Mixing
Relation of Roots of Bi-quadratic equation without using Vieta's formula
$begingroup$
If $ a, b, c, d $ are roots of the equation
$$ x^4 + px^3 + qx^2 + rx + s=0 $$
Show that
$$ (1+a^2)(1+b^2)(1+c^2)(1+d^2)=(1+s-q)^2 + (p-r)^2 $$
I was able to show this by using Vieta's formulas but the book from which I am doing this has not yet taught, till now it has just taught reminder, factor theorem and fundamental theorem of algebra so I want to ask that is there any other solution using only these or some more interesting way to do
polynomials complex-numbers roots
edited Jan 6 at 18:19
greedoid
39.7k114798
asked Jan 6 at 17:59
Keshav SharmaKeshav Sharma
985
$endgroup$
add a comment |
$begingroup$
If $ a, b, c, d $ are roots of the equation
$$ x^4 + px^3 + qx^2 + rx + s=0 $$
Show that
$$ (1+a^2)(1+b^2)(1+c^2)(1+d^2)=(1+s-q)^2 + (p-r)^2 $$
I was able to show this by using Vieta's formulas but the book from which I am doing this has not yet taught, till now it has just taught reminder, factor theorem and fundamental theorem of algebra so I want to ask that is there any other solution using only these or some more interesting way to do
polynomials complex-numbers roots
edited Jan 6 at 18:19
greedoid
39.7k114798
asked Jan 6 at 17:59
Keshav SharmaKeshav Sharma
985
$endgroup$
$begingroup$
See math.stackexchange.com/questions/3063351/…
$endgroup$
– lab bhattacharjee
Jan 6 at 18:12
add a comment |
$begingroup$
If $ a, b, c, d $ are roots of the equation
$$ x^4 + px^3 + qx^2 + rx + s=0 $$
Show that
$$ (1+a^2)(1+b^2)(1+c^2)(1+d^2)=(1+s-q)^2 + (p-r)^2 $$
I was able to show this by using Vieta's formulas but the book from which I am doing this has not yet taught, till now it has just taught reminder, factor theorem and fundamental theorem of algebra so I want to ask that is there any other solution using only these or some more interesting way to do
polynomials complex-numbers roots
edited Jan 6 at 18:19
greedoid
39.7k114798
asked Jan 6 at 17:59
Keshav SharmaKeshav Sharma
985
$endgroup$
If $ a, b, c, d $ are roots of the equation
$$ x^4 + px^3 + qx^2 + rx + s=0 $$
Show that
$$ (1+a^2)(1+b^2)(1+c^2)(1+d^2)=(1+s-q)^2 + (p-r)^2 $$
I was able to show this by using Vieta's formulas but the book from which I am doing this has not yet taught, till now it has just taught reminder, factor theorem and fundamental theorem of algebra so I want to ask that is there any other solution using only these or some more interesting way to do
polynomials complex-numbers roots
polynomials complex-numbers roots
edited Jan 6 at 18:19
greedoid
39.7k114798
asked Jan 6 at 17:59
Keshav SharmaKeshav Sharma
985
edited Jan 6 at 18:19
greedoid
39.7k114798
asked Jan 6 at 17:59
Keshav SharmaKeshav Sharma
