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Condition in terms of b and a if $ax^2+bx+c=0$ has two consecutive odd positive integers as roots
The roots of the equation$$ax^2+bx+c=0$$, where $a geq 0$, are two consecutive odd positive integers, then
(A) $|b|leq 4a$
(B) $|b|geq 4a$
(C) $|b|=2a$
(D) None of these
My attempt
Let p and q be the roots then if they are consecutive positive integers (q>p) then $$ pq=frac{c}{a} geq 0$$
So, $$c geq 0$$ and $$q-p=2$$
So, $$frac{sqrt{b^2-4ac}}{a}=2$$
So,$$|b|>2a$$
$(Since, a>0,c>0)$
But I know that 4ac should be taken in consideration since its not equal to zero. But I don't know how to use it.
Any hints and suggestions are welcome!
algebra-precalculus quadratics self-learning
edited Nov 21 '18 at 21:34
Micah
29.8k1364106
asked Nov 21 '18 at 21:20
jayant98jayant98
476115
add a comment |
The roots of the equation$$ax^2+bx+c=0$$, where $a geq 0$, are two consecutive odd positive integers, then
(A) $|b|leq 4a$
(B) $|b|geq 4a$
(C) $|b|=2a$
(D) None of these
My attempt
Let p and q be the roots then if they are consecutive positive integers (q>p) then $$ pq=frac{c}{a} geq 0$$
So, $$c geq 0$$ and $$q-p=2$$
So, $$frac{sqrt{b^2-4ac}}{a}=2$$
So,$$|b|>2a$$
$(Since, a>0,c>0)$
But I know that 4ac should be taken in consideration since its not equal to zero. But I don't know how to use it.
Any hints and suggestions are welcome!
algebra-precalculus quadratics self-learning
edited Nov 21 '18 at 21:34
Micah
29.8k1364106
asked Nov 21 '18 at 21:20
jayant98jayant98
476115
add a comment |
The roots of the equation$$ax^2+bx+c=0$$, where $a geq 0$, are two consecutive odd positive integers, then
(A) $|b|leq 4a$
(B) $|b|geq 4a$
(C) $|b|=2a$
(D) None of these
My attempt
Let p and q be the roots then if they are consecutive positive integers (q>p) then $$ pq=frac{c}{a} geq 0$$
So, $$c geq 0$$ and $$q-p=2$$
So, $$frac{sqrt{b^2-4ac}}{a}=2$$
So,$$|b|>2a$$
$(Since, a>0,c>0)$
But I know that 4ac should be taken in consideration since its not equal to zero. But I don't know how to use it.
Any hints and suggestions are welcome!
algebra-precalculus quadratics self-learning
edited Nov 21 '18 at 21:34
Micah
29.8k1364106
asked Nov 21 '18 at 21:20
jayant98jayant98
476115
The roots of the equation$$ax^2+bx+c=0$$, where $a geq 0$, are two consecutive odd positive integers, then
(A) $|b|leq 4a$
(B) $|b|geq 4a$
(C) $|b|=2a$
(D) None of these
My attempt
Let p and q be the roots then if they are consecutive positive integers (q>p) then $$ pq=frac{c}{a} geq 0$$
So, $$c geq 0$$ and $$q-p=2$$
So, $$frac{sqrt{b^2-4ac}}{a}=2$$
So,$$|b|>2a$$
$(Since, a>0,c>0)$
But I know that 4ac should be taken in consideration since its not equal to zero. But I don't know how to use it.
Any hints and suggestions are welcome!
algebra-precalculus quadratics self-learning
algebra-precalculus quadratics self-learning
edited Nov 21 '18 at 21:34
Micah
29.8k1364106
asked Nov 21 '18 at 21:20
jayant98jayant98
476115
edited Nov 21 '18 at 21:34
Micah
29.8k1364106
asked Nov 21 '18 at 21:20
jayant98jayant98
476115
edited Nov 21 '18 at 21:34
Micah
29.8k1364106
edited Nov 21 '18 at 21:34
Micah
29.8k1364106
edited Nov 21 '18 at 21:34
Micah
29.8k1364106
29.8k1364106
asked Nov 21 '18 at 21:20
jayant98jayant98
476115
asked Nov 21 '18 at 21:20
jayant98jayant98
476115
asked Nov 21 '18 at 21:20
jayant98jayant98
476115
476115
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
Simply, If $2n-1$ and $2n+1$ are the roots (with $nge 1$) then
$$ax^2+bx+c=a(x-2n+1)(x-2n-1)=a((x-2n)^2-1)=a(x^2-4nx+4n^2-1) $$
so $b=-4na$ and hence $|b|=4nage 4a$.
answered Nov 21 '18 at 21:33
Hagen von EitzenHagen von Eitzen
276k21269496
Oh, you have the easier and more simple way! Thanks.
