rebo1984 said:Hi,
If a = -1, what is the solution set of the equation sqrt (x+ a) = x −7
{5}
{10}
{5, 10}
No solution(5,10) correct?
Thanks
MarkFL said:What do you get when you substitute each potential solution into the original equation?
rebo1984 said:It works out.
What "works out"? Which answer did you get?rebo1984 said:It works out.
The value of x is equal to 3 when a is equal to -1.
To solve this equation, first square both sides to get rid of the square root. This yields x+ a = x^2 - 14x + 49. Then, rearrange the terms to get a quadratic equation in the form of ax^2 + bx + c = 0. In this case, it becomes x^2 - 15x + 49 + a = 0. Finally, use the quadratic formula to solve for x.
Yes, this equation can also be solved using the method of completing the square. This involves manipulating the equation to get it in the form of (x + b/2)^2 = c, where b and c are constants. Then, taking the square root of both sides and solving for x.
The domain of this equation is all real numbers except for -a, as it would result in division by zero. The range is also all real numbers, as the square root of any positive number results in a real number.
Yes, this equation can have multiple solutions depending on the value of a. In this case, when a is equal to -1, there are two solutions: x = 3 and x = 7.