NotThatSmart

Well, every number technically has a perfect square divisor 1. 

This means that the smallest positive integer that has exactly 2 perfect square divisors is the next smallest pefect square. 

Thus, 4 is the smallest. 

We have 

\(1 = 1^2\\ 4=2^2\)

Thus, our answer is just 4. 

Thanks! :)

So, in order to calculate the list, we can use a handy trick. We simply take the largest number, multiply it by that number +1, and divide by 2. 

We have

\(\frac{200 \cdot 201}{2}=20100\)

The prime factorization of \(20100\) is 

\(20100 = 2^2 \cdot 3 \cdot 5^2 \cdot 67\)

Thus, the greatest prime divisor is 67. 

Thanks! :)

Since b is a constant, it must not contain any powers. This means b has no overall effect on the degree. 

Thus, let's take a look at f(x) and g(x). 

The degree of \( f(x) + b \cdot g(x)\) is essentially the larger degree for f(x) and b(x). 

f(x) has a degree of 5, because the leading term is \(2x^5\)

g(x) has a degree of 6, because the leading term is \(x^6\)

Thus, the degree of the polynomial must have a degree of 6. 

So the smallest possible degree is 6. 

Thanks! :)

Cross multiplying, we get

\(12a+24=ab+b\)

Moving ab to the other side and factoring, we have

_{\(a\left(12-b\right)=b-24\)}

Dividng both sides 12-b, we get

\(a=\frac{b-24}{12-b}\)

So, everything works, and we have \(a=\frac{b-24}{12-b};\quad \:b\ne \:12\)

So technically, there are infinite solutions. 

Thanks! :)

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

\(6x^2-21x-31=0\)

Now, we simply use the quadratic equation, we get

\(x_{1,\:2}=\frac{-\left(-21\right)\pm \sqrt{\left(-21\right)^2-4\cdot \:6\left(-31\right)}}{2\cdot \:6}\)

\(x_1=\frac{-\left(-21\right)+\sqrt{1185}}{2\cdot \:6},\:x_2=\frac{-\left(-21\right)-\sqrt{1185}}{2\cdot \:6}\)

\(x=\frac{21+\sqrt{1185}}{12},\:x=\frac{21-\sqrt{1185}}{12}\)

So our final answers are \(x=\frac{21+\sqrt{1185}}{12},\:x=\frac{21-\sqrt{1185}}{12}\)

Thanks! :)

First, let's simplfy our equation. We have

\(2x^2 - 3x - 35 = -43 + jx \\ 2x^2 - 3x - jx + 8 = 0 \\ 2x^2 - (3 + j)x + 8 = 0\)

If there is only one solution, then the descriminant of the quadratic must be 0. So, we can write the equation

\((3 + j)^2 - 4(2)(8) = 0 \\ (j + 3)^2 - 64 = 0 \\ [ (j + 3) - 8 ] [ (j + 3) + 8 ] = 0 \\ [ j - 5 ] [ j + 11] = 0\)

Thus, we have that 

\(j = 5 \\ j = -11\)

So our final answer is 5 and -11. 

Thanks! :)

Let's first combine all like terms. We have

\(x^2+kx-9x+16\\ x^2+(k-9)x+16\)

Since we need this to be a difference of squares, k-9 must be 2 times the square root of 16. 

Thus, we have

\(k-9=8\\ k=17\)

So k is 17.

So our answer is 17. 

Thanks! :) 

The only limit for the problem we have is that

\(6-x-x^2-2x^2\geq0\) so there are no imaginary numbers,

Combining like terms, we have

\(-3x^2-x+6\geq0\)

Applying the quadratic equation, we have

\(\frac{-\sqrt{73}-1}{6}\le \:x\le \frac{\sqrt{73}-1}{6}\)

So our answer is 

\(\frac{-\sqrt{73}-1}{6}\le \:x\le \frac{\sqrt{73}-1}{6}\)

Not the clearest explenation, I can elaborate if needed. 

Thanks! :)

First, let's isolate y in terms of x. We have

\(y=25-x\)

Now, let's subsitute this into the second equation. We have

\(6x+3(25-x)\\ 6x+75-3x\\ 3x+75\)

x can't be negative, so the smallest x can be is 0. Subbing 0 in, we get

\(3(0) + 75\\ 0+75\\ 75\)

So 75 is our answer. 

Thanks! :)

Let's combine all like terms first. We have

\(x^2+18x+3\)

We can't factor this equation yet, so let's complete the square. We get

\(x^2+18x+81+3-81\\ (x+9)^2-78\)

Thus, the first blank is 1, the second blank is 9, and the third blank is -78. 

Thanks! :)