– jayant98
Nov 21 '18 at 21:34
add a comment |
Using the quadratic formula, we have that (using $p<q$ as the roots):
$$p+2=q$$
$$frac{-b-sqrt{b^2-4ac}}{2a}+2=frac{-b+sqrt{b^2-4ac}}{2a}$$
$$frac{-b}{2a}-frac{sqrt{b^2-4ac}}{2a}+2=frac{-b}{2a}+frac{sqrt{b^2-4ac}}{2a}$$
$$2=2frac{sqrt{b^2-4ac}}{2a}$$
$$2a=sqrt{b^2-4ac}$$
$$4a^2=b^2-4ac$$
$$b^2=4a^2+4ac$$
$$frac{b^2}{a^2}=4(1+frac{c}{a})=4(1+pq)$$
$$frac{|b|}{a}=2sqrt{1+pq}$$
Then, given that $p,q$ are different odd integers, you can show that $pqgeq3$, so that $2sqrt{1+pq}geq2sqrt{1+3}=4$
answered Nov 21 '18 at 21:57
MoKo19MoKo19
1914
add a comment |
Since we are given that there are two distinct roots and that $a ge 0$, we must have
$a > 0, tag 0$
since otherwise (i.e., with $a = 0$), the "quadratic" $bx + c$ has at most one root.
Let
$n ge 0, tag 1$
$r = 2n + 1, tag 2$
$s = 2n + 3; tag 3$
suppose for the moment
$a = 1; tag 4$
then
$(x - r)(x - s) = (x - (2n + 1))(x - (2n + 3)) = x^2 -(4n + 4)x + (2n + 1)(2n + 3) = x^2 -(4n + 4)x + (4n^2 + 8n + 3) = 0; tag 5$
here we have
$b = -(4n + 4), tag 6$
whence
$vert b vert = 4n + 4 ge 4 = 4a; tag 7$
thus (B) binds when $a = 1$; now if
$a ne 1, tag 8$
the quadratic of the form
$ax^2 + bx + c = a(x^2 + dfrac{b}{a} x + dfrac{c}{a}) tag 9$
has zeroes $2n + 1$, $2n + 3$ provided
$(x - (2n + 1))(x - (2n + 3)) = x^2 -(4n + 4)x + (4n^2 + 8n + 3) = x^2 + dfrac{b}{a} x + dfrac{c}{a}; tag{10}$
thus,
$dfrac{b}{a} = -(4n + 4), tag{11}$
or
$b = -(4n + 4)a, tag{12}$
whence, with $a > 0$,
$vert b vert = (4n + 4)a ge 4a, tag{13}$
and we see that the correct choice is (B) here as well.
answered Nov 22 '18 at 8:41
Robert LewisRobert Lewis
43.9k22963
add a comment |
Easy way to check:
Take a quadratic equation which has 1 and 3 as roots. $(x-1)(x-3)=x^2-4x+4x$, so you see that it is possible for $|b|=4|a|$, ruling out choice (C).
Do the problem again choosing the roots 3 and 5. $(x-3)(x-5)=x^2-8x+15$, so you can conclude that $|b|>=4a$, and (B) is correct. (If you continue choosing larger pairs of numbers as roots, $frac{|b|}{a}$ will only increase.)
answered Nov 21 '18 at 21:29
MoKo19MoKo19
1914
Sorry @MoKo19 but we have to give answer in subjective way...So we can't just do that...This question given just has the options...but we have to solve it 'properly'.
