Question

X²+8x+15>0x²-4x-5>0​

Answer 1

x²+8x+15>0

x²+8x+15=0, a=1,b=8, c= 15

D=b²-4ac=8²-4×1×15=64-60=4

x1=-b-√D/2 = -8-√4/2=-8-2/2=-10:2=-5

x2=-b+√D /2=-8+√4/2=-8+2/2=-3

Answer 2

To solve the inequalities $x^2 + 8x + 15 > 0$ and $x^2 - 4x - 5 > 0$, we'll break it down into two separate inequalities and solve each one separately.

1. Solve $x^2 + 8x + 15 > 0$: To solve this quadratic inequality, we can use the factored form. However, since the discriminant is positive (in this case, it's $b^2 - 4ac = 8^2 - 4 \cdot 1 \cdot 15 = 64 - 60 = 4$), we know that the parabola opens upwards, and there will be two distinct real solutions. Therefore, we can use the sign of the coefficient of $x^2$ to determine the intervals where the inequality holds.

Let's find the roots of $x^2 + 8x + 15 = 0$ by factoring: $x^2 + 8x + 15 = (x + 3)(x + 5) = 0$

Setting each factor to zero and solving for $x$: $x + 3 = 0 \implies x = -3$ $x + 5 = 0 \implies x = -5$

Now, we have three intervals to check: $(- \infty, -5)$, $(-5, -3)$, and $(-3, \infty)$.

1. For $x < -5$, we can choose $x = -6$: $x^2 + 8x + 15 = (-6)^2 + 8(-6) + 15 = 36 - 48 + 15 = 3$ Since $3 > 0$, the inequality holds for $x < -5$.

2. For $-5 < x < -3$, we can choose $x = -4$: $x^2 + 8x + 15 = (-4)^2 + 8(-4) + 15 = 16 - 32 + 15 = -1$ Since $-1 \nless 0$, the inequality does not hold for $-5 < x < -3$.

3. For $x > -3$, we can choose $x = 0$: $x^2 + 8x + 15 = 0^2 + 8(0) + 15 = 15$ Since $15 > 0$, the inequality holds for $x > -3$.

Thus, the solution to $x^2 + 8x + 15 > 0$ is $x < -5$ or $x > -3$.

2. Solve $x^2 - 4x - 5 > 0$: Similarly, let's find the roots of $x^2 - 4x - 5 = 0$ by factoring: $x^2 - 4x - 5 = (x - 5)(x + 1) = 0$

Setting each factor to zero and solving for $x$: $x - 5 = 0 \implies x = 5$ $x + 1 = 0 \implies x = -1$

Now, we have three intervals to check: $(- \infty, -1)$, $(-1, 5)$, and $(5, \infty)$.

1. For $x < -1$, we can choose $x = -2$: $x^2 - 4x - 5 = (-2)^2 - 4(-2) - 5 = 4 + 8 - 5 = 7$