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## How do I solve 25x^2 - 20x = 11 by completing the square?

AJ Speller
6.652054794520548 years ago

Completing the square is method of solving a quadratic equation that involves finding a value to add to both the left and right side of the equation. This value has the extra benefit of making one side of the equation a perfect trinomial which makes the function easier to identify and/or graph.

Let's begin by factoring $25$ from the left side of the equation.

$25 \left({x}^{2} - \frac{20}{25} x\right) = 11$

Take the coefficient of the $x$ term and divide it by $2$ and square it

${\left(\frac{- \frac{20}{25}}{2}\right)}^{2} = {\left(- \frac{20}{25} \cdot \frac{1}{2}\right)}^{2} = {\left(- \frac{10}{25}\right)}^{2} = \frac{100}{625}$

$\frac{100}{625}$ This the number you add to the left side

$25 \left(\frac{100}{625}\right)$ Is added to the right side because we initially factored out 25 from the left side.

We added these values but the equation remains balanced because they are added to both sides of the equation.

$25 \left({x}^{2} - \frac{20}{25} x + \frac{100}{625}\right) = 11 + 25 \left(\frac{100}{625}\right)$

$25 \left({x}^{2} - \frac{20}{25} x + \frac{100}{625}\right) = 11 + \frac{100}{25}$

$25 \left({x}^{2} - \frac{20}{25} x + \frac{100}{625}\right) = 11 + 4$

$25 \left({x}^{2} - \frac{20}{25} x + \frac{100}{625}\right) = 15$

$\left({x}^{2} - \frac{20}{25} x + \frac{100}{625}\right) = \frac{15}{25}$

${\left(x - \frac{10}{25}\right)}^{2} = \frac{15}{25}$

$\sqrt{{\left(x - \frac{10}{25}\right)}^{2}} = \sqrt{\frac{15}{25}}$

$\left(x - \frac{10}{25}\right) = \sqrt{\frac{15}{25}}$

$x = \sqrt{\frac{15}{25}} + \frac{10}{25}$

$x = \frac{\sqrt{15}}{5} + \frac{10}{25}$

$x = \frac{\sqrt{15}}{5} + \frac{2}{5}$

$x = \frac{\sqrt{15} + 2}{5}$

## How do you determine if -10,20,-40,80 is an arithmetic or geometric sequence?

Don't Memorise
5.865753424657535 years ago

The sequence is a geometric sequence.

#### Explanation:

• In an Arithmetic sequence there is a common difference $d$ between any two consecutive terms
• In a Geometric sequence there is a common ratio $r$ for any two consecutive terms

The sequence given is :

$- 10 , 20 , - 40 , 80$

1) Checking if the sequence is an arithmetic sequence:

 color(blue)(d _1= a_2 - a_1) = 20 - (-10) = color(blue)(30
 d_2 = a_3 - a_2 = -40 - (20) =color(blue)( -60

As observed ${d}_{1} \ne {d}_{2}$ , so it is not an arithmetic sequence.

2) Checking if the sequence is a geometric sequence:

 color(blue)(r _1= a_2/ a_1) = 20 / -10 = color(blue)(-2

 color(blue)(r _2 = a_3/ a_2) = (-40 )/ 20 = color(blue)(-2

 color(blue)(r _3 = a_4/ a_3) = (80 )/-40 = color(blue)(-2

Since ${r}_{1} = {r}_{2} = r 3$ it forms a geometric sequence.

So the sequence is a geometric sequence.

## What is a piecewise continuous function?

George C.
5.7315068493150685 years ago

A piecewise continuous function is a function that is continuous except at a finite number of points in its domain.

#### Explanation:

Note that the points of discontinuity of a piecewise continuous function do not have to be removable discontinuities. That is we do not require that the function can be made continuous by redefining it at those points. It is sufficient that if we exclude those points from the domain, then the function is continuous on the restricted domain.

For example, consider the function:

$s \left(x\right) = \left\{\begin{matrix}- 1 & \text{if x < 0" \\ 0 & "if x = 0" \\ 1 & "if x > 0}\end{matrix}\right.$

graph{(y - x/abs(x))(x^2+y^2-0.001) = 0 [-5, 5, -2.5, 2.5]}

This is continuous for all $x \in \mathbb{R}$ except $x = 0$

The discontinuity at $x = 0$ is not removable. We cannot redefine $s \left(x\right)$ at that point and get a continuous function.

At $x = 0$ the graph of the function 'jumps'. More formally, in the language of limits we find:

${\lim}_{x \to 0 +} s \left(x\right) = 1$

${\lim}_{x \to 0 -} s \left(x\right) = - 1$

So the left limit and right limit disagree with one another and with the value of the function at $x = 0$.

