Worry-able everywhere (1)
How many real roots exist satisfy the above equation?
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1 solution
Moderator note:
I've added explanations to your cases, to make it more explicit what they are and why they work. Note that case 4 is still incomplete, and you have to compare the parity of the exponents.
Note that you also have to show that these roots are distinct from each other. How do you know that the solution in case 1 is not repeated elsewhere?
Since both the LHS and RHS of the equation have the same base, we consider the following cases.
Case 1: Exponents are equal: 3 x 2 + 3 = 1 0 x
By setting the equation to 0, and finding the discriminant, b 2 − 4 a c , we can see that there are two real roots that both work in the original problem.
Case 2: Base is 0, any exponent : x 2 + x − 5 7 = 0
The discriminant of this equation tells us there are also two real roots, but substitution of these roots back into the original problem tells us one of them is not a solution (will be explained later in the set to avoid spoilers).
Case 3: Base is 1, any exponent: x 2 + x − 5 7 = 1
The discriminant of this equation tells us that there are two real roots that work in the original equation.
Case 4: Base is -1, exponents have same parity: x 2 + x − 5 7 = − 1
The discriminant tells us that there are two real roots, but one doesn't work in the original equation (again, no spoilers here!).
So,
2 + 1 + 2 + 1 = 6
6 > 5 , giving us our answer, More\quad than\quad 5 .