Original post by cemfl

How many integer solutions are there to the inequality x^2 - x - 20 < 0?

8 I think

Original post by Amber12100

8 I think

But how do you work it out??

Original post by Amber12100

First factorise the equation pretend it’s x^2 - x -20 = 0 so when factorised it would be (x-5) (x+4) find the values of x which would be +5 and -4 now because it says x^2 - x - 20 < 0 then when you imagine a u shaped graph the values that fits the equation would be under 0 on the y axis(as the equation is less than 0) .Therefore the integers would have to be within the inequality -4<x<5 because it says less than not less than and equal to the integers that fit in the equality are -3,-2,-1,0,1,2,3,4 and that is 8 integers in tota

In this instance, if you happen forget how to solve a quadratic inequality, brute forcing by hand is also not a bad option, since the numbers are small (20 is small enough). That is, we literally just count, "is 0 okay, is 1 okay, is -1 okay..."

Though the systematic way in #3 is recommended, as solving a quadratic inequality should be mechanical.

Though the systematic way in #3 is recommended, as solving a quadratic inequality should be mechanical.

Original post by Amber12100

First factorise the equation pretend it’s x^2 - x -20 = 0 so when factorised it would be (x-5) (x+4) find the values of x which would be +5 and -4 now because it says x^2 - x - 20 < 0 then when you imagine a u shaped graph the values that fits the equation would be under 0 on the y axis(as the equation is less than 0) .Therefore the integers would have to be within the inequality -4<x<5 because it says less than not less than and equal to the integers that fit in the equality are -3,-2,-1,0,1,2,3,4 and that is 8 integers in tota

Please edit - you are breaking forum rules by posting a solutionn.

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can someone please explain what principle domain is and why the answer is a not c?Maths

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