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Algebra II : Solving Quadratic Equations

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Example Questions

Example Question #311 : Quadratic Equations And Inequalities

Solve the equation using the quadratic formula.

$x^2+5x+\frac{9}{4}=0$

Possible Answers:

7,9

$\frac{1}{2}, \frac{9}{2}$

-5,2

$-\frac{1}{2}, -\frac{9}{2}$

Correct answer:

$-\frac{1}{2}, -\frac{9}{2}$

Explanation:

The quadratic formula is

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

Setting a=1 , b=5 , $c=\frac{9}{4}$ yields,

$\\ \frac{-5\pm\sqrt{5^2-4(1)(\frac{9}{4})}}{2(1)}\\ \\=\frac{-5\pm\sqrt{25-9}}{2} \\ \\=-\frac{5}{2}\pm2 \\ \\ = -\frac{1}{2}, -\frac{9}{2}$

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Example Question #481 : Intermediate Single Variable Algebra

Find the roots of: y=x^2-x+100

Possible Answers:

$\pm10$

Since the quadratic cannot be factored, there are no roots.

$\frac{1\pm \sqrt{401}}{2}$

$\frac{1\pm i\sqrt{399}}{2}$

$\frac{1}{2}\pm i\sqrt{399}$

Correct answer:

$\frac{1\pm i\sqrt{399}}{2}$

Explanation:

Identify the values of , , and in the standard form of the parabola.

y=ax^2+bx+c

a= 1

b=-1

c=100

Calculate the discriminant.

b^2-4ac=1-4(1)(100)=-399

Since the discriminant is less than zero, the quadratic is irreducible and there are no real roots. However, there are complex roots. Use the quadratic formula to determine the complex roots.

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

$x= \frac{1\pm \sqrt{-399}}{2(1)}$

$x= \frac{1\pm i\sqrt{399}}{2}$

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Example Question #482 : Intermediate Single Variable Algebra

Use the quadratic formula to find the roots of x^2+5x-14 .

Possible Answers:

x= 0

x=5 and x=-14

x=7 and x=-2

no solution

x=-7 and x=2

Correct answer:

x=-7 and x=2

Explanation:

The parent function of a quadratic is represented as ax^2+bx+c . The quadratic formula is $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ . In this case a=1 , b=5 , and c=-14 . Replacing these values into the quadratic forumula will give you the solutions to the quadratic.

$x=\frac{-5\pm \sqrt{5^2-4(1)(-14)}}{2*1}$

$x=\frac{-5\pm \sqrt{25-(-56)}}{2}$

$x=\frac{-5\pm \sqrt{81}}{2}$

$x=\frac{-5\pm 9}{2}$

x=-7 and x=2

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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 #1621 : Algebra Ii

Solve for .

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

Possible Answers:

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

x=4

x=-2

x=2

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

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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Algebra II : Solving Quadratic Equations

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #311 : Quadratic Equations And Inequalities

Example Question #481 : Intermediate Single Variable Algebra

Example Question #482 : Intermediate Single Variable Algebra

Example Question #483 : Intermediate Single Variable Algebra

Example Question #1621 : Algebra Ii

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

All Algebra II Resources

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