If α and β (α > β) are the zeros of the polynomial x² + ax + 0, then (α - β) is equal to [ ] a) 10 b) 10 c) 8 d) -10 If one zero of the polynomial 6x² + 37x + k is the reciprocal o... If α and β (α > β) are the zeros of the polynomial x² + ax + 0, then (α - β) is equal to [ ] a) 10 b) 10 c) 8 d) -10 If one zero of the polynomial 6x² + 37x + k is the reciprocal of the other, then what is the value of k? a) 6 b) -4 c) -n d) 4 Read more

Steps to Solve

1. Identifying the Zeros of the First Polynomial

The given polynomial is ( x^2 + ax + 0 ). The roots (zeros) can be determined using the quadratic formula:

$$ \alpha, \beta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

Here ( a = 1 ), ( b = a ), and ( c = 0 ). This simplifies to:

$$ \alpha, \beta = \frac{-a \pm \sqrt{a^2}}{2} $$

Since ( c = 0 ), the roots are:

$$ \alpha = 0 \quad \text{and} \quad \beta = -a $$

2. Calculating the Difference Between the Roots

Now find ( \alpha - \beta ):

$$ \alpha - \beta = 0 - (-a) = a $$

Thus, ( \alpha - \beta = a ).

3. Evaluating Conditions for Second Polynomial

For the polynomial ( 6x^2 + 37x + k ), where one root is the reciprocal of the other, denote the roots as ( r ) and ( \frac{1}{r} ).

Using Vieta's formulas:

• The sum of the roots is:

$$ r + \frac{1}{r} = -\frac{b}{a} = -\frac{37}{6} $$

• The product of the roots is:

$$ r \cdot \frac{1}{r} = \frac{c}{a} = \frac{k}{6} $$

This means the product is 1:

$$ \frac{k}{6} = 1 \implies k = 6 $$