Question

A positive root of the equation \[ 18x^3-9x^2-5x+2=0 \] is twice another root of it. If among its three roots, the least value is \(a\) and the greatest value is \(b\), then \(2a+3b=\)

• A. \(0\)
• B. \(1\)
• C. \(-1\)
• D. \(2\)

Hint:

When a relation among roots is given (such as one root being twice another), represent the roots accordingly and use Vieta's formulas to convert the condition into equations involving the unknown root.

Answer 1

The Correct Option is B

Concept: If one root is twice another, let the roots be \(r\), \(2r\), and \(s\). Then use Vieta's formulas to determine the roots.

Step 1: Assume the roots and apply Vieta's formulas. Let the roots be \[ r,\;2r,\;s. \] For \[ 18x^3-9x^2-5x+2=0, \] the sum of roots is \[ r+2r+s=\frac{-(-9)}{18} =\frac12. \] Hence, \[ 3r+s=\frac12. \]

Step 2: Use the product of roots. \[ r(2r)s = -\frac{2}{18} = -\frac19. \] Therefore, \[ 2r^2s=-\frac19. \] Substituting \[ s=\frac12-3r, \] we get \[ 2r^2\left(\frac12-3r\right) = -\frac19. \] \[ r^2-6r^3 = -\frac19. \] \[ 54r^3-9r^2-1=0. \]

Step 3: Find the value of \(r\). Checking rational roots, \[ r=\frac13 \] satisfies \[ 54r^3-9r^2-1=0. \] Thus, \[ r=\frac13, \qquad 2r=\frac23. \] Also, \[ s=\frac12-3\left(\frac13\right) = -\frac12. \] Hence, the roots are \[ -\frac12,\; \frac13,\; \frac23. \]

Step 4: Identify \(a\) and \(b\). The least root is \[ a=-\frac12, \] and the greatest root is \[ b=\frac23. \] Therefore, \[ 2a+3b = 2\left(-\frac12\right) + 3\left(\frac23\right). \] \[ =-1+2. \] \[ =1. \] \[ \boxed{2a+3b=1} \] \[ \boxed{\text{Answer = (B)}} \]