Question

(7p-1)(7p +1)<49p (2a+3) (2a+1)>4a(a+2)

Answer 1

4a^2+6a+2a+3>4a^2+6a
4a^2+6a+2a+3-4a^2-6a>0
2a+3>0
2a>-3
a>-3/2

Answer 2

To solve the given inequalities, let's break them down and simplify step by step:

1. (7p - 1)(7p + 1) < 49p

Expanding the left side: 49p^2 - 1 < 49p

Rearranging the terms: 49p^2 - 49p - 1 < 0

Now, let's solve this quadratic inequality by finding the critical points:

To find the critical points, we set the quadratic expression equal to zero: 49p^2 - 49p - 1 = 0

Using the quadratic formula: p = (-(-49) ± √((-49)^2 - 4 * 49 * (-1))) / (2 * 49) p = (49 ± √(2401 + 196)) / 98 p = (49 ± √2597) / 98

The critical points are p = (49 + √2597) / 98 and p = (49 - √2597) / 98.

Now, we need to determine the sign of the expression (7p^2 - 49p - 1) in each interval formed by the critical points.

Considering the intervals: Interval 1: (-∞, (49 - √2597) / 98) Interval 2: ((49 - √2597) / 98, (49 + √2597) / 98) Interval 3: ((49 + √2597) / 98, ∞)

We can pick test points within each interval to determine the sign of the expression: For Interval 1, let's pick p = 0: Substituting p = 0: 7(0)^2 - 49(0) - 1 = -1

Since -1 is negative, the expression (7p^2 - 49p - 1) is negative in Interval 1.

For Interval 2, let's pick p = 1: Substituting p = 1: 7(1)^2 - 49(1) - 1 = -42

Since -42 is negative, the expression (7p^2 - 49p - 1) is negative in Interval 2.

For Interval 3, let's pick p = 2: Substituting p = 2: 7(2)^2 - 49(2) - 1 = -57

Since -57 is negative, the expression (7p^2 - 49p - 1) is negative in Interval 3.

Therefore, the solution to the inequality (7p - 1)(7p + 1) < 49p is: p ∈ ((49 - √2597) / 98, (49 + √2597) / 98)

2. (2a + 3)(2a + 1) > 4a(a + 2)

Expanding the left side: 4a^2 + 2a + 6a + 3 > 4a^2 + 8a

Combining like terms: 4a^2 + 8a + 3 > 4a^2 + 8a

Subtracting (4a^2 + 8a) from both sides: 3 > 0

The inequality 3 > 0 is always true.

Therefore, the solution to the inequality (2a + 3)(2a + 1) > 4a(a + 2) is: a can take any real value.

In summary:

1. For (7p - 1)(7p + 1) <