Question

if p and q are the zeros of a polynomial X square + 4 x minus 3 find the polynomial whose zeros are 1 + P by q and 1 + q by P​

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

The required polynomial is $x^2+6x+2=0$.

Step-by-step explanation:

If a polynomial is $P(x)=ax^2+bx+c$, then

Sum of zeros = -b/a

Product of zeros =c/a

It is given that p and q are the zeros of a polynomial

$P(x)=x^2+4x-3$

So, we get

$p+q=-4$

$pq=-3$

We need to find the polynomial whose zeros are (1 + p)/q and (1 + q)/p​.

$\dfrac{1+p}{q}+\dfrac{1+q}{p}=\dfrac{p+p^2+q+q^2}{pq}=\dfrac{(p+q)+(p^2+q^2+2pq)-2pq}{pq}=\dfrac{(p+q)+(p+q)^2-2pq}{pq}$

$\dfrac{1+p}{q}\cdot \dfrac{1+q}{p}=\dfrac{1+p+q+pq}{pq}$

Substitute p+q=-4 and pq=-3.

$\dfrac{1+p}{q}+\dfrac{1+q}{p}=\dfrac{(-4)+(-4)^2-2(-3)}{-3}=-6$

$\dfrac{1+p}{q}\cdot \dfrac{1+q}{p}=\dfrac{1+(-4)+(-3)}{-3}=2$

Sum of zeros is -6 and product of zeros is 2.

x² - (Sum of zeros)x + Product of zeros = 0

$x^2+6x+2=0$

Therefore, the required polynomial is $x^2+6x+2=0$.

#Learn more

If p(x)=1/3x^2-5x+3/2 then find the sum and product for all its zeroes.

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