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## How to factor #3a^2+4ab+b^2-2ac-c^2# ?

George C.
Featured 3 months ago

$3 {a}^{2} + 4 a b + {b}^{2} - 2 a c - {c}^{2} = \left(3 a + b + c\right) \left(a + b - c\right)$

#### Explanation:

Given:

$3 {a}^{2} + 4 a b + {b}^{2} - 2 a c - {c}^{2}$

Note that all of the terms are of degree $2$.

So if this factors into simpler polynomials then its factors are homogeneous of degree $1$.

If we ignore the terms involving $a$, then we are looking for a factorisation of ${b}^{2} - {c}^{2}$, which can be written:

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

If we ignore the terms involving $c$, then we find a factorisation:

$3 {a}^{2} + 4 a b + {b}^{2} = \left(3 a + b\right) \left(a + b\right)$

If we ignore the terms involving $b$, then we find a factorisation:

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

These various linear binomial factors can be combined as follows:

$\left(b + c\right) , \left(3 a + b\right) , \left(3 a + c\right) \rightarrow \left(3 a + b + c\right)$

$\left(b - c\right) , \left(a + b\right) , \left(a - c\right) \rightarrow \left(a + b - c\right)$

Hence we find the factorisation:

$3 {a}^{2} + 4 a b + {b}^{2} - 2 a c - {c}^{2} = \left(3 a + b + c\right) \left(a + b - c\right)$

## Prove that #if u# is an odd integer, then the equation #x^2+x-u=0# has no solution that is an integer?

Steve
Featured 3 months ago

Proposition
If $u$ is an odd integer, then the equation ${x}^{2} + x - u = 0$ has no solution that is an integer.

Proof
Suppose that there exists a integer solution $m$ of the equation:

${x}^{2} + x - u = 0$

where $u$ is an odd integer. We must examine the two possible cases:

$m$ is odd; or
$m$ is even.

First, let us consider the case where $m$ is odd, then there exists an integer $k$ such that:

$m = 2 k + 1$

Now, since $m$ is a root of our equation, it must be that:

${m}^{2} + m - u = 0$
# :. (2k + 1)^2 + (2k + 1) − u = 0 #
# :. (4k^2 + 4k + 1) + (2k + 1) − u = 0 #
# :. 4k^2 + 6k + 2 − u = 0 #
$\therefore u = 4 {k}^{2} + 6 k + 2$
$\therefore u = 2 \left(2 {k}^{2} + 3 k + 1\right)$

And we have a contradiction, as $2 \left(2 {k}^{2} + 3 k + 1\right)$ is even, but $u$ is odd.

Next, let us consider the case where $m$ is even, then there exists an integer $k$ such that:

$m = 2 k$

Similarly, since $m$ is a root of our equation, it must be that:

${m}^{2} + m - u = 0$
# :. (2k)^2 + (2k) − u = 0 #
# :. 4k^2 + 2k − u = 0 #
$\therefore u = 4 {k}^{2} + 2 k$
$\therefore u = 2 \left(2 {k}^{2} + k\right)$

And, again, we have a contradiction, as $2 \left(2 {k}^{2} + k\right)$ is even, but $u$ is odd.

So we have proved that there is no integer solution of the equation ${x}^{2} + x - u = 0$ where $u$ is an odd integer.

Hence the proposition is proved. QED

## How do you simplify #9.594 div 0.06#?

Tony B
Featured 1 month ago

A way to get round the decimal place in the initial division expression.

$159.9$

#### Explanation:

Note that $9.594$ is the same as $9594 \times \frac{1}{1000}$

Note that $0.06$ is the same as $6 \times \frac{1}{100}$

So $9.594 \div 0.006$ is the same as $\left[9594 \div 6\right] \times \left[\frac{1}{1000} \div \frac{1}{100}\right]$

$\left(9594 \div 6\right) \times \frac{1}{10}$

I will do the division first then multiply by the $\frac{1}{10}$ at the very end

$\text{ } 9594$
$\textcolor{m a \ge n t a}{1000} \times 6 \to \underline{6000} \leftarrow \text{ subtract}$
$\text{ } 3594$
$\textcolor{m a \ge n t a}{\textcolor{w h i t e}{1} 500} \times 6 \to \underline{3000} \leftarrow \text{ subtract}$
$\text{ } 594$
$\textcolor{m a \ge n t a}{\textcolor{w h i t e}{10} 90} \times 6 \to \underline{\textcolor{w h i t e}{0} 540} \leftarrow \text{ subtract}$
$\text{ } \textcolor{w h i t e}{0} 54$
$\textcolor{m a \ge n t a}{\textcolor{w h i t e}{100} 9} \times 6 \to \text{ "ul(54) larr" subtract}$
$\text{ } 0$

As we have 0 we may now stop the division

$9594 \div 6 = \textcolor{m a \ge n t a}{1599}$

Now we multiply by the $\frac{1}{10}$ giving:

