Question

URGENT...If the zeroes of polynomial px^2+qx+r are of the form B/B-1 and B+1/B then the value of (p+q+r)^2 is
Answer is q^2-4pr
But how...plz explain

Answer 1

The value of $(p+q+r)^2=q^2-4pr$

Step-by-step explanation:

The given quadratic poynomial:

$px^2+qx+r$

To find, the value of $(p+q+r)^2=?$

The sum of the roots,

$\dfrac{B}{B-1}+\dfrac{B+1}{B}=-\dfrac{q}{p}$

The product of the roots,

$\dfrac{B}{B-1}.\dfrac{B+1}{B}=\dfrac{r}{p}$

∴ $(\dfrac{B}{B-1}-\dfrac{B+1}{B})^2=(\dfrac{B}{B-1}+\dfrac{B+1}{B})^2-4(\dfrac{B}{B-1}.\dfrac{B+1}{B})$

⇒ $(\dfrac{B}{B-1}-\dfrac{B+1}{B})^2=(-\dfrac{q}{p} )^2-4(\dfrac{r}{p})$

⇒ $(\dfrac{B}{B-1}-\dfrac{B+1}{B})^2=\dfrac{q^2-4pr}{p^2}$

⇒ $(p+q+r)^2=q^2-4pr$

Hence, the value of $(p+q+r)^2=q^2-4pr$