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Let your number be x, a variable. x has the property that x^2 = 5x + 270; x^2 - 5x - 270 = 0

By quadratic formula, \(x = {5\pm5\sqrt{55}\over2}\), which are your two numbers that you may be thinking of.

x + y + 3z = 10                    Equation 1

x - 2y + 8z = 7                     Equation 2

kx + 5z = 3                          Equation 3

Solve via elimination: 2*Equation 1 + Equation 2:

\(2x + 2y + 6z + x - 2y + 8z=20+7\); \(3x + 14z = 27\)                             Equation 1.2

Now we want to eliminate the z variable so we can get an equation only for x (which k is correlated to):

5*Equation 1.2 - 14*Equation 3:

\(15x + 70z-14kx-70z = 135-42\); \(x(15-14k)=93\)

To make this equation false for all real x, x(0) = 93, so 15 - 14k = 0, 14k = 15, and thus k = 15/14. 

Combine like terms: 5x^2 + 4x - 6 = 0.

(a - 4) + (b - 4) = a + b - 8, since a and b are the roots of the quadratic, then by vieta's, their sum is -1 * the second term of the quadratic over the leading coefficient = -4/5 = a + b. a + b - 8 = -4/5 - 8 = -44/5.

Combine like terms: \(2x^2 + 11x - 1 = 0\)

Through manipulation, \({1\over{u}}+{1\over{v}}={u\over{uv}}+{v\over{uv}}={u + v\over{uv}}\)

By vieta's, the sum of the roots u + v = -11/2, and the product of the roots uv = -1/2. Thus, (u + v)/(uv) = \({{-11\over2}\over{-1\over2}}={-11\over{-1}}=11\).

Combine like terms: \(5x^2-11x+4=9x^2-21x-2\); \(4x^2-10x-6=0\); \(2x^2-5x-3=0\), which factors to \((2x + 1)(x - 3)=0\).

Our roots would then be -1/2 and 3, so 1/a + 1/b = -2 + 1/3 = -5/3.

However, in many competition math problems, note that \({1\over{a}}+{1\over{b}}={a + b\over{ab}}\), and then apply vieta's formula, where the sum of the roots is the second coefficient/first coefficient * -1, and the product of the roots is the constant term/leading coefficient.

Expand (a + 2)(3a^2 - 2a - 12) first, get that expression, then distribute it to (a^2 - 2a + 1) so you don't drown in your work.

Additionally, 3a^2 - 2a - 12 is not (nicely) factorable, so we can't try to factor it and get difference of squares...

Try it out for yourself! At the end, your answer should be \(3a^5-2a^4-21a^3+12a^2+32a-24\)

\({(1 + i)\over{(2-i)}}*(3+i)=2i\)

Or, if your problem specifies the order of operations differently:

\({(1 + i)\over{(2 - i)(3 + i)}}={1 + i\over{7-i}}={8+6i\over50}={4\over25} + {3\over25}i\)

\(\frac{1}{\sqrt{36}} - \sqrt{27} - \frac{1}{\sqrt{27}} - \sqrt{18} + \frac{1}{\sqrt{18}} - \sqrt{9}\)

Begin by simplifying the radicals: \({1\over6}-3\sqrt{3}-{1\over{3\sqrt{3}}}-3\sqrt{2}+{1\over{3\sqrt{2}}}-3\)

Then, rationalize the denominators: for example, \({1\over\sqrt{x}}={1\over\sqrt{x}}*{\sqrt{x}\over\sqrt{x}}={\sqrt{x}\over{x}}\):

\({1\over{6}}-3\sqrt{3}-{3\sqrt{3}\over27}-3\sqrt{2}+{3\sqrt{2}\over18}-3\)

Simplify the denominators, then combine like terms: \({1\over{6}}-3\sqrt{3}-{\sqrt{3}\over9}-3\sqrt{2}+{\sqrt{2}\over6}-3\)

Put everyone under the same common denominator, 18: \({3\over{18}}-{54\sqrt{3}\over18}-{2\sqrt{3}\over18}-{54\sqrt{2}\over18}+{3\sqrt{2}\over18}-{54\over18}\)

Combine like terms: \({-51-56\sqrt{3}-51\sqrt{2}\over18}\), I trust that you can split up the fraction if the answer specifies so.

Expanding: \(2x^2+16x+30<3x^2+19x+6\)

Combine like terms: \(x^2+3x-24>0\)

Quadratic formula: \(x={-3\pm\sqrt{105}\over2}\), which are the two roots r and s, used in the factoring of the quadratic (x - r)(x - s) = x^2 + 3x - 24.

To be positive, we take the extreme values in which both (x - r) and (x - s) are positive or both negative, so our solution interval is:

\((-\inf, {-3-\sqrt{105}\over2})\) U \(({-3 + \sqrt{105}\over2}, \inf)\)

What is the greatest integer n such that \(n^2 - 11n +24 \leq 17n - 8\)?

Combining like terms: \(n^2-28n+32\leq0\)

Using quadratic formula n = \({28\pm4\sqrt{41}\over2}=14\pm2\sqrt{41}\). Since our quadratic must be nonnegative, when factored (x - r)(x - s) where r and s are our roots, they must both be positive or both negative, so we take the extreme intervals: \((-\inf,14-2\sqrt{41}]\) U \([14 + 2\sqrt{41}, \inf)\)

To make a positive with 3 terms, you can have positive * positive * positive = positive, or positive * negative * negative = positive. To ensure x is the smallest, we probably take the latter option. To minimize x, (x - 5) and (x - 3) would be logical to be negative, whereas (x + 5) would be positive. If x + 5 > 0, subtract 5 from both sides and you get x > -5. The smallest integer greater than -5 is x = -4.

