Question

If a , b, c and d are natural numbers such that $a^2+b^2=41$ and $c^2+d^2=25$ , then the polynomial whose zeroes are (a+b) and (c+d) can be

• A. $x^2-9x+12$
• B. $x^2-16x+63$
• C. $x^2-2x+14$
• D. $x^2-7x+9$

Answer 1

The correct option is B $x^2-16x+63$

Given that,

$a^2+b^2=41.......\left(1\right)$

$c^2+d^2=25......\left(2\right)$

By equation (1)

Let put $a=4,b=5$ and we get,

$4^2+5^2=41$

$16+25=41$

$41=41$

Then,$a=4,b=5$ is satisfied this equation.

Similarly that,

By equation (2)

Let put,$c=3,d=4$ and we get,

$c^2+d^2=25$

$3^2+4^2=25$

$25=25$

Then, $c=3,d=4$ is satisfied this equation.

According to given question.

$(a+b)=4+5=9$

$(c+d)=3+4=7$

Then, sum of roots $=9+7=16$

Product of roots $9\times 7=63$

Now, the polynomial is

$x^2-\left(sumofroots\right)x+productofroots=0$

$x^2-16x+63=0$

Hence, this is the answer.