If we exclude the finite set of discontinuities from the domain, then the function restricted to this new domain will be continuous.

In our example, the definition of $s \left(x\right)$ as a function from $\left(- \infty , 0\right) \cup \left(0 , \infty\right) \to \mathbb{R}$ is continuous.

If we graph $s \left(x\right)$ restricted to this domain, it still looks like it is discontinuous at $0$, but $0$ is not part of the domain, so the 'jump' there is irrelevant. At any point, arbitrarily close to $0$, we can choose a little open interval around it in which the function is (constant and therefore) continuous.

Slightly confusingly, the function $\tan \left(x\right)$ is considered continuous - rather than piecewise continuous, because the asymptotes at $x = \frac{\pi}{2} + n \pi$ are excluded from the domain.

graph{tan(x) [-10.06, 9.94, -4.46, 5.54]}

Meanwhile, the sawtooth function $f \left(x\right) = x - \left\lfloor x \right\rfloor$ is not considered piecewise continuous as a function from $\mathbb{R}$ to $\mathbb{R}$, but is piecewise continuous on any finite open interval.

graph{3/5(abs(sin(x * pi/2))-abs(cos(x * pi/2))-abs(sin(x * pi/2)^3)/6+abs(cos(x * pi/2)^3)/6) * tan(x * pi/2)/abs(tan(x * pi/2))+1/2 [-2.56, 2.44, -0.71, 1.79]}

## What is the formula for time from a changing velocity?

Costas C.
5.33972602739726 years ago

$t = \frac{u - {u}_{0}}{a}$

$s = {u}_{0} \cdot t + \frac{1}{2} a {t}^{2}$ (Need to solve quadratic)

#### Explanation:

Via changing velocity I pressume you mean an object that accelerates or decelerates.

If acceleration is constant

If you have initial and final speed:

a=(Δu)/(Δt)

$a = \frac{u - {u}_{0}}{t - {t}_{0}}$

Usually ${t}_{0} = 0$, so:

$t = \frac{u - {u}_{0}}{a}$

If the above method does not work because you are missing some values, you can use the equation below. The distance traveled $s$ can be given from:

$s = {u}_{0} \cdot t + \frac{1}{2} a {t}^{2}$

where ${u}_{0}$ is the initial speed
$t$ is the the time
$a$ is the acceleration (note this value is negative if the case is a deceleration)

Therefore, if you know the distance, initial speed and acceleration you can find the time by solving the quadratic equation that is formed. However, if acceleration if not given, you will need the final speed of the object $u$ and can use the formula:

$u = {u}_{0} + a t$

$u - {u}_{0} = a t$

$a = \frac{u - {u}_{0}}{t}$

and substitute to the distance equation, making it:

$s = {u}_{0} \cdot t + \frac{1}{2} \cdot \frac{u - {u}_{0}}{t} \cdot {t}^{2}$

$s = {u}_{0} \cdot t + \frac{1}{2} \cdot \left(u - {u}_{0}\right) \cdot t$

Factor $t$:

$s = t \cdot \left({u}_{0} + \frac{1}{2} \cdot \left(u - {u}_{0}\right)\right)$

$t = \frac{s}{{u}_{0} + \frac{1}{2} \cdot \left(u - {u}_{0}\right)}$

So you got 2 equations. Pick one of them, which will help you solve with the data you are given:

$s = {u}_{0} \cdot t + \frac{1}{2} a {t}^{2}$

$t = \frac{s}{{u}_{0} + \frac{1}{2} \cdot \left(u - {u}_{0}\right)}$

Below are two other cases where acceleration is not constant. FEEL FREE TO IGNORE THEM if acceleration in your case is constant, since you placed it in the Precalculus category and the below contain calculus.

If acceleration is a function of time $a = f \left(t\right)$

The definition of acceleration:

$a \left(t\right) = \frac{\mathrm{du}}{\mathrm{dt}}$

$a \left(t\right) \mathrm{dt} = \mathrm{du}$

${\int}_{0}^{t} a \left(t\right) \mathrm{dt} = {\int}_{{u}_{0}}^{u} \mathrm{du}$

${\int}_{0}^{t} a \left(t\right) \mathrm{dt} = u - {u}_{0}$

$u = {u}_{0} + {\int}_{0}^{t} a \left(t\right) \mathrm{dt}$

If you still don't have enough to solve, that means you have to go to distance. Just use the definition of speed and move on, as if I analyze it further it will only confuse you:

$u \left(t\right) = \frac{\mathrm{ds}}{\mathrm{dt}}$

The second part of this equation means integrading acceleration with respect to time. Doing that gives an equation with only $t$ as the unknown value.

