Question

If one of the zeroes of the quadratic polynomial (p-1)x^2+px+1 is -3,then the value of p is.

Answer 1

Answer:

$p = \cfrac{4}{3}$

Step-by-step explanation:

Let given quadratic equation is

$p(x) = (p-1)x^2 + px +1$

Since $-3$ is the zero of p(x), so p(-3) = 0

$\therefore (p-1)(-3)^2 +p(-3) + 1 = 0$

$\Rightarrow (p-1)(9) -3p + 1 = 0$

$\Rightarrow 9p - 9 -3p + 1 = 0$

$\Rightarrow 6p-8=0$

$\Rightarrow p = \cfrac{8}{6} = \cfrac{4}{3}$

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Answer 2

Step-by-step explanation:

p=

3

4

Step-by-step explanation:

Let given quadratic equation is

p(x) = (p-1)x^2 + px +1p(x)=(p−1)x

2

+px+1

Since -3−3 is the zero of p(x), so p(-3) = 0

\therefore (p-1)(-3)^2 +p(-3) + 1 = 0∴(p−1)(−3)

2

+p(−3)+1=0

\Rightarrow (p-1)(9) -3p + 1 = 0⇒(p−1)(9)−3p+1=0

\Rightarrow 9p - 9 -3p + 1 = 0⇒9p−9−3p+1=0

\Rightarrow 6p-8=0⇒6p−8=0

\Rightarrow p = \cfrac{8}{6} = \cfrac{4}{3}⇒p=

6

8

= 3

4

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