$159.9$
~~~~~~~~~~~~~~~~~~~~~~~~~~~

If the division had not finished with a 0 at that point then we would have gone into decimal values.

## How do you find the vertex and intercepts for #y=2(x-3)^2 +4#?

Jim G.
Featured 1 month ago

see explanation.

#### Explanation:

The equation of a parabola in $\textcolor{b l u e}{\text{vertex form}}$ is.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = a {\left(x - h\right)}^{2} + k} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
where (h ,k) are the coordinates of the vertex and a is a constant.

$y = 2 {\left(x - 3\right)}^{2} + 4 \text{ is in this form}$

$\text{with " a=2, h=3" and } k = 4$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(3 , 4\right)$

$\text{since " a>0" then min. turning point } \bigcup$

$\textcolor{b l u e}{\text{Intercepts}}$

#• " let x = 0, in equation, for y-intercept"#

#• " let y = 0, in equation, for x-intercepts"#

$x = 0 \to y = 2 {\left(- 3\right)}^{2} + 4 = 22 \leftarrow \textcolor{red}{\text{ y-intercept}}$

$y = 0 \to 2 {\left(x - 3\right)}^{2} + 4 = 0$

$\Rightarrow {\left(x - 3\right)}^{2} = - 2$

This has no real solutions hence f(x) has no x-intercepts.
graph{2(x-3)^2+4 [-31.56, 31.67, -15.8, 15.8]}

## How do you find the x and y intercepts of #2x-y=10#?

Catherine T. Mohs
Featured 1 month ago

$x = 5 \mathmr{and} y = - 10$
graph{2x-10 [-5, 10, -15, 5]}
(the graph has been scaled to show the intercepts and does not accurately depict the slope)

#### Explanation:

Put the equation into double-intercept form.
$\left(\frac{x}{a}\right) + \left(\frac{y}{b}\right) = 1$
where $a$ is the x-intercept and $b$ is the y-intercept

Original equation:
$2 x - y = 10$
Divide all parts by 10:
$\left(\frac{2 x}{10}\right) + \left(\frac{- 1 y}{10}\right) = \left(\frac{10}{10}\right)$

And here is the finished product!
$\left(\frac{x}{5}\right) + \left(\frac{y}{-} 10\right) = 1$

$x = 5 \mathmr{and} y = - 10$

## R is the midpoint of segment PS. Q is the midpoint of segment RS. P is located at (8, 10), and S is located at (12, 6). − What are the coordinates of Q?

Anthony R.
Featured 4 days ago

$\left(11 , 7\right)$

#### Explanation:

We can begin by plotting the points of the line segment $P S$. Since we know that $R$ is the midpoint of the $P S$ line segment, we can use the midpoint formula to find the coordinate of $R$

The midpoint formula: $\left(\frac{{x}_{1} + {x}_{2}}{2} , \frac{{y}_{1} + {y}_{2}}{2}\right)$

Since we have our two coordinates of $P$ and $S$ we can substitute those values into the formula to find the coordinates of $R$

If we let, $\left(8 , 10\right) \to \left(\textcolor{b l u e}{{x}_{1}} , \textcolor{red}{{y}_{1}}\right)$ and $\left(12 , 6\right) \to \left({\textcolor{b l u e}{x}}_{2} , \textcolor{red}{{y}_{2}}\right)$ then.. .

$R = \left(\textcolor{b l u e}{\frac{8 + 12}{2}} , \textcolor{red}{\frac{10 + 6}{2}}\right) = \left(\textcolor{b l u e}{\frac{20}{2}} , \textcolor{red}{\frac{16}{2}}\right) = \left(\textcolor{b l u e}{10} , \textcolor{red}{8}\right)$

Since we now know the coordinates of $R$ and already given $S$ we can find the coordinate of $Q$ since it is the midpoint of $R S$

By the same method above...

If we let, $\left(10 , 8\right) \to \left(\textcolor{b l u e}{{x}_{1}} , \textcolor{red}{{y}_{1}}\right)$ and $\left(12 , 6\right) \to \left({\textcolor{b l u e}{x}}_{2} , \textcolor{red}{{y}_{2}}\right)$ then.. .

$Q = \left(\textcolor{b l u e}{\frac{10 + 12}{2}} , \textcolor{red}{\frac{8 + 6}{2}}\right) = \left(\textcolor{b l u e}{\frac{22}{2}} , \textcolor{red}{\frac{14}{2}}\right) = \left(\textcolor{b l u e}{11} , \textcolor{red}{7}\right)$

So the coordinate of $Q$ lies at $\left(11 , 7\right)$ which is also the midpoint of line segment $R S$

Here is a graph to provide a visual representation of what I just explained.

If the image is to small, you can check the link below to this graph: https://www.desmos.com/calculator/vtaiir6ro8

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