Question

If one root of the quadratic equation \( y^2 + 3y - 18 = 0 \) is -6, then its other root is:

• A. 3
• B. -3
• C. 6
• D. 5

Hint:

The sum and product of the roots of a quadratic equation are given by \( r_1 + r_2 = -\frac{b}{a} \) and \( r_1r_2 = \frac{c}{a} \).

Answer 1

The Correct Option is B

For a quadratic equation \( ay^2 + by + c = 0 \), the sum and product of the roots \( r_1 \) and \( r_2 \) are given by: \[ r_1 + r_2 = -\frac{b}{a}, \quad r_1r_2 = \frac{c}{a}. \] For the equation \( y^2 + 3y - 18 = 0 \), we have \( a = 1 \), \( b = 3 \), and \( c = -18 \). Thus, the sum and product of the roots are: \[ r_1 + r_2 = -\frac{3}{1} = -3, \quad r_1r_2 = \frac{-18}{1} = -18. \] Given that one root is \( r_1 = -6 \), we can find the other root \( r_2 \) by solving: \[ r_1 + r_2 = -3 \quad \Rightarrow \quad -6 + r_2 = -3 \quad \Rightarrow \quad r_2 = 3. \] Thus, the other root is \( \boxed{-3} \).