Patrick's equation

We must consider three cases:

• The exponent equals to $0$ .
  Solving $x^2 - 10x + 21$ we get as a solutions $7$ and $3$ .
  

• The base is $1$ .
  Solving $x^2 + 5x + 4$ we get as a solutions $-1$ and $-4$
  

• The base is $-1$ and the exponent is even.
  Solving $x^2 + 5x + 6$ we get as a solutions $-3$ and $-2$
  We put the values on the exponent equation and only $-3$ gives an even exponent.
  

So the sum of all solutions is: $7 + 3 - 1 - 4 - 3 = 2$

Consider the cases when a^b = 1

This can only occur when a=1, b=0, and a=-1 where b is even.

Consider case one, where a=1

This implies that: x^2 + 5x + 5 = 1 x^2 + 5x + 4 = 0 (x + 4)(x + 1) = 0 So we now have two solutions, x=-4 and x=-1

Consider case two, where b=0

This implies that: x^2 - 10x +21 = 0 (x - 7)(x - 3) = 0 This gives two solutions, x=7, and x=3

Consider case three, where a=-1, and b is even

This implies that: x^2 + 5x + 5 = -1 x^2 + 5x + 6 = 0 (x + 3)(x + 2) = 0 Which gives x=-3 and x=-2 However, we discard x=-2 as it gives rise to an odd value of b (45)

This leaves us with the final solutions: -4, -1, 7, 3 and -3 which sum to 2

$x^{2}+5x+5 =1$ yields solutions -1 and -4

$x^{2}+5x+5 =-1$ and since the exponent must even, this yields solution: -3 (not -2)

$x^{2}-10x+21 =0$ yields solutions 7 and 3

Thus, $-1 + -4 + -3 + 7 + 3 = 2$

There are three scenarios which solve the problem:

• The exponent is 0, as anything (except 0 itself) raised to 0 is 1
• The base is 1, as positive 1 raised to any value is always 1
• The base is -1 and the exponent is even

Case 1:

$x^2 -10x +21 = 0$

$(x -7)(x-3) = 0$

$x = 7, x = 3$

The base is non-zero for both of these roots, so these are two solutions to the problem.

Case 2:

$x^2 + 5x + 5 = 1$

$x^2 + 5x + 4 = 0$

$(x + 4)(x + 1) = 0$

$x = -4, x = -1$

Case 3:

$x^2 + 5x + 5 = -1$

$x^2 + 5x + 6 = 0$

$(x + 3)(x + 2) = 0$

$x = -3, x = -2$

We need to test these two roots in the exponent. Exponent must be even.

$(-3)^2 - 10 (-3) + 21 = 9 + 30 + 21 = 60$

This one works!

$(-2)^2 - 10 (-2) + 21 = 4 + 20 + 21 = 45$

This does not work.

So our only valid solution in Case 3 is $x = -3$

Result

$-4 -1 + 7 + 3 - 3 = \boxed{2}$

we have three choices choice one: x2+5x+5 =1 choice two: x2−10x+21=0 choice three: x2+5x+5=-1 and x2−10x+21 is even the first one have two solutions: 7 , 3 the second one have two solutions: -1,-4 the third one have one solution: -3 7+3 + -1 + -4 + -3 = 2 the solution is 2

$x^{2} + 5x + 5$ can take values 1 and -1(if $x^{2}-10x+21$ is even). $x^{2} + 5x + 5$ is 1 at -4 and -1 and -1 at -3 and -2 ,but $x^{2}-10x+21$ is not even at -2 so the solution from here are -4,-3,-1.

$x^{2}-10x+21$ can be zero $\Rightarrow$ at 7 and 3

sum = $7+3-4-3-1 = 2$

(x^{2}+5 \times x+5)^{x^{2}−10 \times x+21}=1

There 2 cases in which the equation would be zero. That is either, the power {x^{2}−10 \times x+21} is zero or when the base number (x^{2}+5 \times x+5) is one or negative one.

When x^{2}−10 \times x+21= 0, x= 7,3 When x^{2}+5 \times x+5= 1, x= -4, -1 When x^{2}+5 \times x+5= -1, x= -3, -2

However, in the last case above ( when the base is equal to negative one) -2 is invalid. This is due to the fact that when -2 is substituted into the power, the power would be an odd number. If the power was a odd number, the whole equation would be -1^{odd}= -1.

This leaves 5 solutions. They are: 7, 3, -4, -1 and -3. 7+3-4-1-3=2

see jordi bosch's answer.

I had to plug in the equation into wolfram to see what the answer was and realized I was missing 1^any number=1 and (-1)^even number=1 solutions.

Put the power as 0 because the answer on the is one.. so, 2 is the answer..

Missed the third case in first attempt but got it in the second

This is the second problem of the same type I think with in 9 months! This time I did it correctly, using -1, and solving for 0, +1, -1.

I solved the problem the standard way by taking natural logs of both sides. Eventually this comes out to: $(x-3)(x-7)\ln\left(x^{2}+5x+5\right) = 0$ . We want part of this to be zero. Either we can accomplish that by either making $(x-3)$ or $(x-7)$ zero, or we could make the natural log equal to zero, by making the argument of the logarithm equal to 1. That becomes another quadratic equation for the latter, and you can obtain 1 as the argument with $x = -1$ or $x = -4$ . That gives $3, 7, -1, -4$ . Also, if we make the . However, we're not done yet. What if $(x^{2}+5x+5)$ gives -1, and $x^{2}-10x+21)$ is even? Then, we can still satisfy the equation. Set $(x^{2}+5x+5)$ equal to $1$ and factor it. Here, you can get either $-2$ or $-3$ . Only $-3$ works because it makes $x^{2}-10x+21$ even, and $-1$ raised to an even exponent results in $1$ . $-2$ makes $x^{2}-10x+21$ odd. Finally, the sum of all solutions $3+7-1-4-3=2$ .

First make x^2-10x+21=0.then find solutions of this which are 7 and 3 Next make x^2+5x+5=1 find solutions of this which are -1 and -4 At last make x^2+5x+5=-1 find solutions which are -2 and -3. -2 doesn't satisfy this condition. Add 7, 3,-1,-4 and -3 you will get the answer I.e. 2