Fill in the blanks to make a quadratic whose roots are $1$ and $-1.$  
x^2 + ___ x + ___    

If the roots are +1 and –1 then the factors are (x – 1)(x + 1)  

Multiplied out, the quadratic is x² – 1  

So, the first blank holds a 0 (zero) and  

the second blank holds a –1 (negtive 1)   

_.

Rewrite this arithmetic sequence as \(a_{1}, a_{1}+r, a_{1}+2r, \dots, a_{1}+10r, a_{1}+11r\), where a₁ is the first term, and r is the common difference.

\(a_{1} + a_{3} + a_{5} + a_{7} + a_{9} + a_{11}\) is now \(a_{1} + a_{1}+2r + a_{1}+4r + a_{1} + 6r + a_{1} + 8r + a_{1} + 10r\)

\(a_{2} + a_{4} + a_{6} + a_{8} + a_{10} + a_{12}\)is now \(a_{1} + r + a_{1} + 3r \dots+ a_{1} + 11r\)

Each term in the second one has one more r than the first one, so subtracting:

\(6r=0, r=0\).

Sub this back in to equation 1,

\(6a_{1}=0\)

\(a_{1}=0\).

We get this first term is zero. Note, since the common ratio is 0, this sequence is just twelve zeros, but it satisfies the conditions, and is the only sequence to do so. 

C)

We know a-b = 1, and a+b = 1 or a+b = -1.

Consider two cases:

Case 1:

\(\begin{cases} a-b = 1\\ a+b=1 \end{cases}\)

\(2a = 2, a=1\)

\(\begin{cases} a=1 \\ b=0 \end{cases}\)

Case 2:

\(\begin{cases} a-b=1 \\ a+b = -1 \end{cases}\)

\(2a=0, a=0\)

\(\begin{cases} a=0 \\ b=-1 \end{cases}\)

Our two solutions are \(a, b = (1, 0)\) and \(a, b = (0, -1)\).

B)

\({(a+b)}^2={(a-b)}^{2}+4ab\)

From previously, ab=0

\((a+b)^2={1}^{2}\)

\(a+b=\pm1\).

A)

\({(a-b)}^{3}= {a}^{3}-3{a}^{2}b+3a{b}^{2}-{b}^{3} = 1\)

\(1-3{a}^{2}b+3a{b}^{2}=1\)

\(3ab(b-a)=0\)

\(-3ab=0\)

\(ab=0\)

\(d=\frac{24km}{t}\\ e=\frac{(24-10)km}{t}=\frac{14km}{t}\\ f=\frac{(24-15)km}{t}= \frac{9km}{t}\\ s_e=\frac{14km}{t}x=24km\\ s_f=\frac{9km}{t}x=24km\cdot \frac{9}{14}\\ \color{blue}s_e-s_f=24km-24km\cdot \frac{9}{14}=8.571km\)

d,e,f are the running speeds of the runners.

t is the running time of Dirk.

x is the running time of Edith.

s_e, s_f are the routes to x.

Foley will be 8.571 km behind when Edith finishes the race.

 !

If a number is a square of a binomial, it's discriminant is zero.

\(\Delta = {17}^2-4p=0\).

\(4p = {17}^2\)
\(p=\frac{{17}^{2}}{4}=\frac{289}{4}\).

The probability that Alice wins a game is the probability she will win on her first turn + the probability that she will win on her second turn + the probability she will win on her turn turn, and so on...

Alice has a \(\frac{1}{2}\) probability of winning on their first turn, finishing the game.

Since whenever someone wins, the game is finished Bob has a \(\frac{1}{2}\cdot\frac{1}{2}={(\frac{1}{2})}^{2}\) probability of winning on their first turn.

Alice has a \({(\frac{1}{2})}^{2}*\frac{1}{2}={(\frac{1}{2})}^{3}\) probability of winning on her their second turn.

Bob has a \({(\frac{1}{2})}^{3}*\frac{1}{2}={(\frac{1}{2})}^{4}\) probability of winning on their second turn.

Alice has a \({(\frac{1}{2})}^{4}*\frac{1}{2}={(\frac{1}{2})}^{5}\) probability of winning on her thrid turn.

We see Alice has a \({(\frac{1}{2})}^{1}+{(\frac{1}{2})}^{3}+{(\frac{1}{2})}^{5}+{(\frac{1}{2})}^{7}...\) chance of winning.

We can use the infinite geometric series fomula,\(\frac{\frac{1}{2}}{1-\frac{1}{4}}\), we get \(\frac{2}{3}\).

\(\)

Let r=OA

Since central angles AOB and COD are equal, the length of each arc is proportional to the corresponding radius. Thus, 4/r=7/(r+2)
Cross-multiplying, we get 4r+8=7r so 3r=8 and r=8/3 Then the area of the circle with radius r is pir^2=64/9pi

The proportion of AOB to 360 is equal to the proportion of arc AB to the circumference of the whole circle, which is

4/(2rpi)=4(2pi*8/3)=3/(4pi)
so the area of circular sector OAB is 3/(4pi)*64/9 (pi)=16/3
 

\(S_n=\frac{n}{2}(a_1+(n-1)d)\)
So using this and plugging in values we get:
\(\frac{1}{5}=\frac{20}{2}(a_1+19d)\\ 0=\frac{10}{2}(a_1+9d)\) solving we get:
\(a=-\frac{9}{500}, d=\frac{1}{500}\)

Therefore:
\(S_{70}=\frac{70}{2}(a_1+69d)\\ \boxed{S_{70}=\frac{21}{5}}\)