– jayant98
Nov 21 '18 at 21:32
@jayant98: Not sure what you mean by 'properly'. If you mean analytically, then generate the quadratic equation where $a=1$ with roots $p,p+2$. Because This will give you $y=(x-p)(x-p-2)=x^2-2px-2x+p^2+2p$, and $frac{|b|}{a}=|b|=|2p+2|=2p+2$. For odd integer roots, $p=1,3,5,...$, so the ratio will be $4,8,12,16,...$. I will write a different answer showing how to solve via the quadratic formula.
– MoKo19
Nov 21 '18 at 21:42
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
Simply, If $2n-1$ and $2n+1$ are the roots (with $nge 1$) then
$$ax^2+bx+c=a(x-2n+1)(x-2n-1)=a((x-2n)^2-1)=a(x^2-4nx+4n^2-1) $$
so $b=-4na$ and hence $|b|=4nage 4a$.
answered Nov 21 '18 at 21:33
Hagen von EitzenHagen von Eitzen
276k21269496
Oh, you have the easier and more simple way! Thanks.
– jayant98
Nov 21 '18 at 21:34
add a comment |
Simply, If $2n-1$ and $2n+1$ are the roots (with $nge 1$) then
$$ax^2+bx+c=a(x-2n+1)(x-2n-1)=a((x-2n)^2-1)=a(x^2-4nx+4n^2-1) $$
so $b=-4na$ and hence $|b|=4nage 4a$.
answered Nov 21 '18 at 21:33
Hagen von EitzenHagen von Eitzen
276k21269496
Oh, you have the easier and more simple way! Thanks.
– jayant98
Nov 21 '18 at 21:34
add a comment |
Simply, If $2n-1$ and $2n+1$ are the roots (with $nge 1$) then
$$ax^2+bx+c=a(x-2n+1)(x-2n-1)=a((x-2n)^2-1)=a(x^2-4nx+4n^2-1) $$
so $b=-4na$ and hence $|b|=4nage 4a$.
answered Nov 21 '18 at 21:33
Hagen von EitzenHagen von Eitzen
276k21269496
Simply, If $2n-1$ and $2n+1$ are the roots (with $nge 1$) then
$$ax^2+bx+c=a(x-2n+1)(x-2n-1)=a((x-2n)^2-1)=a(x^2-4nx+4n^2-1) $$
so $b=-4na$ and hence $|b|=4nage 4a$.
answered Nov 21 '18 at 21:33
Hagen von EitzenHagen von Eitzen
276k21269496
answered Nov 21 '18 at 21:33
Hagen von EitzenHagen von Eitzen
276k21269496
answered Nov 21 '18 at 21:33
Hagen von EitzenHagen von Eitzen
276k21269496
answered Nov 21 '18 at 21:33
Hagen von EitzenHagen von Eitzen
276k21269496
276k21269496
Oh, you have the easier and more simple way! Thanks.
– jayant98
Nov 21 '18 at 21:34
add a comment |
Oh, you have the easier and more simple way! Thanks.
– jayant98
Nov 21 '18 at 21:34
Oh, you have the easier and more simple way! Thanks.
– jayant98
Nov 21 '18 at 21:34
Oh, you have the easier and more simple way! Thanks.
