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First, let's combine all like terms and move all terms to one side. 

We get

\(x^2 + (17 - m)x + 4 > 0 \)

Let's note that if roots are real, then the descriminant must be greater or equal to 0. 

We have

\((17 - m)^2 - 4(1)(4) > 0 \\ (17 - m)^2 > 16 \)

From here, there are 2 ways to go. First, we have

\(17 - m \geq 4 \\ 17 - 4 \geq m \\ m \leq 13 \)

and, we have

\( 17 - m \leq -4 \\ 21 \leq m \\ m \geq 21\)

This means our final answer is

\(m = [-\infty , 13]\bigcup [21, \infty]\)

Thanks! :)

Thanks! It's funny because I'm taking an intermediate algebra course, and this is what we are learning. 

Also, NTS is such a wonderful abbreviation, LOL. Sounds quite proffesional. 

Thanks! :)

Combining all like terms and moving all terms to one side, we get 

\(x^{2}+12x-4=0\)

The two roots of this are

\(x=2\sqrt{10}-6\\ x=-2\sqrt{10}-6\)

Squaring them we get 

\(4\left(\sqrt{10}\right)^{2}-12\sqrt{10}-6\left(2\sqrt{10}-6\right)+\left(-2\sqrt{10}-6\right)^{2}\)

\(4\cdot 10-12\sqrt{10}-6\left(2\sqrt{10}-6\right)+\left(-2\sqrt{10}-6\right)^{2}\)

This simplifies to 152, which is our answer.

Thanks! :)

First, let's subtract 10 from both sides. \(x^2-mx+14\)

The answer to this question lies in the number of possible multiples 14 has. 

For every mutliple, there are 4 possibles. 

Say the two multiples are \(a, b\)

We have \((x-a)(x-b), (x+a)(x+b), (x+a)(x-b), (x-a)(x+b)\)

For example, 2 and 7. 

\((x-7)(x+2), (x-2)(x+7), (x-2)(x-7), (x+7)(x+2)\)These all give different values for m. 

We have only 2 possible groups. 1 and 14, 2 and 7. 

This means there are 4*2=8 different values for m

Thanks! :)

You are correct, NTS !!!

This can't be a quadratric  because of the Conjugate Property

If a + bi is a  root so is a - bi

If c + di is a root so is c - di

We would have 4 complex roots, not  just two

So, I might have forgotten to consider than b and c could include z. I'll solve the problem like that, 

Like I stated in post 1, there are 4 roots

\(4+2i, 4-2i, -5+8i, -5-8i\)

This means we have the polynomial, 

\((z-4-2i)(z-4+2i)(z+5-8i)(z+5+8i)\)

Simplfying, we have 

\(\left(z^2-8z+20\right)\left(z^2+10z+89\right)\\z^4+2z^3+29z^2-512z+1780\)

So, in order for this to be valid, b would have to be z^3 and c would have to be \(2z^3+28z^2-512z+1780\)

So our final answer is \(z^4+2z^3+28z^2-512z+1780\)

Thanks! :)

\( \left(\sin(a)\right)^7 + \left(\cos ( a)\right)^7< 2*sin(a)^4*cos(a)^4. \)

\(0 < a < \frac{\pi}{2} \)

This doesn't appear to be true

There's a problem with this question, though. 

According some theorem I totally forgot, 

If \(4 + 2i\) is a root, then \(4-2i\) must also be a root. 

If \(-5 + 8i\) is a root, then \(-5-8i\) must also be a root.

With 4 distinct roots, then the polynomial would have degree 4, not 2, 

So it is impossible to have a quadratic with these two roots. 

I'm not sure if I'm mistaken, but that should be true. 

Thanks! :)

I'm sure it was a coincidence, but the problem posted before this was the exact same!

First, we want to get rid of the deniminators. We can do this by multiplying both sides by the LCM of the denominators. 

The LCM is (x-5)(x-2). 

We have

\(x\left(x-2\right)=2\left(x-5\right)+2x\left(x-5\right)\left(x-2\right)\)

We now get a quadratic. 

We get 

\(x^2-2x=2x^3-14x^2+22x-10\\ 2x^3-15x^2+24x-10=0\)

This where it gets tricky. There are no integer roots for this equation. 

We could use the Newton-Raphson equation, and we eventually get

\(x\approx \:0.67798\dots ,\:x\approx \:1.34697\dots ,\:x\approx \:5.47503\dots \)

I hope this was clear enough!

