Question

if -2 and 3 are the zeroes of the quadratic polynomial x²+(p+1)x+q, then find the values of p and q​

Answer 1

$\large\underline{\sf{Given- }}$

• -2 and 3 are the zeroes of the quadratic polynomial x² + (p + 1) x + q

$\large\underline{\sf{To\:Find - }}$

$\: \: \: \: \: \: \: \: \bull \sf \: values \: of \: p \: and \: q$

$\large\underline{\sf{Solution-}}$

We know that

$\sf \: If \: \alpha \: and \: \beta \: are \: zeroes \: of \: f(x) = {ax}^{2} + bx + c \: \: then$

$\boxed{\red{\sf Sum\ of\ the\ zeroes \: = \alpha + \beta =\dfrac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

OR

$\boxed{\purple{\tt Sum\ of\ the\ zeroes = \alpha + \beta =\dfrac{-b}{a}}}$

And

$\boxed{\red{\sf Product\ of\ the\ zeroes = \alpha \beta =\dfrac{Constant}{coefficient\ of\ x^{2}}}}$

OR

$\boxed{\purple{\tt Product\ of\ the\ zeroes = \alpha \beta =\dfrac{c}{a}}}$

Now,

Given quadratic polynomial is

$\: \: \: \: \: \: \: \: \bull \: \: \sf \: f(x) \: = \: {x}^{2} + (p + 1)x + q$

↝ On Comparing with ax² + bx + c, we get

$\: \: \: \: \: \: \: \: \bull \sf \: a \: = \: 1$

$\: \: \: \: \: \: \: \: \bull \sf \: b \: = \: p + 1$

$\: \: \: \: \: \: \: \: \bull \sf \: c \: = \: q$

↝ Aɢᴀɪɴ,

↝ -2 and 3 are the zeroes of the quadratic polynomial x² + (p + 1)x + q,

$\: \: \: \: \: \: \: \: \bull \sf \: \alpha = - \: 2$

$\: \: \: \: \: \: \: \: \bull \sf \: \beta \: = \: 3$

Now,

$\rm :\longmapsto\: \alpha + \beta = - \: \dfrac{b}{a}$

$\rm :\longmapsto\: - 2 + 3 = - \: \dfrac{(p + 1)}{1}$

$\rm :\longmapsto\:1 = - p - 1$

$\bf\implies \:p \: = \: - \: 2$

Also,

$\rm :\longmapsto\: \alpha \beta = \dfrac{c}{a}$

$\rm :\longmapsto\:( - 2) \times 3 = \dfrac{q}{1}$

$\bf\implies \:q = - \: 6$

$\overbrace{ \underline { \boxed { \bf \therefore \: The \: value \: of \:p = - 2 \: \: and \: \: q = - 6}}}$

Answer 2

Answer:

p = -2   and q = - 6

Step-by-step explanation:

If -2 and 3 are the zeroes of the given polynomial(and be 0), these must satisfy the equation for x = -2 and x = 3.

 For x = - 2 :  f(-2) = 0

⇒ (-2)² +  (p + 1)(-2) + q = 0

⇒ 4 - 2p - 2 + q = 0

⇒ 2p - 2 = q            

For x = 3 :   f(3) = 0

⇒ (3)² + (p + 1)(3) + q = 0

⇒ 9 + 3p + 3 + q = 0

⇒ 12 + 3p + (2p - 2) = 0

⇒ p = - 2

  thus. 2(-2) - 2 = q = - 6