Question

The zeroes of the polynomial p(y) = 5√5y² + 30y + 8√5 are a) -3/√5 , -7/√11 b) -2/√5 , -4/√5 c) 3/√5 , 3/√11 d) 2/√5 , 4√5​

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{P(y)=5\sqrt{5}y^2+30y+8\sqrt{5}}$

$\underline{\textbf{To find:}}$

$\textsf{Zeroes of P(y)}$

$\underline{\textbf{Solution:}}$

$\textsf{We apply factorization method to find the zeroes of P(y)}$

$\mathsf{Consider,}$

$\mathsf{P(y)=5\sqrt{5}y^2+30y+8\sqrt{5}}$

$\textsf{This can be written as}$

$\mathsf{P(y)=5\sqrt{5}y^2+20y+10y+8\sqrt{5}}$

$\mathsf{P(y)=5y(\sqrt{5}y+4)+2\sqrt{5}(\sqrt{5}y+4)}$

$\mathsf{P(y)=(\sqrt{5}y+4)(5y+2\sqrt{5})}$

$\mathsf{Now,\;P(y)=0}$

$\implies\mathsf{(\sqrt{5}y+4)(5y+2\sqrt{5})=0}$

$\implies\mathsf{\sqrt{5}y+4=0\;\;(or)\;\;5y+2\sqrt{5}=0}$

$\implies\mathsf{\sqrt{5}y=-4\;\;(or)\;\;5y=-2\sqrt{5}}$

$\implies\mathsf{y=\dfrac{-4}{\sqrt5}\;\;(or)\;\;y=\dfrac{-2\sqrt{5}}{5}}$

$\implies\mathsf{y=\dfrac{-4}{\sqrt5}\;\;(or)\;\;y=\dfrac{-2}{\sqrt{5}}}$

$\therefore\mathsf{Zeroes\;are\;\dfrac{-2}{\sqrt{5}},\dfrac{-4}{\sqrt{5}}}$

$\underline{\textbf{Answer:}}$

$\mathsf{Option\;(b)\;is\;correct}$

$\underline{\textbf{Find more:}}$

if α & β are the zeros of the quadratic polynomial p(x)= 3X²-6x+4,find the value of α\β+β+α+2(1/α+1/β)+3αβ​

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