Question

5.34. 1) x2 – 3x - 4 < 0; 3) x2 - 8x - 9 < 0;5) -x" + x + 6 >0;2) x² – 3x - 4>0;4) - x2 - 2x + 48 < 0;6) 2x2 + 3x - 5>0.только 1.3.5​

Answer 1

Ответ:

Объяснение:

1)x²-3x-4<0

x²-3x-4=0

Δ=15  , √Δ=5

x1=(3-5)/2=-2/2=-1

x2=(3+5)/2=8/2=4

   x ∈ (-1;4)

3) x²-8x-9 <0

x²-8x-9=0

Δ=100   ;  √Δ=10

x1=(8-10)/2=-2/2=-1     ;  x2=(8+10)/2=18/2=9

       x ∈ (-1 ; 9)

5) -x²+x+6 > 0

    -x²+x+6=0

Δ=25   ; √Δ=5

x1=(-1-5)/-2=-6/-2=3    ;  x2=(-1+5)/-2=3/-2=-1,5

  x ∈ (-1,5;3)

Answer 2

Sure, let's solve these inequalities:

1. $x^2 - 3x - 4 < 0$

This factors as $(x-4)(x+1) < 0$. The solution is $-1 < x < 4$.

3. $x^2 - 8x - 9 < 0$

This factors as $(x-9)(x+1) < 0$. The solution is $-1 < x < 9$.

5. $-x^2 + x + 6 > 0$

This factors as $-(x-3)(x+2) > 0$. The solution is $x < -2$ or $x > 3$.

So, the solutions for 1), 3), and 5) are $-1 < x < 4$, $-1 < x < 9$, and $x < -2$ or $x > 3$, respectively.

Looks like I've made an error in my interpretation of the question, though. Which of the provided options would you like to discuss?