Thanks! :)

First, let's combine all like terms. We get

\(-15x^2-kx+45\)

In order for the solutions not to be real, the descriminant must be less than 0 (negative number).

We have 

\((-15)^2 - 4(-k)(45) < 0 \\ 225+180k<0\\ k<-1.25\)

Therefore, the largest integer must be -2. 

Thanks! :)

Simplify as

8x^2 + 8x + 28 = 0

2x^2 +  2x + 7 = 0

x =  [-2 + sqrt (-52) ] / 4  =  [-2 + 2sqrt (13) i ] / 4  =  -1/2 + (1/2) sqrt (13) i

"a"  is not positive

a * b =  (-1/2) (1/2)sqrt 13  =   -sqrt (13)  /  4

First, let's combine all like terms to get a real quadratic. 

We get \(x^{2}+8x+9\).

The descriminant of the quadratic is \(b^2-4ac\) for those in \(ax^2+bx+c\)

We have \(8^2-4(1)(9) = 64-36=28\)

So 28 is our answer!

Thanks! :)

The  roots of (x -2) (x + 2)  are  2. -2

sum of these roots = B/A  = 0

product of the roots =  1/A

-4 = 1/A

A = -1/4

A + B  =  -1/4

x + y  = 10

y = 10 -x

x^2 + (10 -x)^2  = 32 + 2x(10-x)

x^2 + x^2 -20x + 100 = 32 + 20x -2x^2

4x^2 -40x + 68 = 0

x^2 -10x + 17  = 0

x^2 -10x  =  -17

x^2 -10x + 25  = -17 + 25

(x -5)^2  = 8

x -5 = +/- sqrt (8) = +/- 2sqrt 2

x = 5 +  2sqrt2      or      x =  5 - 2sqrt 2

y = 10 -x   =    5 -2sqrt 2    or   5 + 2sqrt 2

(x,y)  = (5 + 2sqrt 2 , 5-2sqrt 2)   or ( 5- 2sqrt 2, 5 + 2sqrt 2)

Let a,b be the sides

2 ( a + b)  = 40

a + b    = 20

b = 20  - a

a^2 + b^2  = (10 sqrt 2)^2

a^2  + (20 - a)^2  = 200

a^2 + a^2 -40a + 400 = 200

2a^2 -40a + 200  = 0

a^2 -20a + 100 =  0

(a - 10*2  = 0

a =10

b =20 -10  =10

Area =  a*b =  10 * 10   =  100

Half as many members will take  twice as long = 20 min

S1 = a1 = 1/5

S2  = 2a1 + d

1/10 = 2/5 +  d

1/10 - 2/5 = d

d  =  -3/10

sum of the first n terms of an  arithmetic sequence  =

(n)*a1 +  (n)(n-1)/2 *d

S15  =  (15 ) (1/5)  + (15)(14)/2 * (-3/10)  =   3 - 63/2  = -57/2  = - 28.5

\( \frac{4}{1 - \sqrt[3]{3}} + \sqrt[3]{9} \)

Note that   1 - 3   can  be written as  ( 1 - 3^(1/3)) ( 1 + 3^(1/3) + 3^(2/3))

Multiply num/den  of the first fraction by the expression in re and  we  get

4 (1 + 3^(1/3)  + 3^(2/3) )

_____________________ =   -2 ( 1 + 3^(1/3) + 3^(2/3) )  =   -2 - 2*3^(1/3) -2*3^(2/3)

          1 - 3

And   9^(1/3)  =   3^(2/3)

So we have

-2 -2*3^(1/3) - 2 *3^(2/3) + 3^(2/3)  =

-2 - 2*3^(1/3) - 3^(2/3) 

OA =  OB /  5  =  30 / 5   =   6

alpha = the measure of one of the interior angles  = 

(15 - 2) *180  / 15   =     156°

Simplify as

ab  - 9a - 5b

Take the product  of the two coefficients on a, b  =  (-9) (-5)  = 45

(This is the number that goes in the blank)

Add this

ab - 9a - 5b + 45       factor as

(a - 5) ( b  - 9)

∛-8  = -2   which is rational

l floor ( -2) l  = l -2 l  = 2

-pi  is irrational

(ceiling (-pi) )^2  =  (-3)^2  =  9

√50  is irrationa;

( ceiling (√50) )^2  = (8)^2  = 64

9/2 is rational

l floor (9/2) l  =  l 4 l  =  4

2 + 9 + 64 + 4  =   79

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