High School Math : Finding Roots

Study concepts, example questions & explanations for High School Math

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Example Questions

Example Question #21 : Solving Quadratic Equations

Find the root(s) of the following quadratic polynomial. 

\(\displaystyle f(x) = x^{2} + 4x + 4\)

Possible Answers:

\(\displaystyle x=0\)

\(\displaystyle x=2, x=-2\)

\(\displaystyle x =- 2\)

\(\displaystyle x=2\)

Correct answer:

\(\displaystyle x =- 2\)

Explanation:

\(\displaystyle f(x) = x^{2} + 4x + 4\)

We set the function equal to 0 and factor the equation. By FOIL, we can confirm that \(\displaystyle (x+2)(x+2) = 0\) is equivalent to the given function. Thus, the only zero comes from\(\displaystyle x + 2 = 0\), and \(\displaystyle x = -2\). Thus, \(\displaystyle x = -2\) is the only root. 

Example Question #42 : Quadratic Equations And Inequalities

\(\displaystyle \textup{What are the possible values of }x \textup{ when }\; \; x^{2}-2x-15=0\:\textup{?}\)

Possible Answers:

\(\displaystyle -5. 3\)

\(\displaystyle 3, 5\)\(\displaystyle -5, -3\)

\(\displaystyle 3,5\)

\(\displaystyle -2, -15\)

\(\displaystyle -3, 5\)

Correct answer:

\(\displaystyle -3, 5\)

Explanation:

\(\displaystyle \textup{Factor to solve. The product of the constants in the two roots is -15, while }\)

\(\displaystyle \textup{their sum is -2. }\:\:\left ( x+3\right )\left ( x-5\right )=0\;\;\;\;x=-3 \textup{ or }x=5\)

Example Question #74 : Intermediate Single Variable Algebra

Solve the quadratic equation using any method:

\(\displaystyle bx^2+cx+a=0\)

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

Use the quadratic formula to solve:

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

\(\displaystyle a = b\)

\(\displaystyle b = c\)

\(\displaystyle c = a\)

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

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

Example Question #323 : Algebra Ii

Solve the following equation using the quadratic form:

\(\displaystyle x-4\sqrt{x}-32=0\)

Possible Answers:

\(\displaystyle x=49\)

\(\displaystyle x=36\)

\(\displaystyle x=16\)

\(\displaystyle x=64\)

\(\displaystyle x=25\)

Correct answer:

\(\displaystyle x=64\)

Explanation:

Factor and solve:

\(\displaystyle x-4\sqrt{x}-32=0\)

\(\displaystyle (\sqrt{x}-8)(\sqrt{x}+4)=0\)

\(\displaystyle \sqrt{x}=8\)

\(\displaystyle x=64\)

or

\(\displaystyle \sqrt{x}=-4\)

This has no solutions.

Therefore there is only one solution:

\(\displaystyle x=64\)

Example Question #21 : Finding Roots

Solve the following equation using the quadratic form:

\(\displaystyle x^4-12x^2+27=0\)

Possible Answers:

\(\displaystyle x=\pm 7, \pm \sqrt{7}\)

\(\displaystyle x=\pm 6, \pm \sqrt{6}\)

\(\displaystyle x=\pm 5, \pm \sqrt{5}\)

\(\displaystyle x=\pm 2, \pm \sqrt{2}\)

\(\displaystyle x=\pm 3, \pm \sqrt{3}\)

Correct answer:

\(\displaystyle x=\pm 3, \pm \sqrt{3}\)

Explanation:

Factor and solve:

\(\displaystyle x^4-12x^2+27=0\)

\(\displaystyle (x^2-9)(x^2-3)=0\)

\(\displaystyle x^2-9=0\)

\(\displaystyle (x-3)(x+3)=0\)

\(\displaystyle x= \pm3\)

or

\(\displaystyle x^2-3=0\)

\(\displaystyle x^2=3\)

\(\displaystyle x=\pm \sqrt{3}\)

Therefore the equation has four solutions:

\(\displaystyle x= \pm3, \pm \sqrt{3}\)

Example Question #22 : Solving Quadratic Equations

Solve the following equation using the quadratic form:

\(\displaystyle x^{\frac{2}{3}}-9x^{\frac{1}{3}}+20=0\)

Possible Answers:

\(\displaystyle x=120, 64\)

\(\displaystyle x=125, 64\)

\(\displaystyle x=125, 65\)

\(\displaystyle x=124, 64\)

\(\displaystyle x=124, 65\)

Correct answer:

\(\displaystyle x=125, 64\)

Explanation:

Factor and solve:

\(\displaystyle x^{\frac{2}{3}}-9x^{\frac{1}{3}}+20=0\)

\(\displaystyle (x^{\frac{1}{3}})^2-9x^{\frac{1}{3}}+20=0\)

\(\displaystyle (x^\frac{1}{3}-5)(x^\frac{1}{3}-4)=0\)

\(\displaystyle x^\frac{1}{3}=5\)

\(\displaystyle x=125\)

or

\(\displaystyle x^\frac{1}{3}=4\)

\(\displaystyle x=64\)

Therefore the equation has two solutions.

