SOLUTION: I am stuck on this problem. My answer is different from what the textbook answer says.
Find the domain of each function.
y=√2x^2 + 5x - 12 (The whole right side is und
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-> SOLUTION: I am stuck on this problem. My answer is different from what the textbook answer says.
Find the domain of each function.
y=√2x^2 + 5x - 12 (The whole right side is und
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Question 597389: I am stuck on this problem. My answer is different from what the textbook answer says.
Find the domain of each function.
y=√2x^2 + 5x - 12 (The whole right side is under the square root sign.
This is how I did it:
y=√2x^2 + 5x - 12
y-(x+4)(2x-3)
x+4 is greater or equal to 0
x is greater or equal to -4
2x-3 is greater or equal to 0
2x is greater or equal to 3
x is greater or equal to 3/2
However, the textbook answer says the answer is:
x is less than or equal to -4 OR x is greater or equal to 3/2
Please explain how the textbook got the answer.
Thanks!! Answer by edjones(8007) (Show Source):
You can put this solution on YOUR website! x is less than or equal to -4 OR x is greater or equal to 3/2
Test to see what 2x^2+5x-12 equals when -4 < x < 3/2 and you will find that it is a negative number. The sqrt of a negative number is not allowed in the domain.
.
Ed