– jayant98
Nov 21 '18 at 21:34
add a comment |
Using the quadratic formula, we have that (using $p<q$ as the roots):
$$p+2=q$$
$$frac{-b-sqrt{b^2-4ac}}{2a}+2=frac{-b+sqrt{b^2-4ac}}{2a}$$
$$frac{-b}{2a}-frac{sqrt{b^2-4ac}}{2a}+2=frac{-b}{2a}+frac{sqrt{b^2-4ac}}{2a}$$
$$2=2frac{sqrt{b^2-4ac}}{2a}$$
$$2a=sqrt{b^2-4ac}$$
$$4a^2=b^2-4ac$$
$$b^2=4a^2+4ac$$
$$frac{b^2}{a^2}=4(1+frac{c}{a})=4(1+pq)$$
$$frac{|b|}{a}=2sqrt{1+pq}$$
Then, given that $p,q$ are different odd integers, you can show that $pqgeq3$, so that $2sqrt{1+pq}geq2sqrt{1+3}=4$
answered Nov 21 '18 at 21:57
MoKo19MoKo19
1914
add a comment |
Using the quadratic formula, we have that (using $p<q$ as the roots):
$$p+2=q$$
$$frac{-b-sqrt{b^2-4ac}}{2a}+2=frac{-b+sqrt{b^2-4ac}}{2a}$$
$$frac{-b}{2a}-frac{sqrt{b^2-4ac}}{2a}+2=frac{-b}{2a}+frac{sqrt{b^2-4ac}}{2a}$$
$$2=2frac{sqrt{b^2-4ac}}{2a}$$
$$2a=sqrt{b^2-4ac}$$
$$4a^2=b^2-4ac$$
$$b^2=4a^2+4ac$$
$$frac{b^2}{a^2}=4(1+frac{c}{a})=4(1+pq)$$
$$frac{|b|}{a}=2sqrt{1+pq}$$
Then, given that $p,q$ are different odd integers, you can show that $pqgeq3$, so that $2sqrt{1+pq}geq2sqrt{1+3}=4$
answered Nov 21 '18 at 21:57
MoKo19MoKo19
1914
add a comment |
Using the quadratic formula, we have that (using $p<q$ as the roots):
$$p+2=q$$
$$frac{-b-sqrt{b^2-4ac}}{2a}+2=frac{-b+sqrt{b^2-4ac}}{2a}$$
$$frac{-b}{2a}-frac{sqrt{b^2-4ac}}{2a}+2=frac{-b}{2a}+frac{sqrt{b^2-4ac}}{2a}$$
$$2=2frac{sqrt{b^2-4ac}}{2a}$$
$$2a=sqrt{b^2-4ac}$$
$$4a^2=b^2-4ac$$
$$b^2=4a^2+4ac$$
$$frac{b^2}{a^2}=4(1+frac{c}{a})=4(1+pq)$$
$$frac{|b|}{a}=2sqrt{1+pq}$$
Then, given that $p,q$ are different odd integers, you can show that $pqgeq3$, so that $2sqrt{1+pq}geq2sqrt{1+3}=4$
answered Nov 21 '18 at 21:57
MoKo19MoKo19
1914
Using the quadratic formula, we have that (using $p<q$ as the roots):
$$p+2=q$$
$$frac{-b-sqrt{b^2-4ac}}{2a}+2=frac{-b+sqrt{b^2-4ac}}{2a}$$
$$frac{-b}{2a}-frac{sqrt{b^2-4ac}}{2a}+2=frac{-b}{2a}+frac{sqrt{b^2-4ac}}{2a}$$
$$2=2frac{sqrt{b^2-4ac}}{2a}$$
$$2a=sqrt{b^2-4ac}$$
$$4a^2=b^2-4ac$$
$$b^2=4a^2+4ac$$
$$frac{b^2}{a^2}=4(1+frac{c}{a})=4(1+pq)$$
$$frac{|b|}{a}=2sqrt{1+pq}$$
Then, given that $p,q$ are different odd integers, you can show that $pqgeq3$, so that $2sqrt{1+pq}geq2sqrt{1+3}=4$
answered Nov 21 '18 at 21:57
MoKo19MoKo19
1914
answered Nov 21 '18 at 21:57
MoKo19MoKo19
1914
answered Nov 21 '18 at 21:57
MoKo19MoKo19
1914
answered Nov 21 '18 at 21:57
MoKo19MoKo19
1914
1914
add a comment |
add a comment |
Since we are given that there are two distinct roots and that $a ge 0$, we must have
$a > 0, tag 0$
since otherwise (i.e., with $a = 0$), the "quadratic" $bx + c$ has at most one root.
Let
$n ge 0, tag 1$
$r = 2n + 1, tag 2$
$s = 2n + 3; tag 3$
suppose for the moment
$a = 1; tag 4$
then
$(x - r)(x - s) = (x - (2n + 1))(x - (2n + 3)) = x^2 -(4n + 4)x + (2n + 1)(2n + 3) = x^2 -(4n + 4)x + (4n^2 + 8n + 3) = 0; tag 5$
here we have
$b = -(4n + 4), tag 6$
whence
$vert b vert = 4n + 4 ge 4 = 4a; tag 7$
thus (B) binds when $a = 1$; now if
$a ne 1, tag 8$
the quadratic of the form
$ax^2 + bx + c = a(x^2 + dfrac{b}{a} x + dfrac{c}{a}) tag 9$
has zeroes $2n + 1$, $2n + 3$ provided
$(x - (2n + 1))(x - (2n + 3)) = x^2 -(4n + 4)x + (4n^2 + 8n + 3) = x^2 + dfrac{b}{a} x + dfrac{c}{a}; tag{10}$
thus,
$dfrac{b}{a} = -(4n + 4), tag{11}$
or
$b = -(4n + 4)a, tag{12}$
whence, with $a > 0$,
$vert b vert = (4n + 4)a ge 4a, tag{13}$
and we see that the correct choice is (B) here as well.
