Question 12 Marks
Find the quadratic polynomial whose sum and product of the zeroes are $\frac{21}{8}$ and $\frac{5}{16}$ respectively.
Answer
View full question & answer→Let the polynomial be
$
p(x)=a x^2+b x+c
$
Its given that, Sum of zeroes $=\frac{21}{8}$
$
\Rightarrow \frac{-b}{a}=\frac{21}{8}
$
Assuming $a=1$,
$
\Rightarrow b=-\frac{21}{8}
$
We also know that, Product of zeroes $=\frac{5}{16}$
$
\Rightarrow \frac{c}{a}=\frac{5}{16}
$
Assuming $a=1$,$
\Rightarrow c=\frac{5}{16}
$
Now, $a=1, b=-\frac{21}{8}, c=\frac{5}{16}$
Hence, the required quadratic polynomial
$
=a x^2+b x+c
$
Substituting the values of $a, b$ and $c$ in the above equation, we get,
$
x^2-\frac{21}{8} x+\frac{5}{16}
$
Multiply the equation by 16 .
$
\Rightarrow 16 x^2-42 x+5
$
Hence, the required quadratic polynomial is $16 x^2-42 x+5$
$
p(x)=a x^2+b x+c
$
Its given that, Sum of zeroes $=\frac{21}{8}$
$
\Rightarrow \frac{-b}{a}=\frac{21}{8}
$
Assuming $a=1$,
$
\Rightarrow b=-\frac{21}{8}
$
We also know that, Product of zeroes $=\frac{5}{16}$
$
\Rightarrow \frac{c}{a}=\frac{5}{16}
$
Assuming $a=1$,$
\Rightarrow c=\frac{5}{16}
$
Now, $a=1, b=-\frac{21}{8}, c=\frac{5}{16}$
Hence, the required quadratic polynomial
$
=a x^2+b x+c
$
Substituting the values of $a, b$ and $c$ in the above equation, we get,
$
x^2-\frac{21}{8} x+\frac{5}{16}
$
Multiply the equation by 16 .
$
\Rightarrow 16 x^2-42 x+5
$
Hence, the required quadratic polynomial is $16 x^2-42 x+5$