If acceleration is a function of speed $a = f \left(u\right)$

The definition of acceleration:

$a \left(u\right) = \frac{\mathrm{du}}{\mathrm{dt}}$

$\mathrm{dt} = \frac{\mathrm{du}}{a \left(u\right)}$

${\int}_{0}^{t} \mathrm{dt} = {\int}_{{u}_{0}}^{u} \frac{\mathrm{du}}{a \left(u\right)}$

$t - 0 = {\int}_{{u}_{0}}^{u} \frac{\mathrm{du}}{a \left(u\right)}$

$t = {\int}_{{u}_{0}}^{u} \frac{\mathrm{du}}{a \left(u\right)}$

## How do you write the partial fraction decomposition of the rational expression (5x - 1) / ((x - 2)(x + 1))?

Zack M.
5.336986301369863 years ago

$\frac{5 x - 1}{\left(x - 2\right) \left(x + 1\right)} = \frac{3}{x - 2} + \frac{2}{x + 1}$

#### Explanation:

The partial fraction decomposition suggests that the function can be broken down into the sum of two other functions, or;

$\frac{5 x - 1}{\left(x - 2\right) \left(x + 1\right)} = \frac{A}{x - 2} + \frac{B}{x + 1}$

Where we need to solve for $A$ and $B$. We can cross multiply to combine the terms on the right hand side over a common denominator. We get;

$\frac{5 x - 1}{\left(x - 2\right) \left(x + 1\right)} = \frac{A \left(x + 1\right) + B \left(x - 2\right)}{\left(x - 2\right) \left(x + 1\right)}$

We can now cancel the denominator on each side, leaving;

$5 x - 1 = A \left(x + 1\right) + B \left(x - 2\right)$

Now we can solve for $A$ and $B$. We can make one of the terms cancel out by choosing the right value for $x$. Lets try x=~1.

5(~1)-1 = A(~1+1) + B(~1-2)

The $A$ term goes away since it is multiplied by zero, leaving;

~6 = ~3B

Solving for $B$;

$B = 2$

We can substitute $B$ and solve for $A$, but it would be easier to do the same trick that we used to solve for $B$. Let $x = 2$.

$5 \left(2\right) - 1 = A \left(2 + 1\right) + B \left(2 - 2\right)$

This time, the $B$ term goes away;

$9 = 3 A$

$A = 3$

Now that we have our values for $A$ and $B$ we can plug into our first function and get;

$\frac{5 x - 1}{\left(x - 2\right) \left(x + 1\right)} = \frac{3}{x - 2} + \frac{2}{x + 1}$

## What are the possible number of positive, negative, and complex zeros of f(x) = –3x^4 – 5x^3 – x^2 – 8x + 4?

George C.
5.3342465753424655 years ago

Look at changes of signs to find this has $1$ positive zero, $1$ or $3$ negative zeros and $0$ or $2$ non-Real Complex zeros.

Then do some sums...

#### Explanation:

$f \left(x\right) = - 3 {x}^{4} - 5 {x}^{3} - {x}^{2} - 8 x + 4$

Since there is one change of sign, $f \left(x\right)$ has one positive zero.

$f \left(- x\right) = - 3 {x}^{4} + 5 {x}^{3} - {x}^{2} + 8 x + 4$

Since there are three changes of sign $f \left(x\right)$ has between $1$ and $3$ negative zeros.

Since $f \left(x\right)$ has Real coefficients, any non-Real Complex zeros will occur in conjugate pairs, so $f \left(x\right)$ has exactly $1$ or $3$ negative zeros counting multiplicity, and $0$ or $2$ non-Real Complex zeros.

$f ' \left(x\right) = - 12 {x}^{3} - 15 {x}^{2} - 2 x - 8$

Newton's method can be used to find approximate solutions.

Pick an initial approximation ${a}_{0}$.

Iterate using the formula:

${a}_{i + 1} = {a}_{i} - f \frac{{a}_{i}}{f ' \left({a}_{i}\right)}$

Putting this into a spreadsheet and starting with ${a}_{0} = 1$ and ${a}_{0} = - 2$, we find the following approximations within a few steps:

$x \approx 0.41998457522194$

$x \approx - 2.19460208831628$

We can then divide $f \left(x\right)$ by $\left(x - 0.42\right)$ and $\left(x + 2.195\right)$ to get an approximate quadratic $- 3 {x}^{2} + 0.325 x - 4.343$ as follows:

Notice the remainder $0.013$ of the second division. This indicates that the approximation is not too bad, but it is definitely an approximation.

