Question

if alpha and beta are the zeros of the quadratic polynomial 25x^2+25x+5 then find the value of 1 upon alpha ^2+1 upon beta^2.

Answer 1

Hi.....plz find the answer in the following attachment below.!

( 15 is the final answer ).

Answer 2

Answer:

$\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}=15$

Step-by-step explanation:

$Given\:\alpha \:\beta \:are \\the\:zeroes \:of \:the\\quadratic\: polynomial\\25x^{2}+25x+5,$

Compare this polynomial with ax²+bx+c , we get

a = 25, b=25, c = 5

$i)Sum\:of\:the\: zeroes\\=-\frac{b}{a}$

$\implies \alpha+\beta=-\frac{25}{25}\\=-1--(1)$

$ii) Product\:of\: the\:zeroes\\=\frac{c}{a}$$\implies \alpha\beta=\frac{5}{25}\\=\frac{1}{5}--(2)$

$\alpha^{2}+\beta^{2}\\=(\alpha+\beta)^{2}-2\alpha\beta\\=(-1)^{2}-2\times \frac{1}{5}\\=1-\frac{2}{5}\\=\frac{5-2}{5}\\=\frac{3}{5}--(3)$

$Now,\\\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}\\=\frac{\alpha^{2}+\beta^{2}}{\alpha^{2}\beta^{2}}\\=\frac{\alpha^{2}+\beta^{2}}{\left(\alpha\beta\right)^{2}}\\=\frac{\left(\frac{3}{5}\right)}{\left(\frac{1}{5}\right)^{2}}$

$=\frac{\left(\frac{3}{5}\right)}{\left(\frac{1}{25}\right)}\\=\frac{3}{5}\times \frac{25}{1}\\=15$

Therefore,

$\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}=15$

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