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# Algebra

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For an integer n, the inequality
x^2 + nx + 15 < -89 - 3x^2
has no real solutions in $x$. Find the number of different possible values of $n$.

Jun 24, 2024

#1
+1230
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First, let's move all terms to one side and simplify. We have

$$4x^2 + nx + 104 < 0$$

Let's set this up as an equation. We have

$$4x^2 + nx + 104 = 0$$

This will have real solutions when

$$n^2 - 4 (4) (104) ≥ 0 \\ n^2 ≥ 1664$$

$$n ≥ 40.79 \\ n ≤ -40.79$$

This is when there is a real value. Which means that when

$$n < 40.79 \\ n > -40.79$$

there will be no real solutions. Thus, we have

$$n = [ -40 , 40 ]$$

There are 81 values of n.

Thus our answer is 81 !

Thanks! :)

Jun 24, 2024
#2
+85
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$$x^2 + nx + 15 < -89 - 3x^2$$

so i first brought everything to the left side to get $$4x^2 + nx + 104 < 0$$

then i divided by 4 to get$$x^2 + \frac{n}{4}x + 26 < 0$$

for x to have no real solutions the discriminant must be negative

so $$\frac{n^2}{16}-4(26)$$ must be negative. the less than sign is useless since smaller right hand values make c greater

the greatest value of n that works is 40

but we also can have negatives, least of which is -40

so -40 to 40 is $$\boxed{81}$$ numbers

Jun 24, 2024