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Quadratic/polynomial problem
What is a short way to deal with this cubic polynomial problem?Condition $|x_1x_2+1|<x_1+x_2$ in quadratic polynomialFinding roots of quartic polynomial using a quadratic polynomial of similar form.Elementary number theory and quadraticFind a nonzero polynomial $f(X)$ with integer coefficients such that $f(sqrt3 + sqrt7)=0$quadratic equation with integer rootsCoefficients nature in a quadratic polynomialPolynomial p(a)=1, why does it have at most 2 integer roots?Is a polynomial of degree 3 with irrational roots possible?Algebra of quadratic equation
$begingroup$
The equation in $x: x^2+px+q=0$ has two nonzero integer roots, and $p+q=198$. What is $p$?
algebra-precalculus elementary-number-theory
edited 2 hours ago
Brian Tung
26.7k32657
asked 2 hours ago
IMbADdAtMathIMbADdAtMath
372
$endgroup$
add a comment |
$begingroup$
The equation in $x: x^2+px+q=0$ has two nonzero integer roots, and $p+q=198$. What is $p$?
algebra-precalculus elementary-number-theory
edited 2 hours ago
Brian Tung
26.7k32657
asked 2 hours ago
IMbADdAtMathIMbADdAtMath
372
$endgroup$
add a comment |
$begingroup$
The equation in $x: x^2+px+q=0$ has two nonzero integer roots, and $p+q=198$. What is $p$?
algebra-precalculus elementary-number-theory
edited 2 hours ago
Brian Tung
26.7k32657
asked 2 hours ago
IMbADdAtMathIMbADdAtMath
372
$endgroup$
The equation in $x: x^2+px+q=0$ has two nonzero integer roots, and $p+q=198$. What is $p$?
algebra-precalculus elementary-number-theory
algebra-precalculus elementary-number-theory
edited 2 hours ago
Brian Tung
26.7k32657
asked 2 hours ago
IMbADdAtMathIMbADdAtMath
372
edited 2 hours ago
Brian Tung
26.7k32657
asked 2 hours ago
IMbADdAtMathIMbADdAtMath
372
edited 2 hours ago
Brian Tung
26.7k32657
edited 2 hours ago
Brian Tung
26.7k32657
edited 2 hours ago
Brian Tung
26.7k32657
26.7k32657
asked 2 hours ago
IMbADdAtMathIMbADdAtMath
372
asked 2 hours ago
IMbADdAtMathIMbADdAtMath
372
asked 2 hours ago
IMbADdAtMathIMbADdAtMath
372
372
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$$ x^2+px+198-p=0$$ so $$x-1mid 198+x^2$$ Since we always have $x-1mid 1-x^2$ we have also $$x-1mid (198+x^2) + (1-x^2)=199$$
Can you finish now?
answered 2 hours ago
Maria MazurMaria Mazur
51.8k1363131
$endgroup$
add a comment |
$begingroup$
We call $x_1$, $x_2$ are two integer roots, $x_1 leq x_2$. One then has:
$$x_1 + x_2 = -p; x_1x_2 = q$$
So, $(x_1-1)(x_2 -1) = q + p + 1 = 199.$
So, we have
Case 1: $x_1 - 1 = 1$ and $x_2 - 1 = 199$. Then $x_1 = 2$, $x_2 = 200$.
Finally, $p = -x_1 - x_2 = -202$.
Case 2: $x_1 - 1 = -1$ and $x_2 - 1 = -199$.
answered 2 hours ago
GAVDGAVD
6,67811129
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
$$ x^2+px+198-p=0$$ so $$x-1mid 198+x^2$$ Since we always have $x-1mid 1-x^2$ we have also $$x-1mid (198+x^2) + (1-x^2)=199$$
Can you finish now?
answered 2 hours ago
Maria MazurMaria Mazur
51.8k1363131
$endgroup$
add a comment |
$begingroup$
$$ x^2+px+198-p=0$$ so $$x-1mid 198+x^2$$ Since we always have $x-1mid 1-x^2$ we have also $$x-1mid (198+x^2) + (1-x^2)=199$$
Can you finish now?
