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# C1 watch

1. Find the set of values of x for which:

x2-7x-18>0

I had factorised it, which got me (x-9) (x+2) > 0

Therefore shouldn't the answer be -2<x<9?
2. Because of the > 0 sign, you want all the x values that will give corresponding y values that are above 0. Imagine a positive quadratic graph that intersects on the left at -2 and the right at +9. The corresponding y values lie on either side of these two values, i.e. less than -2 and greater than +9

x<-2
x>9

because 0 > , y values must be positive

3. (Original post by rm_27)
Find the set of values of x for which:

x2-7x-18>0

I had factorised it, which got me (x-9) (x+2) > 0

Therefore shouldn't the answer be -2<x<9?
Susbtitute a value of x between -2 and 9 into the inequality - that will show that your answer is incorrect.

Sketch the graph and it should become clearer.
4. (Original post by rm_27)
Find the set of values of x for which:

x2-7x-18>0

I had factorised it, which got me (x-9) (x+2) > 0

Therefore shouldn't the answer be -2<x<9?

Either graph the function as already suggested, or note that you need either both brackets positive or both brackets negative in order for the product to be positive and see what these possibilities lead to.

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