answered Nov 22 '18 at 8:41
Robert LewisRobert Lewis
43.9k22963
add a comment |
Since we are given that there are two distinct roots and that $a ge 0$, we must have
$a > 0, tag 0$
since otherwise (i.e., with $a = 0$), the "quadratic" $bx + c$ has at most one root.
Let
$n ge 0, tag 1$
$r = 2n + 1, tag 2$
$s = 2n + 3; tag 3$
suppose for the moment
$a = 1; tag 4$
then
$(x - r)(x - s) = (x - (2n + 1))(x - (2n + 3)) = x^2 -(4n + 4)x + (2n + 1)(2n + 3) = x^2 -(4n + 4)x + (4n^2 + 8n + 3) = 0; tag 5$
here we have
$b = -(4n + 4), tag 6$
whence
$vert b vert = 4n + 4 ge 4 = 4a; tag 7$
thus (B) binds when $a = 1$; now if
$a ne 1, tag 8$
the quadratic of the form
$ax^2 + bx + c = a(x^2 + dfrac{b}{a} x + dfrac{c}{a}) tag 9$
has zeroes $2n + 1$, $2n + 3$ provided
$(x - (2n + 1))(x - (2n + 3)) = x^2 -(4n + 4)x + (4n^2 + 8n + 3) = x^2 + dfrac{b}{a} x + dfrac{c}{a}; tag{10}$
thus,
$dfrac{b}{a} = -(4n + 4), tag{11}$
or
$b = -(4n + 4)a, tag{12}$
whence, with $a > 0$,
$vert b vert = (4n + 4)a ge 4a, tag{13}$
and we see that the correct choice is (B) here as well.
answered Nov 22 '18 at 8:41
Robert LewisRobert Lewis
43.9k22963
add a comment |
Since we are given that there are two distinct roots and that $a ge 0$, we must have
$a > 0, tag 0$
since otherwise (i.e., with $a = 0$), the "quadratic" $bx + c$ has at most one root.
Let
$n ge 0, tag 1$
$r = 2n + 1, tag 2$
$s = 2n + 3; tag 3$
suppose for the moment
$a = 1; tag 4$
then
$(x - r)(x - s) = (x - (2n + 1))(x - (2n + 3)) = x^2 -(4n + 4)x + (2n + 1)(2n + 3) = x^2 -(4n + 4)x + (4n^2 + 8n + 3) = 0; tag 5$
here we have
$b = -(4n + 4), tag 6$
whence
$vert b vert = 4n + 4 ge 4 = 4a; tag 7$
thus (B) binds when $a = 1$; now if
$a ne 1, tag 8$
the quadratic of the form
$ax^2 + bx + c = a(x^2 + dfrac{b}{a} x + dfrac{c}{a}) tag 9$
has zeroes $2n + 1$, $2n + 3$ provided
$(x - (2n + 1))(x - (2n + 3)) = x^2 -(4n + 4)x + (4n^2 + 8n + 3) = x^2 + dfrac{b}{a} x + dfrac{c}{a}; tag{10}$
thus,
$dfrac{b}{a} = -(4n + 4), tag{11}$
or
$b = -(4n + 4)a, tag{12}$
whence, with $a > 0$,
$vert b vert = (4n + 4)a ge 4a, tag{13}$
and we see that the correct choice is (B) here as well.
answered Nov 22 '18 at 8:41
Robert LewisRobert Lewis
43.9k22963
Since we are given that there are two distinct roots and that $a ge 0$, we must have
$a > 0, tag 0$
since otherwise (i.e., with $a = 0$), the "quadratic" $bx + c$ has at most one root.