Check the discriminant of the approximate quotient polynomial:

$- 3 {x}^{2} + 0.325 x - 4.343$

$\Delta = {b}^{2} - 4 a c = {0.325}^{2} - \left(4 \cdot - 3 \cdot - 4.343\right) = 0.105625 - 52.116 = - 52.010375$

Since this is negative, this quadratic has no Real zeros and we can be confident that our original quartic has exactly $2$ non-Real Complex zeros, $1$ positive zero and $1$ negative one.

## What is the standard form of the equation of a circle with centre and radius of the circle x^2 + y^2 - 4x + 8y - 80?

Alan P.
5.331506849315068 years ago

${\left(x - 2\right)}^{2} + {\left(y - \left(- 4\right)\right)}^{2} = {10}^{2}$

#### Explanation:

The general standard form for the equation of a circle is
$\textcolor{w h i t e}{\text{XXX}} {\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$
for a circle with center $\left(a , b\right)$ and radius $r$

Given
$\textcolor{w h i t e}{\text{XXX")x^2+y^2-4x+8y-80 (=0)color(white)("XX}}$(note: I added the $= 0$ for the question to make sense).

We can transform this into the standard form by the following steps:

Move the $\textcolor{\mathmr{and} a n \ge}{\text{constant}}$ to the right side and group the $\textcolor{b l u e}{x}$ and $\textcolor{red}{y}$ terms separately on the left.
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b l u e}{{x}^{2} - 4 x} + \textcolor{red}{{y}^{2} + 8 y} = \textcolor{\mathmr{and} a n \ge}{80}$

Complete the square for each of the $\textcolor{b l u e}{x}$ and $\textcolor{red}{y}$ sub-expressions.
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b l u e}{{x}^{2} - 4 x + 4} + \textcolor{red}{{y}^{2} + 8 y + 16} = \textcolor{\mathmr{and} a n \ge}{80} \textcolor{b l u e}{+ 4} \textcolor{red}{+ 16}$

Re-write the $\textcolor{b l u e}{x}$ and $\textcolor{red}{y}$ sub-expressions as binomial squares and the constant as a square.
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b l u e}{{\left(x - 2\right)}^{2}} + \textcolor{red}{{\left(y + 4\right)}^{2}} = \textcolor{g r e e n}{{10}^{2}}$

Often we would leave it in this form as "good enough",
but technically this wouldn't make the $y$ sub-expression into the form ${\left(y - b\right)}^{2}$ (and might cause confusion as to the y component of the center coordinate).

So more accurately:
color(white)("XXX")color(blue)((x-2)^2)+color(red)((y-(-4))^2=color(green)(10^2)
with center at $\left(2 , - 4\right)$ and radius $10$

## How do you find an exponential function given the points are (-1,8) and (1,2)?

mason m
5.328767123287672 years ago

$y = 4 {\left(\frac{1}{2}\right)}^{x}$

#### Explanation:

An exponential function is in the general form

$y = a {\left(b\right)}^{x}$

We know the points $\left(- 1 , 8\right)$ and $\left(1 , 2\right)$, so the following are true:

$8 = a \left({b}^{-} 1\right) = \frac{a}{b}$

$2 = a \left({b}^{1}\right) = a b$

Multiply both sides of the first equation by $b$ to find that

$8 b = a$

Plug this into the second equation and solve for $b$:

$2 = \left(8 b\right) b$

$2 = 8 {b}^{2}$

${b}^{2} = \frac{1}{4}$

$b = \pm \frac{1}{2}$

Two equations seem to be possible here. Plug both values of $b$ into the either equation to find $a$. I'll use the second equation for simpler algebra.

If $b = \frac{1}{2}$:

$2 = a \left(\frac{1}{2}\right)$

$a = 4$

Giving us the equation: color(green)(y=4(1/2)^x

If $b = - \frac{1}{2}$:

$2 = a \left(- \frac{1}{2}\right)$

$a = - 4$

Giving us the equation: $y = - 4 {\left(- \frac{1}{2}\right)}^{x}$

However! In an exponential function, $b > 0$, otherwise many issues arise when trying to graph the function.

The only valid function is

color(green)(y=4(1/2)^x

## How do you convert (-4, 3) into polar coordinates?

Tony B
5.328767123287672 years ago

Given that a point $\textcolor{b r o w n}{P \to \left(x , y\right) \to \left(- 4 , 3\right) \text{ Cartesian}}$

Then $\textcolor{b l u e}{P \to \left(5 , {143.13}^{o}\right) \text{ Polar }}$ to 2 decimal places

#### Explanation:

This is not a polar graph!!!