Example Question #21 : Solving Quadratic Equations

Solve the following equation using the quadratic form:

\(\displaystyle n^6+9n^3+8=0\)

Possible Answers:

\(\displaystyle n=-3,-2,2 \pm i\sqrt{3}, \frac{1 \pm i\sqrt{3}}{2}\)

\(\displaystyle n=-2,-1,1 \pm i\sqrt{3}, \frac{1 \pm i\sqrt{3}}{2}\)

\(\displaystyle n=-3,-1,1 \pm i\sqrt{3}, \frac{3 \pm i\sqrt{3}}{2}\)

\(\displaystyle n=-3,-1,1 \pm i\sqrt{3}, \frac{1 \pm i\sqrt{3}}{2}\)

\(\displaystyle n=-3,-2,1 \pm i\sqrt{3}, \frac{1 \pm i\sqrt{3}}{2}\)

Correct answer:

\(\displaystyle n=-2,-1,1 \pm i\sqrt{3}, \frac{1 \pm i\sqrt{3}}{2}\)

Explanation:

Factor and solve:

\(\displaystyle n^6+9n^3+8=0\)

\(\displaystyle (n^3)^2+9n^3+8=0\)

\(\displaystyle (n^3+8)(n^3+1)=0\)

\(\displaystyle (n+2)(n^2-2n+4)(n+1)(n^2-n+1)=0\)

Each of these factors gives solutions to the equation:

\(\displaystyle n=-2\)

\(\displaystyle n=-1\)

\(\displaystyle n=1 \pm i\sqrt{3}\)

\(\displaystyle n=\frac{1 \pm i\sqrt{3}}{2}\)

Example Question #22 : Finding Roots

The product of two consecutive positive numbers is \(\displaystyle 156\).  What is the sum of the two numbers?

Possible Answers:

\(\displaystyle 31\)

\(\displaystyle 23\)

\(\displaystyle 25\)

\(\displaystyle 27\)

\(\displaystyle 29\)

Correct answer:

\(\displaystyle 25\)

Explanation:

Let \(\displaystyle x=\) the first number and \(\displaystyle x+1=\) the second number.

The equation to sovle becomes \(\displaystyle x(x+1)=156\), or \(\displaystyle x^{2}+x-156=0\).

Factoring we get \(\displaystyle (x+13)(x-12)=0\), so the solution is \(\displaystyle x=-13 \; \; or\; \; x=12\).  The problem states that the numbers are positive, so the correct numbers are \(\displaystyle 12\) and \(\displaystyle 13\), which sum to \(\displaystyle 25\).

Example Question #332 : Algebra Ii

Two positive, consecutive odd numbers have a product of \(\displaystyle 143\).  What is their sum?

Possible Answers:

\(\displaystyle 28\)

\(\displaystyle 24\)

\(\displaystyle 32\)

\(\displaystyle 40\)

\(\displaystyle 36\)

Correct answer:

\(\displaystyle 24\)

Explanation:

Let \(\displaystyle x=\) first odd number and \(\displaystyle x+2=\) second odd number. Then:

\(\displaystyle x(x+2)=143\)

Use the distributive property and subtract \(\displaystyle 143\) from both sides to get \(\displaystyle x^{2}+2x-143=0\).

Factoring we get \(\displaystyle (x+13)(x-11)=0\).

Solving we get \(\displaystyle x+13=0\; or\; x-11=0\), so \(\displaystyle x=-13\; or\; x=11\).

The problem stated that the numbers were positive so the answer becomes \(\displaystyle 11+13=24\).

Example Question #51 : Quadratic Equations And Inequalities

Find the sum of the solutions to:

 \(\displaystyle \frac{-6}{x^2} + \frac{1}{x} + 1 = 0\)

Possible Answers:

\(\displaystyle -3\)

\(\displaystyle -1\)

\(\displaystyle 0\)

\(\displaystyle 2\)

\(\displaystyle 11\)

Correct answer:

\(\displaystyle -1\)

Explanation:

Multiply both sides of the equation by \(\displaystyle x^2\), to get

 \(\displaystyle -6 + x + x^2 = 0\)

 

This can be factored into the form

\(\displaystyle (x+3)(x-2) = 0\)

 

So we must solve 

\(\displaystyle x+3 = 0\) 

and

\(\displaystyle x-2 = 0\)

to get the solutions. 

 

The solutions are:

\(\displaystyle x =-3, 2\)

and their sum is  \(\displaystyle -1\) .

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