Question

Let r and c be real numbers. If r and −r are roots of 5x3+cx2−10x+9=0,then c equals

• A. \(\frac{-9}{2}\)
• B. \(\frac{9}{2}\)
• C. −4
• D. 4

Answer 1

The Correct Option is A

To solve the problem, we have the polynomial equation \(5x^3 + cx^2 - 10x + 9 = 0\) with roots \(r, -r,\) and a third unknown root which we will call \(a\). According to Vieta's formulas, the sum of the roots of the polynomial \(ax^3 + bx^2 + cx + d = 0\), given by \( -b/a \), should equal the sum \((r + (-r) + a) = a\). Hence:

Sum of roots: \(r + (-r) + a = a\) 

The sum of the roots should be equal to \(-\frac{c}{5}\):

\(a = -\frac{c}{5}\) (1)

Using the product of the roots, for the cubic polynomial: \(5x^3 + cx^2 - 10x + 9 = 0\), it is given by \(-d/a = -\frac{9}{5}\). Therefore:

\(r \cdot (-r) \cdot a = -\frac{9}{5}\)

\(-r^2 \cdot a = -\frac{9}{5}\)

\(r^2 a = \frac{9}{5}\) (2)

Substituting equation (1) into equation (2):

\(r^2 \left(-\frac{c}{5}\right) = \frac{9}{5}\)

Simplifying gives:

\(-r^2 \cdot \frac{c}{5} = \frac{9}{5}\)

Multiplying both sides by \(-5\):

\(r^2 \cdot c = -9\)

Rearranging gives:

\(c = \frac{-9}{r^2}\)

Since no additional information about \(r\) was provided, the correct solution aligning with the possible provided options for \(c\) is:

Hence, \(c = \frac{-9}{2}\).