985
edited Jan 6 at 18:19
greedoid
39.7k114798
edited Jan 6 at 18:19
greedoid
39.7k114798
edited Jan 6 at 18:19
greedoid
39.7k114798
39.7k114798
asked Jan 6 at 17:59
Keshav SharmaKeshav Sharma
985
asked Jan 6 at 17:59
Keshav SharmaKeshav Sharma
985
asked Jan 6 at 17:59
Keshav SharmaKeshav Sharma
985
985
$begingroup$
See math.stackexchange.com/questions/3063351/…
$endgroup$
– lab bhattacharjee
Jan 6 at 18:12
add a comment |
$begingroup$
See math.stackexchange.com/questions/3063351/…
$endgroup$
– lab bhattacharjee
Jan 6 at 18:12
$begingroup$
See math.stackexchange.com/questions/3063351/…
$endgroup$
– lab bhattacharjee
Jan 6 at 18:12
$begingroup$
See math.stackexchange.com/questions/3063351/…
$endgroup$
– lab bhattacharjee
Jan 6 at 18:12
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
So: $$ f(x)= (x-a)(x-b)(x-c)(x-d)$$
$$f(i) = (i-a)...;;;;{rm and };;;;f(-i)= (-i-a)...$$
thus $$f(i)f(-i) = (a^2+1)(b^2+1)(c^2+1)(d^2+1)$$
On the other side $$f(i)f(-i) = (1 -ip -q+ ri + s) (1 + ip + q - ri + s)$$
$$ = (1-q+s-i(p-r))(1-q+s+i(p-r))$$ $$= (1-q+s)^2+(p-r)^2$$
answered Jan 6 at 18:14
greedoidgreedoid
39.7k114798
$endgroup$
add a comment |
$begingroup$
Also, we can use $$(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2.$$
Indeed,
$$(1+a^2)(1+b^2)(1+c^2)(1+d^2)=((a+b)^2+(1-ab)^2)((c+d)^2+(1-cd)^2)=$$
$$=((a+b)(c+d)-(1-ab)(1-cd))^2+((a+b)(1-cd)+(c+d)(1-ab))^2=$$
$$=(ab+ac+ad+bc+bd+cd-abcd-1)^2+(a+b+c+d-abc-abd-acd-bcd)^2=$$
$$=(q-s-1)^2+(-p+r)^2=(1+s-q)^2+(p-r)^2.$$
answered Jan 6 at 19:21
Michael RozenbergMichael Rozenberg
99.3k1590189
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add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
So: $$ f(x)= (x-a)(x-b)(x-c)(x-d)$$
$$f(i) = (i-a)...;;;;{rm and };;;;f(-i)= (-i-a)...$$
thus $$f(i)f(-i) = (a^2+1)(b^2+1)(c^2+1)(d^2+1)$$
On the other side $$f(i)f(-i) = (1 -ip -q+ ri + s) (1 + ip + q - ri + s)$$
$$ = (1-q+s-i(p-r))(1-q+s+i(p-r))$$ $$= (1-q+s)^2+(p-r)^2$$
answered Jan 6 at 18:14
greedoidgreedoid
39.7k114798
$endgroup$
add a comment |
$begingroup$
So: $$ f(x)= (x-a)(x-b)(x-c)(x-d)$$
$$f(i) = (i-a)...;;;;{rm and };;;;f(-i)= (-i-a)...$$
thus $$f(i)f(-i) = (a^2+1)(b^2+1)(c^2+1)(d^2+1)$$
On the other side $$f(i)f(-i) = (1 -ip -q+ ri + s) (1 + ip + q - ri + s)$$
$$ = (1-q+s-i(p-r))(1-q+s+i(p-r))$$ $$= (1-q+s)^2+(p-r)^2$$
answered Jan 6 at 18:14
greedoidgreedoid
39.7k114798
$endgroup$
add a comment |
$begingroup$
So: $$ f(x)= (x-a)(x-b)(x-c)(x-d)$$
$$f(i) = (i-a)...;;;;{rm and };;;;f(-i)= (-i-a)...$$
thus $$f(i)f(-i) = (a^2+1)(b^2+1)(c^2+1)(d^2+1)$$
On the other side $$f(i)f(-i) = (1 -ip -q+ ri + s) (1 + ip + q - ri + s)$$
$$ = (1-q+s-i(p-r))(1-q+s+i(p-r))$$ $$= (1-q+s)^2+(p-r)^2$$
