Solving Equations

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Problems about solving equations.

Jess is solving the equation $6x+13 = 25$ . Here is their work.

$25 - 13 = 12 \div 6 = 2$

What is the issue with this work?

The algebra is incorrect. The equals sign does not mean equal here. The solution is not related to the original equation. There is no issue with this work.

Complete the following sentences:

(a): Expressions have solutions values zeros roots .
(b): Equations have solutions values zeros roots .
(c): Polynomials have solutions values zeros roots .
(d): Functions have solutions zeros roots .

Correct!

Solutions of equations are the values of the variables that make the equation $\answer [format=string]{true}$ .

And zeros of functions are input values that give output values of $\answer {0}$ .

And these ideas are connected: The $\answer [format=string]{zeros}$ of a function $f$ are the $\answer [format=string]{solutions}$ to the equation $f(x)=0$ . And the $\answer [format=string]{roots}$ of a polynomial $p(x)$ are $\answer [format=string]{zeros}$ of the polynomial function $p$ .

Give a polynomial $p(x)$ whose leading coefficient is $1$ , and which has $x=12$ and $x=-1$ as roots (and no other roots).

$p(x) = \answer [given]{(x-12)(x+1)}$

Give a polynomial $p(x)$ whose leading coefficient is $3$ , and which has $x=\frac {2}{3}$ and $x=2$ as roots (and no other roots).

$p(x) = \answer [given]{(3x-2)(x-2)}$

Give a polynomial $p(x)$ of degree 3 whose leading coefficient is $1$ , and which has $x=-5$ as a root (and no other roots).

$p(x) = \answer [given]{(x+5)^3}$

Give a polynomial $p(x)$ of degree 4 whose leading coefficient is $1$ , and which has $x=8$ , $x=1+\sqrt {3}$ and $x = 1-\sqrt {3}$ as roots (and no other roots).

$p(x) = \answer [given]{(x-8)^2(x-(1+\sqrt {3}))(x-(1-\sqrt {3}))}$

Solve the problem below by completing the square. Practice drawing a diagram to help! Enter your answers from least to greatest. $x^2 + 8x = 20$ To complete the square, add $\answer {16}$ to both sides.

Then take the $\answer [format=string]{square root}$ (two words) of both sides, to yield $\answer {x+4} = \pm \answer {\sqrt {20+16}}$ .

So the solutions are, from least to greatest, $\answer [given]{-10}$ , $\answer [given]{2}$ .

Solve the problem below by completing the square. Practice drawing a diagram to help! Enter your answers from least to greatest. $x^2 + 3x = 5$ To complete the square, add $\answer {\frac {9}{4}}$ to both sides.

Then take the $\answer [format=string]{square root}$ (two words) of both sides, to yield $\answer {x+\frac {3}{2}} = \pm \answer {\sqrt {5+\frac {9}{4}}}$ .

So the solutions are, from least to greatest, $\answer [given]{-\frac {3}{2}-\frac {\sqrt {29}}{2}}$ , $\answer [given]{-\frac {3}{2}+\frac {\sqrt {29}}{2}}$ .

According to the Fundamental Theorem of Algebra, how many roots should the polynomial $p(x) = x^4 - 3x^3 + x - 2$ have? $\answer [given]{4}$

Remember that the Fundamental Theorem of Algebra counts complex roots (real and imaginary), and also repeated roots.

According to the Fundamental Theorem of Algebra, how many roots should the polynomial $p(x) = x^{16} - 1$ have? $\answer [given]{16}$

State the Polynomial Division Theorem:

Given a polynomial $p(x)$ and a divisor $d(x)\ne 0$ , there exist polynomials $q(x)$ and $r(x)$ such that $p(x) = \answer {q(x)}\cdot d(x)+\answer {r(x)}$ with the degree of $r(x)$ the degree of $d(x)$ .

By the division theorem for polynomials, given $p(x)$ and a linear polynomial $(x-a)$ , there exists a quotient polynomial $q(x)$ and a remainder $r$ so that $p(x)= q(x)\cdot \left (\answer {x-a}\right ) + \answer {r}$ .

From this, the Remainder Theorem states that $p(a)=\answer {r}$ . And the Factor Theorem states that $(x-a)$ is a factor of $p(x)$ exactly when $p(a)=\answer {0}$ .

The Rational Root Theorem says that if $\pm \frac {a}{b}$ (written in lowest terms) is a root of a polynomial with counting number integer rational real coefficients, then $a$ must be a factor of the leading quadratic linear constant coefficient, and $b$ must be a factor of the leading quadratic linear constant coefficient.

By the Rational Root Theorem, which of the following could be rational roots of $p(x) = x^3 + 2x^2 - 8x + 4$ ? (Do not solve this problem by plugging the values into the polynomial!)

$1$ $-1$ $\frac {3}{2}$ $-\frac {2}{3}$ $-4$ $\frac {1}{4}$

By the Rational Root Theorem, which of the following could be rational roots of $p(x) = 12x^8 + 5x^7 + 3x^5 + 14x^3 - x + 20$ ? (Do not solve this problem by plugging the values into the polynomial!)

$3$ $-1$ $\frac {5}{7}$ $-\frac {5}{3}$ $-4$ $\frac {1}{4}$

Find all solutions to the equation $2x^3-6x^2+x+6=0$ .

The Rational Root Theorem combined with some division of polynomials might help!

Enter your answers in order from least to greatest. $\answer [given]{(1-\sqrt {7})/2}$ , $\answer [given]{(1+\sqrt {7})/2}$ , $\answer [given]{2}$ .

Solve the equation $a^2+3ba = 4b - c$ for $b$ .

How many solutions? $\answer {1}$ .

Correct! The equation is a $\answer [format=string]{linear}$ equation in $b$ .

The solution is $b=\answer {\frac {a^2 + c}{4 - 3 a}}$ , as long as $a\ne \answer {\frac {4}{3}}$ .

Solve the equation $a^2+3ba = 4b - c$ for $a$ .

How many solutions? $\answer {2}$ .

Correct! There are two solutions because the equation is $\answer [format=string]{quadratic}$ in $a$ .

Here the solutions are $a=\answer {-\frac {3b}{2}}\pm \answer {\frac {\sqrt {(3b)^2 + 4(4b-c)}}{2}}$ .