What do find all the real solutions mean and how do I find them?

Question

Examples: 10/x-12/x-3+4=0, sqr5-x+1=x-2 and x^6-2x^3=0

Answer 1

Sometimes a term in an equation equates to the nth root of a negative number, which is imaginary (not real).

10/x - 12/x -3 + 4 = 0
First, subtract the x terms, since they already have a common denominator:
((10 - 12) / x) - 3 + 4 = 0
Then simplify:
-2/x + 1 = 0
Subtract 1 from both sides:
-2/x = -1
Divide both sides by -2:
-2 / -2x = -1 / -2
1/x = -1 / -2
Reciprocate both sides:
x / 1 = -2 / -1
x = 2, a real solution

√5 - x + 1 = x - 2
Subtract x from both sides:
√5 - x + 1 - x = -2
√5 - x - x + 1 = -2
Simplify the x's and subtract 1 from both sides:
√5 - 2x = -3
Subtract √5 from both sides:
-2x = -3 - √5
Divide both sides by -2:
x = (-3 - √5) / -2
x = (3/2) + (1/2√5)
x = 1½ + (1/2√5)
That's about as far as you can go without computing √5, but it is also a real solution.

x^6 - 2x^3 = 0
First factor out x^3:
x^3(x^3 - 2) = 0
If the product of two quantities is zero, then one of the quantities is zero.
If x^3 = 0, then x = 0
If (x^3 - 2) = 0, then
x^3 = 2
Taking the cube root of both sides:
x = ³√2
Therefore, x = 0 and ³√2, both real solutions.

Answer 2

For the first two you need some parentheses.

Here's the last one.
x^6 - 2x^3 = 0
Factor out x^3
x^3 (x^3 -2) = 0
x^3 = 0 or x^3 -2 = 0
x = 0 or x = cube root (2)

I'll take a guess at the first one (when I say "guess" I mean I'm guessing at where the parentheses should be).
10/x - 12/(x-3) +4 = 0
Multiply both sides by x(x-3)
10(x-3) - 12x + 4x(x-3) = 0
10x - 30 - 12x + 4x^2 - 12x = 0
4x^2 - 14x - 30 = 0
Divide both sides by 2
2x^2 -7x -15 = 0
Factor
(x-5)(2x+3) = 0
x-5 = 0 or 2x+3 = 0
x = 5 or x = -3/2

I'll guess where the parentheses should be on the second one too.
sqrt(5-x) + 1 = x - 2
Subtract 1 from each side
sqrt(5-x) = x - 3
Square both sides
5 - x = x^2 - 6x + 9
Move everything to one side
x^2 - 5x + 4 = 0
Factor
(x-4)(x-1) = 0
x-4 = 0 or x-1 = 0
x = 4 or x = 1
Be sure to CHECK both the answers. One of them is NOT a solution.

Answer 3

Real solutions means when you substitute the value of x (real number) in the equation the equation is eq.

Like x -5 = 1 then x = 6 is a real solution.

As for your eqs listed I think the parenthesis are missing or something. Try asking another question with the eq type out correctly.

Answer 4

The real soln. means that when u put a real ( +ve ) value of x in the given equation , the equations holds good in each condition and for all values of x .