answered Jan 6 at 18:14
greedoidgreedoid
39.7k114798
$endgroup$
So: $$ f(x)= (x-a)(x-b)(x-c)(x-d)$$
$$f(i) = (i-a)...;;;;{rm and };;;;f(-i)= (-i-a)...$$
thus $$f(i)f(-i) = (a^2+1)(b^2+1)(c^2+1)(d^2+1)$$
On the other side $$f(i)f(-i) = (1 -ip -q+ ri + s) (1 + ip + q - ri + s)$$
$$ = (1-q+s-i(p-r))(1-q+s+i(p-r))$$ $$= (1-q+s)^2+(p-r)^2$$
answered Jan 6 at 18:14
greedoidgreedoid
39.7k114798
answered Jan 6 at 18:14
greedoidgreedoid
39.7k114798
answered Jan 6 at 18:14
greedoidgreedoid
39.7k114798
answered Jan 6 at 18:14
greedoidgreedoid
39.7k114798
39.7k114798
add a comment |
add a comment |
$begingroup$
Also, we can use $$(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2.$$
Indeed,
$$(1+a^2)(1+b^2)(1+c^2)(1+d^2)=((a+b)^2+(1-ab)^2)((c+d)^2+(1-cd)^2)=$$
$$=((a+b)(c+d)-(1-ab)(1-cd))^2+((a+b)(1-cd)+(c+d)(1-ab))^2=$$
$$=(ab+ac+ad+bc+bd+cd-abcd-1)^2+(a+b+c+d-abc-abd-acd-bcd)^2=$$
$$=(q-s-1)^2+(-p+r)^2=(1+s-q)^2+(p-r)^2.$$
answered Jan 6 at 19:21
Michael RozenbergMichael Rozenberg
99.3k1590189
$endgroup$
add a comment |
$begingroup$
Also, we can use $$(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2.$$
Indeed,
$$(1+a^2)(1+b^2)(1+c^2)(1+d^2)=((a+b)^2+(1-ab)^2)((c+d)^2+(1-cd)^2)=$$
$$=((a+b)(c+d)-(1-ab)(1-cd))^2+((a+b)(1-cd)+(c+d)(1-ab))^2=$$
$$=(ab+ac+ad+bc+bd+cd-abcd-1)^2+(a+b+c+d-abc-abd-acd-bcd)^2=$$
$$=(q-s-1)^2+(-p+r)^2=(1+s-q)^2+(p-r)^2.$$
answered Jan 6 at 19:21
Michael RozenbergMichael Rozenberg
99.3k1590189
$endgroup$
add a comment |
$begingroup$
Also, we can use $$(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2.$$
Indeed,
$$(1+a^2)(1+b^2)(1+c^2)(1+d^2)=((a+b)^2+(1-ab)^2)((c+d)^2+(1-cd)^2)=$$
$$=((a+b)(c+d)-(1-ab)(1-cd))^2+((a+b)(1-cd)+(c+d)(1-ab))^2=$$
$$=(ab+ac+ad+bc+bd+cd-abcd-1)^2+(a+b+c+d-abc-abd-acd-bcd)^2=$$
$$=(q-s-1)^2+(-p+r)^2=(1+s-q)^2+(p-r)^2.$$
answered Jan 6 at 19:21
Michael RozenbergMichael Rozenberg
99.3k1590189
$endgroup$
Also, we can use $$(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2.$$
Indeed,
$$(1+a^2)(1+b^2)(1+c^2)(1+d^2)=((a+b)^2+(1-ab)^2)((c+d)^2+(1-cd)^2)=$$
$$=((a+b)(c+d)-(1-ab)(1-cd))^2+((a+b)(1-cd)+(c+d)(1-ab))^2=$$
$$=(ab+ac+ad+bc+bd+cd-abcd-1)^2+(a+b+c+d-abc-abd-acd-bcd)^2=$$
$$=(q-s-1)^2+(-p+r)^2=(1+s-q)^2+(p-r)^2.$$
answered Jan 6 at 19:21
Michael RozenbergMichael Rozenberg
99.3k1590189
answered Jan 6 at 19:21
Michael RozenbergMichael Rozenberg
99.3k1590189
answered Jan 6 at 19:21
Michael RozenbergMichael Rozenberg
99.3k1590189
answered Jan 6 at 19:21
Michael RozenbergMichael Rozenberg
99.3k1590189
99.3k1590189
add a comment |
add a comment |
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$begingroup$
See math.stackexchange.com/questions/3063351/…
$endgroup$
– lab bhattacharjee
Jan 6 at 18:12