What is the solution for 2x^2-7x > always greater than -3 ? (2 marks) (Show your work)
To find the solution for the inequality 2x^2 - 7x > -3, we need to manipulate the equation to isolate the variable x. Here's how you can solve it step by step:
1. Subtract (-3) from both sides:
2x^2 - 7x + 3 > 0
Now we have a quadratic inequality.
2. To solve this, we will factorize the quadratic expression:
2x^2 - 6x - x + 3 > 0
2x(x - 3) - 1(x - 3) > 0
(2x - 1)(x - 3) > 0
3. Set each factor equal to zero to find the critical points:
2x - 1 = 0 and x - 3 = 0
2x = 1 and x = 3
x = 1/2 and x = 3
Now we have two critical points, x = 1/2 and x = 3.
4. Create a number line with the critical points (1/2 and 3) and test a value from each interval to determine the sign of the inequality:
Test x = 0: (2(0) - 1)(0 - 3) > 0 => (-1)(-3) > 0 => 3 > 0 (True)
Test x = 2: (2(2) - 1)(2 - 3) > 0 => (3)(-1) > 0 => -3 > 0 (False)
Test x = 4: (2(4) - 1)(4 - 3) > 0 => (7)(1) > 0 => 7 > 0 (True)
Using the test values, we discover that the inequality is true when x is less than 1/2 or when x is greater than 3, and false when x is between 1/2 and 3.
5. Express the solution using interval notation:
(-∞, 1/2) U (3, +∞)
Therefore, the solution to the inequality 2x^2 - 7x > -3 is (-∞, 1/2) U (3, +∞).