Plug in - 5 where X is.
f(- 5) = (- 5)^2 - ( - 5).
(- 5)( - 5) = 25 so now we have 25 - (- 5). Because there are two negatives back to back, the number becomes a positive so the equation becomes 25 + 5 = 30.
f(- 5) = 30.
An integer may be a multiple of 3.
An integer may be 1 greater than a multiple of 3.
An integer may be 2 greater than a multiple of 3.
It is redundant to say an integer is 3 greater than a multiple of 3 (that's just a multiple of 3, we've got it covered). Same for 4, 5, 6, 7...
Let's consider a number which is a multiple of 3. Clearly, we can write 3+3+3+3+... until we reach the number. It can be written as only 3's.
Let's consider a number which is 2 greater than a multiple of 3. If we subtract 5 from that number, it'll be a multiple of 3. That means we can write the number as 5+3+3+3+3+... Of course, the number must be at least 8.
Let's consider a number which is 1 greater than a multiple of 3. If we subtract 5 from that number, it'll be 2 greater than a multiple of 3. If we subtract another 5, it'll be a multiple of 3. That means we can write the number as 5+5+3+3+3+3+... Of course, the number must be at least 13.
That's it. We considered all the numbers. We forgot 9, 10, 11, and 12, but these are easy peasy.
Beautiful question.
Answer:
no solution
Step-by-step explanation:
For getting the nature of solution of the quadratic equation of the form:
ax² + bx + c = 0
We need to find Discriminant which is:
Discriminant (D) = b² - 4ac
- If D < 0, there is no solution of equation.
- If D = 0, there are two equal and real solution of equation
- If D < 0, there are two real and distinct solution of equation
Here we have equation is:
2x² - 9x + 12 = 0
∴ a=2, b = -9, c = 12
⇒ D = 81 - 4 × 2 × 12 = -16 < 0
Hence, there is no solution of given equation.