Let
$n ge 0, tag 1$
$r = 2n + 1, tag 2$
$s = 2n + 3; tag 3$
suppose for the moment
$a = 1; tag 4$
then
$(x - r)(x - s) = (x - (2n + 1))(x - (2n + 3)) = x^2 -(4n + 4)x + (2n + 1)(2n + 3) = x^2 -(4n + 4)x + (4n^2 + 8n + 3) = 0; tag 5$
here we have
$b = -(4n + 4), tag 6$
whence
$vert b vert = 4n + 4 ge 4 = 4a; tag 7$
thus (B) binds when $a = 1$; now if
$a ne 1, tag 8$
the quadratic of the form
$ax^2 + bx + c = a(x^2 + dfrac{b}{a} x + dfrac{c}{a}) tag 9$
has zeroes $2n + 1$, $2n + 3$ provided
$(x - (2n + 1))(x - (2n + 3)) = x^2 -(4n + 4)x + (4n^2 + 8n + 3) = x^2 + dfrac{b}{a} x + dfrac{c}{a}; tag{10}$
thus,
$dfrac{b}{a} = -(4n + 4), tag{11}$
or
$b = -(4n + 4)a, tag{12}$
whence, with $a > 0$,
$vert b vert = (4n + 4)a ge 4a, tag{13}$
and we see that the correct choice is (B) here as well.
answered Nov 22 '18 at 8:41
Robert LewisRobert Lewis
43.9k22963
answered Nov 22 '18 at 8:41
Robert LewisRobert Lewis
43.9k22963
answered Nov 22 '18 at 8:41
Robert LewisRobert Lewis
43.9k22963
answered Nov 22 '18 at 8:41
Robert LewisRobert Lewis
43.9k22963
43.9k22963
add a comment |
add a comment |
Easy way to check:
Take a quadratic equation which has 1 and 3 as roots. $(x-1)(x-3)=x^2-4x+4x$, so you see that it is possible for $|b|=4|a|$, ruling out choice (C).
Do the problem again choosing the roots 3 and 5. $(x-3)(x-5)=x^2-8x+15$, so you can conclude that $|b|>=4a$, and (B) is correct. (If you continue choosing larger pairs of numbers as roots, $frac{|b|}{a}$ will only increase.)
answered Nov 21 '18 at 21:29
MoKo19MoKo19
1914
Sorry @MoKo19 but we have to give answer in subjective way...So we can't just do that...This question given just has the options...but we have to solve it 'properly'.
– jayant98
Nov 21 '18 at 21:32
@jayant98: Not sure what you mean by 'properly'. If you mean analytically, then generate the quadratic equation where $a=1$ with roots $p,p+2$. Because This will give you $y=(x-p)(x-p-2)=x^2-2px-2x+p^2+2p$, and $frac{|b|}{a}=|b|=|2p+2|=2p+2$. For odd integer roots, $p=1,3,5,...$, so the ratio will be $4,8,12,16,...$. I will write a different answer showing how to solve via the quadratic formula.
– MoKo19
Nov 21 '18 at 21:42
add a comment |
Easy way to check:
Take a quadratic equation which has 1 and 3 as roots. $(x-1)(x-3)=x^2-4x+4x$, so you see that it is possible for $|b|=4|a|$, ruling out choice (C).
Do the problem again choosing the roots 3 and 5. $(x-3)(x-5)=x^2-8x+15$, so you can conclude that $|b|>=4a$, and (B) is correct. (If you continue choosing larger pairs of numbers as roots, $frac{|b|}{a}$ will only increase.)
answered Nov 21 '18 at 21:29
MoKo19MoKo19
1914
Sorry @MoKo19 but we have to give answer in subjective way...So we can't just do that...This question given just has the options...but we have to solve it 'properly'.
– jayant98
Nov 21 '18 at 21:32
@jayant98: Not sure what you mean by 'properly'. If you mean analytically, then generate the quadratic equation where $a=1$ with roots $p,p+2$. Because This will give you $y=(x-p)(x-p-2)=x^2-2px-2x+p^2+2p$, and $frac{|b|}{a}=|b|=|2p+2|=2p+2$. For odd integer roots, $p=1,3,5,...$, so the ratio will be $4,8,12,16,...$. I will write a different answer showing how to solve via the quadratic formula.
