Find minimum value of $a^2+b^2$
$begingroup$
Given that $a$ and $b$ are real constants and that the equation $x^4+ax^3+2x^2+bx+1=0$ has at least one real root, find the minimum possible value of $a^2+b^2$.
I began this way: Let the polynomial be factorized as $(x^2+alpha x + 1)(x^2+beta x +1)$. Then expanding and comparing coefficients we get $alphabeta=0$, meaning either $alpha=0$ or $beta=0$. Suppose $alpha=0$. Then we see that $(x^2+beta x+1)$ should have real roots, from which we get $beta^2 geq 4$. But we get $a=b=beta$ from the comparison above. So $a^2+b^2 = 2beta^2 geq 8$.
Is it correct? Or is there any mistake? Any other solution is also welcome.
polynomials
asked Jan 10 at 5:24
YellowYellow
16011
$endgroup$
|
show 2 more comments
$begingroup$
Given that $a$ and $b$ are real constants and that the equation $x^4+ax^3+2x^2+bx+1=0$ has at least one real root, find the minimum possible value of $a^2+b^2$.
I began this way: Let the polynomial be factorized as $(x^2+alpha x + 1)(x^2+beta x +1)$. Then expanding and comparing coefficients we get $alphabeta=0$, meaning either $alpha=0$ or $beta=0$. Suppose $alpha=0$. Then we see that $(x^2+beta x+1)$ should have real roots, from which we get $beta^2 geq 4$. But we get $a=b=beta$ from the comparison above. So $a^2+b^2 = 2beta^2 geq 8$.
Is it correct? Or is there any mistake? Any other solution is also welcome.
polynomials
asked Jan 10 at 5:24
YellowYellow
16011
$endgroup$
2
$begingroup$
How do you know the polynomial has such a factorisation?
$endgroup$
– Lord Shark the Unknown
Jan 10 at 5:27
$begingroup$
Any fourth degree polynomial can be factorised into product of two quadratic expressions, no matter what the nature of its roots are, right?
$endgroup$
– Yellow
Jan 10 at 5:32
$begingroup$
@Any Radha, yes but not in that way you write...
$endgroup$
– dmtri
Jan 10 at 5:38
$begingroup$
I can't find anything wrong in your solution.
$endgroup$
– Anik Bhowmick
Jan 10 at 5:43
1
$begingroup$
Wouldn't this imply that $alpha + beta = a = b$ which need not be the case? If $x^4 + ax^3 + 2x^2 +bx +1 = (m x^2 + nx + p)(jx^2 + kx + q)$ we know $j = frac 1m$ and wolog it factors to $(x^2 + alpha x + gamma)(x^2 + beta x + delta)$ but we can't assume $gamma = delta = 1$
$endgroup$
– fleablood
Jan 10 at 6:24
|
show 2 more comments
$begingroup$
Given that $a$ and $b$ are real constants and that the equation $x^4+ax^3+2x^2+bx+1=0$ has at least one real root, find the minimum possible value of $a^2+b^2$.
I began this way: Let the polynomial be factorized as $(x^2+alpha x + 1)(x^2+beta x +1)$. Then expanding and comparing coefficients we get $alphabeta=0$, meaning either $alpha=0$ or $beta=0$. Suppose $alpha=0$. Then we see that $(x^2+beta x+1)$ should have real roots, from which we get $beta^2 geq 4$. But we get $a=b=beta$ from the comparison above. So $a^2+b^2 = 2beta^2 geq 8$.
Is it correct? Or is there any mistake? Any other solution is also welcome.
polynomials
asked Jan 10 at 5:24
YellowYellow
16011
$endgroup$
Given that $a$ and $b$ are real constants and that the equation $x^4+ax^3+2x^2+bx+1=0$ has at least one real root, find the minimum possible value of $a^2+b^2$.
I began this way: Let the polynomial be factorized as $(x^2+alpha x + 1)(x^2+beta x +1)$. Then expanding and comparing coefficients we get $alphabeta=0$, meaning either $alpha=0$ or $beta=0$. Suppose $alpha=0$. Then we see that $(x^2+beta x+1)$ should have real roots, from which we get $beta^2 geq 4$. But we get $a=b=beta$ from the comparison above. So $a^2+b^2 = 2beta^2 geq 8$.
Is it correct? Or is there any mistake? Any other solution is also welcome.
polynomials
polynomials
asked Jan 10 at 5:24
YellowYellow
16011
asked Jan 10 at 5:24
YellowYellow
16011
asked Jan 10 at 5:24
YellowYellow
16011
asked Jan 10 at 5:24
YellowYellow
16011
asked Jan 10 at 5:24
YellowYellow
16011
16011
2
$begingroup$
How do you know the polynomial has such a factorisation?
$endgroup$
– Lord Shark the Unknown
Jan 10 at 5:27
$begingroup$
Any fourth degree polynomial can be factorised into product of two quadratic expressions, no matter what the nature of its roots are, right?
$endgroup$
– Yellow
Jan 10 at 5:32
$begingroup$
@Any Radha, yes but not in that way you write...
$endgroup$
– dmtri
Jan 10 at 5:38
$begingroup$
I can't find anything wrong in your solution.
