Question

if a and b are the zeroes of quadratic polynomial ax2+bx+c then evaluate a-b

Answer 1

$since \: \alpha \: and \: \beta \: are \: the \: roots \\ \alpha + \beta = \frac{ - b}{a} \: and \: \alpha \beta = \frac{c}{a} \\ using \: the \: identity \\ {( \alpha + \beta )}^{2} - {( \alpha - \beta )}^{2} = 4 \alpha \beta \\ {( \alpha - \beta )}^{2} = {( \frac{ - b}{a} )}^{2} - 4 \times \frac{c}{a} \\ = \frac{ {b}^{2} - 4ac}{ {a}^{2} } \\ therefore \\ \alpha - \beta = \frac{ \sqrt{ {b}^{2} - 4ac } }{a}$

Answer 2

If p and q are the zeroes of quadratic polynomial ax²+bx+c then evaluate p-q.

(this is the correct form of the question.)

Given:

p and q are the zeroes of quadratic polynomial ax²+bx+c.

To Find:

The value of p-q.

Solution:

1) If p and q are the zeros of the polynomial the there is a relation between the zeros and the coefficient of the quadratic polynomial.

2) Sum of the roots = p+q= −b/a = −(coefficient of x) / (coefficient of x²)

Product of the toots = pq = c/a = (coefficient of x) / (coefficient of x²)

3) To find p-q we have to find the square of the first expression.

• (p+q)²= (−b/a)²                                                          
• p² + q² + 2pq = b²/a²
• p² + q² + 2pq - 4pq = b²/a² - 4pq (Subtract 4pq form both side)
• p² + q² - 2pq = b²/a² - 4c/a
• (p-q)² = (b² - 4ac) / a²

• (p-q) = √[(b² - 4ac) / a²]
• (p-q) = √(b² - 4ac) / a

The value of  (p-q) = √(b² - 4ac) / a