$\textcolor{b l u e}{\text{Where it all comes from}}$

We are give the coordinates of (-4,3)
Suppose we viewed this in the context of Cartesian form and use $y = m x + c$

Then $c = 0$ and $m = \frac{y}{x} = - \frac{3}{4}$

So we would have $y = - \frac{3}{4} x$

Suppose the graph was only plotted over the range $x \to \left(0 , - 4\right)$

Then the above graph would not be continuous but be a line from
$\left(0 , 0\right) \to \left(3 , - 4\right)$

All we need now is the angle that that line makes to the x-axis and the length of that line.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Finding the Polar r value}}$

$\text{Length } = \sqrt{{x}^{2} + {y}^{2}} = \sqrt{{3}^{2} + {4}^{2}} = 5$

$\textcolor{b l u e}{\text{So the polar } r = 5}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Finding the Polar "theta" value}}$
The Polar angle $\theta$ is measured from the positive x-axis counterclockwise.

Let the angle from the line to the negative x-axis be $\phi$

Then $\phi = {\tan}^{- 1} \left(\frac{y}{x}\right) = {\tan}^{- 1} \left(\frac{3}{4}\right)$

But $\textcolor{w h i t e}{. .} \phi + \theta = 180$

so $\textcolor{w h i t e}{. .} \theta = 180 - {\tan}^{- 1} \left(\frac{3}{4}\right)$

color(blue)( theta = 143.13 to 2 decimal places
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Putting at all together}}$

Given that a point $P \to \left(x , y\right) \to \left(- 4 , 3\right) \text{Cartesian}$

Then $P \to \left(5 , {143.13}^{o}\right) \text{ Polar }$ to 2 decimal places

## How do you write the partial fraction decomposition of the rational expression  (8x-1)/(x^3 -1)?

Jim G.
5.328767123287672 years ago

$\frac{\frac{7}{3}}{x - 1} + \frac{- \frac{7}{3} x - \frac{10}{3}}{{x}^{2} + x + 1}$

#### Explanation:

The first thing that has to be done is to factorise the denominator.
${x}^{3} - 1$ is a difference of cubes and is factorised as follow :

${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

here then $a = x$ and $b = 1$ .

So ${x}^{3} - 1 = \left(x - 1\right) \left({x}^{2} + x + 1\right)$

Now $\left(x - 1\right)$ is of degree 1 and so numerator will be of degree 0 ie. a constant. Similarly $\left({x}^{2} + x + 1\right)$ is of degree 2 and so numerator will be of degree 1 ie of the form $a x + b$. Now the expression can be written as :

$\frac{8 x - 1}{{x}^{3} - 1} = \frac{A}{x - 1} + \frac{B x + C}{{x}^{2} + x + 1}$

Multiplying through by $\left({x}^{3} - 1\right)$

8x - 1 = A(x^2 + x + 1 ) + (Bx + C )(x - 1 )" " " " color(red)(("*"))

We now have to find the values of $A$ , $B$ and $C$.

Note that if we use $x = 1$ then the term with A will be 0.

Substitute x = 1 in equation $\textcolor{red}{\left(\text{*}\right)}$

$7 = 3 A + 0 \Rightarrow A = \frac{7}{3}$

To find B and C it will be necessary to compare the coefficients on both sides of the equation $\textcolor{red}{\left(\text{*}\right)}$. Multiplying out the brackets on the right hand side to begin with.

$\Rightarrow 8 x - 1 = A {x}^{2} + A x + A + B {x}^{2} + C x - B x - C$

This can be 'tidied up' by collecting like terms and letting $A = \frac{7}{3}$

$8 x - 1 = \frac{7}{3} {x}^{2} + \frac{7}{3} x + \frac{7}{3} + B {x}^{2} + C x - B x - C$

$\Rightarrow 8 x - 1 = \left(\frac{7}{3} + B\right) {x}^{2} + \left(\frac{7}{3} + C - B\right) x + \left(\frac{7}{3} - C\right)$

Compare ${x}^{2}$ terms

$0 = \frac{7}{3} + B \Rightarrow B = - \frac{7}{3}$

Now compare constant terms.

$- 1 = A - C = - \frac{7}{3} - C \Rightarrow C = - \frac{10}{3}$

Finally

$\frac{8 x - 1}{{x}^{3} - 1} = \frac{\frac{7}{3}}{x - 1} + \frac{- \frac{7}{3} x - \frac{10}{3}}{{x}^{2} + x + 1}$