answered 2 hours ago
Maria MazurMaria Mazur
51.8k1363131
$endgroup$
add a comment |
$begingroup$
$$ x^2+px+198-p=0$$ so $$x-1mid 198+x^2$$ Since we always have $x-1mid 1-x^2$ we have also $$x-1mid (198+x^2) + (1-x^2)=199$$
Can you finish now?
answered 2 hours ago
Maria MazurMaria Mazur
51.8k1363131
$endgroup$
$$ x^2+px+198-p=0$$ so $$x-1mid 198+x^2$$ Since we always have $x-1mid 1-x^2$ we have also $$x-1mid (198+x^2) + (1-x^2)=199$$
Can you finish now?
answered 2 hours ago
Maria MazurMaria Mazur
51.8k1363131
answered 2 hours ago
Maria MazurMaria Mazur
51.8k1363131
answered 2 hours ago
Maria MazurMaria Mazur
51.8k1363131
answered 2 hours ago
Maria MazurMaria Mazur
51.8k1363131
51.8k1363131
add a comment |
add a comment |
$begingroup$
We call $x_1$, $x_2$ are two integer roots, $x_1 leq x_2$. One then has:
$$x_1 + x_2 = -p; x_1x_2 = q$$
So, $(x_1-1)(x_2 -1) = q + p + 1 = 199.$
So, we have
Case 1: $x_1 - 1 = 1$ and $x_2 - 1 = 199$. Then $x_1 = 2$, $x_2 = 200$.
Finally, $p = -x_1 - x_2 = -202$.
Case 2: $x_1 - 1 = -1$ and $x_2 - 1 = -199$.
answered 2 hours ago
GAVDGAVD
6,67811129
$endgroup$
add a comment |
$begingroup$
We call $x_1$, $x_2$ are two integer roots, $x_1 leq x_2$. One then has:
$$x_1 + x_2 = -p; x_1x_2 = q$$
So, $(x_1-1)(x_2 -1) = q + p + 1 = 199.$
So, we have
Case 1: $x_1 - 1 = 1$ and $x_2 - 1 = 199$. Then $x_1 = 2$, $x_2 = 200$.
Finally, $p = -x_1 - x_2 = -202$.
Case 2: $x_1 - 1 = -1$ and $x_2 - 1 = -199$.
answered 2 hours ago
GAVDGAVD
6,67811129
$endgroup$
add a comment |
$begingroup$
We call $x_1$, $x_2$ are two integer roots, $x_1 leq x_2$. One then has:
$$x_1 + x_2 = -p; x_1x_2 = q$$
So, $(x_1-1)(x_2 -1) = q + p + 1 = 199.$
So, we have
Case 1: $x_1 - 1 = 1$ and $x_2 - 1 = 199$. Then $x_1 = 2$, $x_2 = 200$.
Finally, $p = -x_1 - x_2 = -202$.
Case 2: $x_1 - 1 = -1$ and $x_2 - 1 = -199$.
answered 2 hours ago
GAVDGAVD
6,67811129
$endgroup$
We call $x_1$, $x_2$ are two integer roots, $x_1 leq x_2$. One then has:
$$x_1 + x_2 = -p; x_1x_2 = q$$
So, $(x_1-1)(x_2 -1) = q + p + 1 = 199.$
So, we have
Case 1: $x_1 - 1 = 1$ and $x_2 - 1 = 199$. Then $x_1 = 2$, $x_2 = 200$.
Finally, $p = -x_1 - x_2 = -202$.
Case 2: $x_1 - 1 = -1$ and $x_2 - 1 = -199$.
answered 2 hours ago
GAVDGAVD
6,67811129
answered 2 hours ago
GAVDGAVD
6,67811129
answered 2 hours ago
GAVDGAVD
6,67811129
answered 2 hours ago
GAVDGAVD
6,67811129
6,67811129
add a comment |
add a comment |
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