I have this problem(S^(2)+(3S)-4 is greater than 0) now i really need help with just one part. I simplified till I got (S^(2)+S is greater than (4/3). I just need to know that when you try to get rid of the square(^2) do you square root all of both sides or just the S^2 and (4/3)?
Calc. homework
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\(\displaystyle s^{2}+3s-4>0\)
First off, you can not divide by 3 that way. It is an addition, not multiplication.
Try completing the square: \(\displaystyle (s+\frac{3}{2})^{2}-\frac{25}{4}=s^{2}+3s-4\)
\(\displaystyle (s+\frac{3}{2})^{2}>\frac{25}{4}\)
Now, you can take the +/- square root of both sides.
You could also just plain factor the quadratic.
First off, you can not divide by 3 that way. It is an addition, not multiplication.
Try completing the square: \(\displaystyle (s+\frac{3}{2})^{2}-\frac{25}{4}=s^{2}+3s-4\)
\(\displaystyle (s+\frac{3}{2})^{2}>\frac{25}{4}\)
Now, you can take the +/- square root of both sides.
You could also just plain factor the quadratic.
mmm4444bot
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galactus said:
I like this idea because the entire exercise can be done mentally.
Clearly, the quadratic polynomial factors as (s - 1)(s + 4).
Clearly, the parabola intersects the x-axis at x = -4 and x = 1.
Clearly, the parabola opens upward, so the parabola must lie below the x-axis between the x-intercepts.
Clearly, the quadratic is greater than zero everywhere else (other than -4 and 1, of course).
Therefore, the inequality solution set is x < -4 U x > 1.