Question

Find an inequality for "K" in its simplest form such that the quadratic equation kx²-(2k+3)x+(k+2)=0 has two distinct solutions

Answer 1

The inequality for 'k' such that the quadratic equation kx² - (2k+3)x + (k+2) = 0 has two distinct solutions is k > -9/4. This is found by ensuring the discriminant (2k+3)² - 4(k)(k+2) is positive and simplifying the resulting inequality.

Step-by-step explanation:

To determine an inequality for 'k' that ensures the quadratic equation kx² - (2k+3)x + (k+2) = 0 has two distinct solutions, we need to consider the discriminant of the quadratic equation, which is given by b² - 4ac from the standard form ax² + bx + c = 0. For two distinct solutions to exist, the discriminant must be positive, which gives us the condition (2k+3)² - 4(k)(k+2) > 0. Simplifying this inequality will give us the simplest form for the range of values that 'k' can take to satisfy the condition.

Let's solve the inequality step by step:

1. Expand the square: (2k+3)² becomes 4k² + 12k + 9.
2. Multiply out the 4ac term: 4(k)(k+2) becomes 4k² + 8k.
3. Subtract 4k² + 8k from 4k² + 12k + 9, which simplifies to 4k + 9.
4. Our inequality thus simplifies to: 4k + 9 > 0.
5. Finally, solving for 'k', we get: k > -9/4.

Therefore, for the equation kx² - (2k+3)x + (k+2) = 0 to have two distinct solutions, it is necessary that k > -9/4.