Question

If \( \alpha \neq 0 \) and zero are the roots of the equation \( x^2 - 5kx + (6k^2-2k) = 0 \), then \( \alpha = \)

• A. \( \frac{1}{3} \)
• B. \( 1 \)
• C. \( \frac{5}{3} \)
• D. \( 5 \)

Hint:

If roots are \(\alpha\) and \(0\), then the product of roots is \(0\). Product from equation: \( (6k^2-2k)/1 = 6k^2-2k \). So, \(6k^2-2k = 0 \implies 2k(3k-1)=0 \implies k=0\) or \(k=1/3\). Sum of roots is \(\alpha+0 = \alpha\). Sum from equation: \(-(-5k)/1 = 5k\). So, \(\alpha = 5k\). Since \(\alpha \neq 0\), \(k \neq 0\). Thus \(k=1/3\). Then \(\alpha = 5(1/3) = 5/3\).

Answer 1

The Correct Option is C

The given quadratic equation is \( x^2 - 5kx + (6k^2-2k) = 0 \).
Let the roots be \(r_1 = \alpha\) and \(r_2 = 0\).
We are given \(\alpha \neq 0\).
From Vieta's formulas for a quadratic \(ax^2+bx+c=0\): Product of roots \(r_1 r_2 = c/a\).
Sum of roots \(r_1+r_2 = -b/a\).
For the given equation, \(a=1, b=-5k, c=6k^2-2k\).
1.
Product of roots: \( \alpha \cdot 0 = \frac{6k^2-2k}{1} \) \( 0 = 6k^2-2k \) \( 0 = 2k(3k-1) \) This implies \(2k=0\) or \(3k-1=0\).
So, \(k=0\) or \(k=\frac{1}{3}\).
2.
Sum of roots: \( \alpha + 0 = - \frac{-5k}{1} \) \( \alpha = 5k \).
3.
Evaluate \(\alpha\) for the possible values of \(k\): If \(k=0\), then \(\alpha = 5(0) = 0\).
This contradicts the given condition \(\alpha \neq 0\).
So, \(k \neq 0\).
If \(k=\frac{1}{3}\), then \(\alpha = 5\left(\frac{1}{3}\right) = \frac{5}{3}\).
This value \(\alpha = 5/3\) is not zero, so it is valid.
Therefore, \(\alpha = \frac{5}{3}\).
\[ \boxed{\frac{5}{3}} \]