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Ram Kushwah 6 years, 7 months ago
f(x)=x2-3x-2
an b are zeros
so a+b=3,ab=-2
now if the zeros are 1/2a+b and 1/2b+a
{tex}\begin{array}{l}\frac1{2a+b}+\frac1{2b+a}=\frac{2a+b+2b+a}{(2a+b)(2b+a)}=\frac{3(a+b)}{(2a+b)(2b+a)}\\=\frac{3(3)}{4ab+ab+2a^2+2b^2}=\frac9{5\ast(-2)+2\lbrack\;(a+b)^2-2ab)}\\=\frac9{-10+2\ast(9-2(-2)\rbrack}=\frac9{-10+2\ast(9+4)}=\frac9{-10+26}=\frac9{16}\\\\\end{array}{/tex}
{tex}\begin{array}{l}Product\;of\;zeros=\frac1{2a+b}\times\frac1{2b+a}=\frac1{(2a+b)(2b+a)}=\frac1{(2a+b)(2b+a)}\\=\frac1{4ab+ab+2a^2+2b^2}=\frac1{5\ast(-2)+2\lbrack\;(a+b)^2-2ab)}\\=\frac1{-10+2\ast(9-2(-2)\rbrack}=\frac1{-10+2\ast(9+4)}=\frac1{-10+26}=\frac1{16}\\\\\end{array}{/tex}
hence the polynomial
=x2-(9/16)x+1/16
or 16x2-9x+1
1Thank You