt{47}}{}$

${-14\pm\sqrt{47}}{}$

None of the above

${-14\pm\sqrt{188}}{}$

Correct answer:

${-7\pm\sqrt{47}}{}$

Explanation:

The quadratic equations is:

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

From here you just plug in your numbers, so:

$x=\frac{-14\pm{\sqrt{14^2-4(1)(2)}}{}}{2(1)}$

Simplify:

$\frac{-14\pm \sqrt{188}}{2}$

Then you need to simplify the inside looking for perfect squares:

$\frac{-14\pm \sqrt{4}*\sqrt{47}}{2}=\frac{-14\pm 2\sqrt{47}}{2}$

Each term is divisible by 2, so your final answer is:

${-7\pm \sqrt{47}$

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Example Question #28 : Quadratic Formula

Use the quadratic formula to solve the equation 3x^2-7x+11=0

Possible Answers:

$x=\frac{7\pm i\sqrt{181}}{6}$

$x=\frac{7\pm \sqrt{181}}{6}$

$x=\frac{7\pm i\sqrt{83}}{6}$

$x=\frac{-7\pm i\sqrt{83}}{6}$

$x=\frac{-7\pm i\sqrt{181}}{6}$

Correct answer:

$x=\frac{7\pm i\sqrt{83}}{6}$

Explanation:

a=3 , b=-7 , c=11

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

$x=\frac{7\pm \sqrt{(-7)^2-4(3)(11)}}{2(3)}$

$x=\frac{7\pm \sqrt{49-132}}{6}$

$x=\frac{7\pm \sqrt{-83}}{6}$

$x=\frac{7\pm i\sqrt{83}}{6}$

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Example Question #29 : Quadratic Formula

Find the zeros of 2x^2+3x+1 ?

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

This specific function cannot be factored, so use the quadratic equation:

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

Our function is in the form ax^2+bx+c where,

a=2, b=3, c=1

Therefore the quadratic equation becomes,

$\frac{-3\pm \sqrt{3^2 -4(2)(1)}}{2(2)}$

$\frac{-3\pm \sqrt{9 -8}}{4}$

$\frac{-3\pm \sqrt{1}}{4}$

$\frac{-3+1}{4}$ OR $\frac{-3-1}{4}$

$\frac{-2}{4}$ OR $\frac{-4}{4}$

$x=-\frac{1}{2}$ OR x=-1

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Example Question #30 : Quadratic Formula

Find the roots of 2x^2+4x+4 .

Possible Answers:

no real solutions

x=0,-1

x=-1,1

x=1,2

x=0,1

Correct answer:

no real solutions

Explanation:

Use the quadratic equation:

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

Since the original equation is in standard form, ax^2+bx+c where

a=2, b=4, c=4 .

Therefore,

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

$x=\frac{-4\pm\sqrt{16-32}}{4}$

$x=\frac{-4\pm\sqrt{-16}}{4}$

$x=\frac{-4\pm4i}{4}$

$x=-1 \pm i$

because the value of the discriminant (the component beneath the square root) is negative, this function has no real solutions.

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Example Question #31 : Quadratic Formula

Solve $-3x^2+6x-12{}$ .

Possible Answers:

$x=6;x=-\frac{2}{3}$

x=1;x=-1

x=3;x=2

x=-3

No real solutions

Correct answer:

No real solutions

Explanation:

-3x^2+6x-12=0

This function cannot be factor therefore, use the quadratic equation.

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

Since the original equation is in the form ax^2+bx+c where

a=-3, b=6, c=-12 .

Therefore,

$\frac{-6\pm\sqrt{6^2-4(-3)(-12)}}{2(-3)}$

$\frac{-6\pm\sqrt{36-144}}{-6}$

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

Since the value of the discriminant (the value beneath the square root symbol) is negative, this function has no real solutions.

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Example Question #172 : Solving Quadratic Equations

Solve x^2+8x-2=0 .

Possible Answers:

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

$x={4\pm {3\sqrt{2}}}$

$x={-4\pm {3\sqrt{6}}}$

$x={-4\pm {6\sqrt{2}}}$

$x={-8\pm {3\sqrt{2}}}$

Correct answer:

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

Explanation:

x^2+8x-2=0

This particular function cannot be factored therefore, use the quadratic formula to solve.

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

Since the function is in the form ax^2+bx+c where

a=1,b=8, c=-2

the quadratic formula becomes as follows.

$\frac{-8\pm {\sqrt{8^2-4(1)(-2)}}}{2(1)}$

$\frac{-8\pm {\sqrt{64+8}}}{2}$

$\frac{-8\pm {\sqrt{72}}}{2}$

$\frac{-8\pm {\sqrt{9}\times\sqrt{4}\times\sqrt{2}}}{2}$

$\frac{-8\pm {3\times2\times\sqrt{2}}}{2}$

${-4\pm {3\sqrt{2}}}$

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Example Question #1621 : Algebra Ii

Use the quadratic formula to solve for x:

4x+x^2+2

Possible Answers:

$-4\pm\sqrt{8}$

$-4\pm2\sqrt{2}$

$-4\pm\sqrt{2}$

$-2\pm\sqrt{2}$

${-2\pm2\sqrt{2}}$

Correct answer:

$-2\pm\sqrt{2}$

Explanation:

To solve this problem, you must first rewrite the equation into ax^2+bx+c form (quadratic form).

After this you plug the numbers into the following quadratic equation:

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

Which upon doing you get:

$x=\frac{-4\pm \sqrt{16-8}}{2}$

This simplifies to:

$x=-2\pm\sqrt{2}$

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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #483 : Intermediate Single Variable Algebra

Example Question #161 : Solving Quadratic Equations

Example Question #484 : Intermediate Single Variable Algebra

Example Question #1621 : Algebra Ii

Example Question #28 : Quadratic Formula

Example Question #29 : Quadratic Formula

Example Question #30 : Quadratic Formula

Example Question #31 : Quadratic Formula

Example Question #172 : Solving Quadratic Equations

Example Question #1621 : Algebra Ii

All Algebra II Resources

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