Question

If p and q are the zeroes of the polynomial x2 - 9 , then pq( p +q ) is

Answer 1

Step-by-step explanation:

x²-9=0

(x+3)(x-3)=0

x=3 or -3

p=3 and q= -3

pq(p+q)=3(-3)(3-3)=-9(0)

pq(p+q)=0

Answer 2

The value of pq(p + q) = 0

Given :

p and q are the zeroes of the polynomial x² - 9

To find :

The value of pq(p + q)

Concept :

If α and β are the zeroes of the quadratic polynomial ax² + bx + c , then

$\displaystyle \sf Sum \: of \: the \: zeroes = \alpha + \beta = - \frac{b}{a}$

$\displaystyle \sf Product \: of \: the \: zeroes = \alpha \beta = \frac{c}{a}$

Solution :

Step 1 of 2 :

Find the lvalue of pq and (p + q)

Here it is given that p and q are the zeroes of the polynomial x² - 9

Comparing the polynomial x² - 9 with the polynomial ax² + bx + c we get a = 1 , b = 0 , c = - 9

$\displaystyle \sf Sum \: of \: the \: zeroes = p + q = - \frac{b}{a} = 0$

$\displaystyle \sf Product \: of \: the \: zeroes = pq = \frac{c}{a}= - 9$

Step 2 of 2 :

Find value of pq(p + q)

$\displaystyle \sf pq(p + q)$

$\displaystyle \sf = - 9 \times 0$

$\displaystyle \sf = 0$

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