If natural numbers $a, b, c, d$ $a, b, c, d$ satisfy $a^{2} + b^{2} = 41$ $a^2+b^2 = 41$ and $c^{2} + d^{2} = 25$ $c^2+d^2=25$ then what monic quadratic in $x$ $x$ has zeros $(a + b)$ $(a+b)$ and $(c + d)$ $(c+d)$ ?

Question

If natural numbers $a, b, c, d$ $a, b, c, d$ satisfy $a^{2} + b^{2} = 41$ $a^2+b^2 = 41$ and $c^{2} + d^{2} = 25$ $c^2+d^2=25$ then what monic quadratic in $x$ $x$ has zeros $(a + b)$ $(a+b)$ and $(c + d)$ $(c+d)$ ?

1 Answer

Impact of this question

2740 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-18T22:46:25

$x^{2} - 14 x + 45$ $x^2-14x+45" "$ or $x^{2} - 16 x + 63$ $" "x^2-16x+63$

Explanation:

I will assume that "natural number" includes $0$ $0$ .

We may suppose that $a \leq b$ $a <= b$ and $c \leq d$ $c <= d$ since swapping $a$ $a$ and $b$ $b$ or $c$ $c$ and $d$ $d$ does not change any of the expressions $a^{2} + b^{2}$ $a^2+b^2$ , $c^{2} + d^{2}$ $c^2+d^2$ , $(a + b)$ $(a+b)$ or $(c + d)$ $(c+d)$ .

Hence:

$a^{2} \leq \frac{41}{2}$ $a^2 <= 41/2" "$ so $a = 0, 1, 2, 3$ $" "a = 0, 1, 2, 3$ or $4$ $4" "$ and $a^{2} = 0, 1, 4, 9$ $" "a^2 = 0, 1, 4, 9$ or $16$ $16$ .

$b^2 = 41 - a^2 = color(red)(cancel(color(black)(41))), color(red)(cancel(color(black)(40))), color(red)(cancel(color(black)(37))), color(red)(cancel(color(black)(32)))$ or $25$ .

$c^2 <= 25/2" "$ so $" "c = 0, 1, 2$ or $3" "$ and $" "c^2 = 0, 1, 4$ or $9$ .

$d^2 = 25-c^2 = 25, color(red)(cancel(color(black)(24))), color(red)(cancel(color(black)(21)))$ or $16$ .

So ${a, b} = {4, 5}$ and ${c, d} = {0, 5}$ or ${3, 4}$ .

So $a+b = 9$ and $c+d = 5$ or $7$ .

So the polynomial we are looking for is one of:

$(x-9)(x-5) = x^2-14x+45$

$(x-9)(x-7) = x^2-16x+63$

Science

Math

Humanities

... and beyond

If natural numbers $a, b, c, d$ $a, b, c, d$ satisfy $a^{2} + b^{2} = 41$ $a^2+b^2 = 41$ and $c^{2} + d^{2} = 25$ $c^2+d^2=25$ then what monic quadratic in $x$ $x$ has zeros $(a + b)$ $(a+b)$ and $(c + d)$ $(c+d)$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

If natural numbers a, b, c, da,b,c,da, b, c, d satisfy a^2+b^2 = 41a2+b2=41a^2+b^2 = 41 and c^2+d^2=25c2+d2=25c^2+d^2=25 then what monic quadratic in xxx has zeros (a+b)(a+b)(a+b) and (c+d)(c+d)(c+d) ?

1 Answer

Explanation:

Related questions

Impact of this question

If natural numbers $a, b, c, d$ $a, b, c, d$ satisfy $a^{2} + b^{2} = 41$ $a^2+b^2 = 41$ and $c^{2} + d^{2} = 25$ $c^2+d^2=25$ then what monic quadratic in $x$ $x$ has zeros $(a + b)$ $(a+b)$ and $(c + d)$ $(c+d)$ ?