Answer:
Proved (See Explanation Below)
Explanation:
Given that one root being thrice the other root
Let the first root = m
And the second root = n
From the question, we have that m = 3n
So, x = m or x = n
Recall that m = 3n
x = 3n or x = n
Equate both expressions to 0
x - 3n = 0 or x - n = 0
Multiply both expressions together
(x - 3n)(x - n) = 0 * 0
(x - 3n)(x - n) = 0
Open bracket
x² - nx - 3nx + 3n² = 0
x² - 4nx + 3n² = 0
According to the question, the quadratic equation is ax² - bx - c = 0.
By comparison,
a = 1
-b = -4n
b = 4n
-c = 3n²
c = -3n²
So, a = 1; b = 4n and c = -3n²
To check if 16ac + 3b² = 0, we plug in the above values
So, 16ac + 3b² = 0 becomes
16 * 1 * -3n² + 3 * (-4n)² = 0
-48n² + 3(16n²) = 0
-48n² + 48n² = 0
0 = 0
Since the value at the right hand side (0) equals the value at left hand side (0)
Then, 16ac + 3b² = 0 is proved