– MoKo19
Nov 21 '18 at 21:42
add a comment |
Easy way to check:
Take a quadratic equation which has 1 and 3 as roots. $(x-1)(x-3)=x^2-4x+4x$, so you see that it is possible for $|b|=4|a|$, ruling out choice (C).
Do the problem again choosing the roots 3 and 5. $(x-3)(x-5)=x^2-8x+15$, so you can conclude that $|b|>=4a$, and (B) is correct. (If you continue choosing larger pairs of numbers as roots, $frac{|b|}{a}$ will only increase.)
answered Nov 21 '18 at 21:29
MoKo19MoKo19
1914
Easy way to check:
Take a quadratic equation which has 1 and 3 as roots. $(x-1)(x-3)=x^2-4x+4x$, so you see that it is possible for $|b|=4|a|$, ruling out choice (C).
Do the problem again choosing the roots 3 and 5. $(x-3)(x-5)=x^2-8x+15$, so you can conclude that $|b|>=4a$, and (B) is correct. (If you continue choosing larger pairs of numbers as roots, $frac{|b|}{a}$ will only increase.)
answered Nov 21 '18 at 21:29
MoKo19MoKo19
1914
answered Nov 21 '18 at 21:29
MoKo19MoKo19
1914
answered Nov 21 '18 at 21:29
MoKo19MoKo19
1914
answered Nov 21 '18 at 21:29
MoKo19MoKo19
1914
1914
Sorry @MoKo19 but we have to give answer in subjective way...So we can't just do that...This question given just has the options...but we have to solve it 'properly'.
– jayant98
Nov 21 '18 at 21:32
@jayant98: Not sure what you mean by 'properly'. If you mean analytically, then generate the quadratic equation where $a=1$ with roots $p,p+2$. Because This will give you $y=(x-p)(x-p-2)=x^2-2px-2x+p^2+2p$, and $frac{|b|}{a}=|b|=|2p+2|=2p+2$. For odd integer roots, $p=1,3,5,...$, so the ratio will be $4,8,12,16,...$. I will write a different answer showing how to solve via the quadratic formula.
– MoKo19
Nov 21 '18 at 21:42
add a comment |
Sorry @MoKo19 but we have to give answer in subjective way...So we can't just do that...This question given just has the options...but we have to solve it 'properly'.
– jayant98
Nov 21 '18 at 21:32
@jayant98: Not sure what you mean by 'properly'. If you mean analytically, then generate the quadratic equation where $a=1$ with roots $p,p+2$. Because This will give you $y=(x-p)(x-p-2)=x^2-2px-2x+p^2+2p$, and $frac{|b|}{a}=|b|=|2p+2|=2p+2$. For odd integer roots, $p=1,3,5,...$, so the ratio will be $4,8,12,16,...$. I will write a different answer showing how to solve via the quadratic formula.
– MoKo19
Nov 21 '18 at 21:42
Sorry @MoKo19 but we have to give answer in subjective way...So we can't just do that...This question given just has the options...but we have to solve it 'properly'.
– jayant98
Nov 21 '18 at 21:32
Sorry @MoKo19 but we have to give answer in subjective way...So we can't just do that...This question given just has the options...but we have to solve it 'properly'.
– jayant98
Nov 21 '18 at 21:32
@jayant98: Not sure what you mean by 'properly'. If you mean analytically, then generate the quadratic equation where $a=1$ with roots $p,p+2$. Because This will give you $y=(x-p)(x-p-2)=x^2-2px-2x+p^2+2p$, and $frac{|b|}{a}=|b|=|2p+2|=2p+2$. For odd integer roots, $p=1,3,5,...$, so the ratio will be $4,8,12,16,...$. I will write a different answer showing how to solve via the quadratic formula.
– MoKo19
Nov 21 '18 at 21:42
@jayant98: Not sure what you mean by 'properly'. If you mean analytically, then generate the quadratic equation where $a=1$ with roots $p,p+2$. Because This will give you $y=(x-p)(x-p-2)=x^2-2px-2x+p^2+2p$, and $frac{|b|}{a}=|b|=|2p+2|=2p+2$. For odd integer roots, $p=1,3,5,...$, so the ratio will be $4,8,12,16,...$. I will write a different answer showing how to solve via the quadratic formula.
– MoKo19
Nov 21 '18 at 21:42
add a comment |
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