Worry-able everywhere - 4

Real solutions consist of rational and irrational solutions. We have already figured these out in previous problems. (If you already know the solutions for these, then skip the explanation).

Looking at our 4 cases and the discriminant of each from the solution to the first problem:

$3{ x }^{ 2 }+3=10x$

Discriminant: $64$

${ x }^{ 2 }+x-57=0$

Discriminant: $229$

${ x }^{ 2 }+x-57=1$

Discriminant: $233$

${ x }^{ 2 }+x-57=-1$

Discriminant: $225$

$Rational\quad Solutions$

We see that the first case and the fourth case yield discriminants that are perfect squares, and will therefore yield rational solutions.

Factoring the first equation gives us

$(3x-1)(x-3)=0$

So, $x=3,x=\frac { 1 }{ 3 }$

Substituting these values in the original problem, we see that they both work.

Factoring the fourth equation gives us

$(x-7)(x+8)=0$

So, $x=7,\quad x=-8$

However, since this case sets the base as $-1$ , we must check if these values make both $3{ x }^{ 2 }+3$ and $10x$ even or both odd.

Substituting $x=7$ , we get $3{ (7) }^{ 2 }+3=150,\quad 10(7)=70$

These are both even, so the end equation is $1=1$

Substituting $x=-8$ , we get $3{ (-8) }^{ 2 }+3=195,\quad 10(-8)=-80$

Since one is odd and one is even, the end result in $-1=1$ , which is not true.

We disregard the solution $x=-8$ , so adding up the other solutions gives us $3+\frac { 1 }{ 3 } +7=10\frac { 1 }{ 3 } ,\quad or\quad 10.\overline { 3 }$ .

$Irrational\quad Solutions$

We see that the second and third equations have positive but non perfect square roots, so they will yield irrational solutions.

Using the quadratic formula on the second equation, $\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2 }$ , gives us the solutions $x=\frac { -1\pm \sqrt { 229 } }{ 2 }$ .

These solutions will make the base $0$ , but we must check to see what they make the powers, ${ 3{ x }^{ 2 }+3\quad and\quad 10x }$ , because if the power(s) is negative, then it would equal $\frac { 1 }{ { 0 }^{ n } }$ , which is undefined.

Substituting in $x=\frac { -1+\sqrt { 229 } }{ 2 }$ ,

$3{ (\frac { -1+\sqrt { 229 } }{ 2 } ) }^{ 2 }+3\cong 24.199$

$10{ (\frac { -1+\sqrt { 229 } }{ 2 } ) }\cong 70.664$

Therefore, this solution works.

Substituting in $x=\frac { -1-\sqrt { 229 } }{ 2 }$ ,

$3{ (\frac { -1-\sqrt { 229 } }{ 2 } ) }^{ 2 }+3\cong 198.199$

$10{ (\frac { -1+\sqrt { 229 } }{ 2 } ) }\cong -80.664$

The $10x$ power becomes negative, and therefore, this is not a valid solution. Using the quadratic equation on the third equation, we get $x=\frac { -1\pm \sqrt { 233 } }{ 2 }$

Since 1 raised to any power, positive or negative, is 1, these solutions work.

Adding up the solutions, we get $\frac { -1+\sqrt { 233 } }{ 2 } { +\frac { -1-\sqrt { 233 } }{ 2 } +\frac { -1+\sqrt { 229 } }{ 2 } \cong }$ $6.066372975$ .

Now, adding up our sums of rational and irrational solutions, we can get the sum of all real solutions. Note that we want to add the not rounded sums of each first, then round.

$6.066372975+10.3333333333=16.39970631$ , which rounds to $\boxed {16.400}$