If the quadratic equation ax2 - bx - c = 0 has one root being thrice the other root, show that 16ac + 3b2 = 0.

Proved (See Explanation Below)

Step-by-step 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