Question

If p and q are zeroes of 3x2 + 2x - 9, then value of p-q is?​

Answer 1

Answer:

p-q =

Step by vieta's rule:

we know,

If A & B be roots of ax² + bx + c = 0,

Then,

A+B = -b/a ,. A.B = c/a

For given Equation 3x²+2x-9 =0 , p=A & q=B

Now,

p+q= -2/3 ,. p.q = -9/3 => pq = -3

Therefore,

(p - q)² = (p+q)²- 4pq

= ( -2/3)² - 4(-3)

= 4/9 + 12, » (4 + 108)/9

(p-q)² = 112/9

(p-q) = ±√(112/9), » ± 4√7/3

.°. (p - q) = ± 4√7/3

Answer 2

Given: p and q are zeroes of $3 {x}^{2} + 2x - 9$

To find: Value of p-q

Explanation: Any quadratic equation is in the form of

$a {x}^{2} + bx + c$

where sum of roots= -b/a

and product of roots= c/a

In this question, a = 3 ,b =2 and c = -9

Therefore, p+q = -2/3

pq = -9/3

= -3

Now, for finding p-q the formula used is:

$({p - q)}^{2} = {(p + q)}^{2} - 4pq$

$({p - q)}^{2} = {( \frac{ - 2}{3} )}^{2} - (4 \times - 3)$

$({p - q)}^{2} = \frac{4}{9} + 12$

$({p - q)}^{2} = \frac{4 + 108}{9}$

$({p - q)}^{2} = \frac{112}{9}$

$p - q = \sqrt{ \frac{112}{9} }$

$p - q = \frac{4 \sqrt{7} }{3}$

The value of p-q is $\frac{4 \sqrt{7} }{3}$