|x^2-10|=4
x^2-10=4 or x^2-10=-4
X^2-14=0 x^2-6=0
(x-7)(x+2)=0 (x-2)(x+3)=0
x=7 x=-2 x=2 x=-3
solution set {-3,-2,2,7} (?????)
Not quite. You cannot factor x^2-14=0 and get (x-7)(x+2)=0 nor can you factor x^2-6=0 and get (x-2)(x+3). Try to expand them and you'll see they are not the same.
(x-7)(x+2) = x^2-5x-14
(x-2)(x+3) = x^2+x-6
Your work was find until you tried to factor them. There is a much simpler(?) solution. Try to do it and I'll be happy to check your answer if needed.
Is this right?
x^2-14=0 x^2-6=0
x^2=14 x^2=6
sqrtx^2=sqrt14 sqrtx^2=sqrt6
x=+-sqrt14 x=+-sqrt6
solution set is:
{+-sqrt6, +-sqrt14}
Yes.
To solve the equation |x^2-10|=4, you need to consider two cases - one with the positive value inside the absolute value brackets and one with the negative value inside.
Case 1: x^2-10=4
To get the positive value case, you can remove the absolute value by setting the expression inside the brackets to be equal to the given positive value. So, x^2-10=4.
To solve this equation, add 10 to both sides: x^2 = 14.
Taking the square root of both sides gives you two possibilities:
x = √14 or x = -√14.
Case 2: x^2-10=-4
Now, consider the negative value case, where the expression inside the absolute value brackets becomes negative. So, x^2-10=-4.
To solve this equation, add 10 to both sides: x^2 = 6.
Taking the square root of both sides gives you two more possibilities:
x = √6 or x = -√6.
Combining the solutions from both cases, you have:
x = √14, -√14, √6, -√6.
So, the solution set for the equation |x^2-10|=4 is {-√14, √14, -√6, √6}.