$endgroup$
– Anik Bhowmick
Jan 10 at 5:43
1
$begingroup$
Wouldn't this imply that $alpha + beta = a = b$ which need not be the case? If $x^4 + ax^3 + 2x^2 +bx +1 = (m x^2 + nx + p)(jx^2 + kx + q)$ we know $j = frac 1m$ and wolog it factors to $(x^2 + alpha x + gamma)(x^2 + beta x + delta)$ but we can't assume $gamma = delta = 1$
$endgroup$
– fleablood
Jan 10 at 6:24
|
show 2 more comments
2
$begingroup$
How do you know the polynomial has such a factorisation?
$endgroup$
– Lord Shark the Unknown
Jan 10 at 5:27
$begingroup$
Any fourth degree polynomial can be factorised into product of two quadratic expressions, no matter what the nature of its roots are, right?
$endgroup$
– Yellow
Jan 10 at 5:32
$begingroup$
@Any Radha, yes but not in that way you write...
$endgroup$
– dmtri
Jan 10 at 5:38
$begingroup$
I can't find anything wrong in your solution.
$endgroup$
– Anik Bhowmick
Jan 10 at 5:43
1
$begingroup$
Wouldn't this imply that $alpha + beta = a = b$ which need not be the case? If $x^4 + ax^3 + 2x^2 +bx +1 = (m x^2 + nx + p)(jx^2 + kx + q)$ we know $j = frac 1m$ and wolog it factors to $(x^2 + alpha x + gamma)(x^2 + beta x + delta)$ but we can't assume $gamma = delta = 1$
$endgroup$
– fleablood
Jan 10 at 6:24
2
2
$begingroup$
How do you know the polynomial has such a factorisation?
$endgroup$
– Lord Shark the Unknown
Jan 10 at 5:27
$begingroup$
How do you know the polynomial has such a factorisation?
$endgroup$
– Lord Shark the Unknown
Jan 10 at 5:27
$begingroup$
Any fourth degree polynomial can be factorised into product of two quadratic expressions, no matter what the nature of its roots are, right?
$endgroup$
– Yellow
Jan 10 at 5:32
$begingroup$
Any fourth degree polynomial can be factorised into product of two quadratic expressions, no matter what the nature of its roots are, right?
$endgroup$
– Yellow
Jan 10 at 5:32
$begingroup$
@Any Radha, yes but not in that way you write...
$endgroup$
– dmtri
Jan 10 at 5:38
$begingroup$
@Any Radha, yes but not in that way you write...
$endgroup$
– dmtri
Jan 10 at 5:38
$begingroup$
I can't find anything wrong in your solution.
$endgroup$
– Anik Bhowmick
Jan 10 at 5:43
$begingroup$
I can't find anything wrong in your solution.
$endgroup$
– Anik Bhowmick
Jan 10 at 5:43
1
1
$begingroup$
Wouldn't this imply that $alpha + beta = a = b$ which need not be the case? If $x^4 + ax^3 + 2x^2 +bx +1 = (m x^2 + nx + p)(jx^2 + kx + q)$ we know $j = frac 1m$ and wolog it factors to $(x^2 + alpha x + gamma)(x^2 + beta x + delta)$ but we can't assume $gamma = delta = 1$
$endgroup$
– fleablood
Jan 10 at 6:24
$begingroup$
Wouldn't this imply that $alpha + beta = a = b$ which need not be the case? If $x^4 + ax^3 + 2x^2 +bx +1 = (m x^2 + nx + p)(jx^2 + kx + q)$ we know $j = frac 1m$ and wolog it factors to $(x^2 + alpha x + gamma)(x^2 + beta x + delta)$ but we can't assume $gamma = delta = 1$
$endgroup$
– fleablood
Jan 10 at 6:24
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
$$x^4+ax^3+2x^2+bx+1=$$
$$=(x^2+frac{a}{2}x)^2-frac{a^2}{4}x^2+2x^2+bx+1=$$
$$=(x^2+frac{a}{2}x)^2-frac{a^2}{4}x^2+2x^2-frac{b^2}{4}x^2+(frac{b}{2}x+1)^2=$$
$$=(x^2+frac{a}{2}x)^2+x^2(2-frac{a^2+b^2}{4})+(frac{b}{2}x+1)^2$$
If $2-frac{a^2+b^2}{4}>0$, then we have a problem; the sum of three nonnegative terms equals zero (at our real root), which can only happen if $x=0$ to make the second term zero. But then $(frac{b}{2}x+1)^2>0$. So if $a^2+b^2<8$ there is no real root. Hence $a^2+b^2ge 8$.
Now, if $a^2+b^2=8$, then the middle term is zero. Can we make the other two terms zero as well? We need $x=-frac{a}{2}$ and $x=-frac{2}{b}$, to make the first and last terms zero. This can be achieved, with $a=b=2$ and $x=-1$.
Hence, the minimum possible value of $a^2+b^2$ is indeed $8$.
answered Jan 10 at 5:38
vadim123vadim123
75.9k897189
$endgroup$
2
$begingroup$
Whoa, that was quick..
$endgroup$
– Yellow
Jan 10 at 5:42
2
$begingroup$
@AnuRadha, I've seen similar problems before, involving completing the square repeatedly.
$endgroup$
– vadim123
Jan 10 at 5:49
add a comment |
$begingroup$
I would write first $(x^2+cx+d) (x^2+ex+f) $ and then find $c, d, e, f$ in terms of $a, b$.
answered Jan 10 at 5:53
dmtridmtri
1,4542521
$endgroup$
add a comment |
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2 Answers
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$begingroup$
$$x^4+ax^3+2x^2+bx+1=$$
$$=(x^2+frac{a}{2}x)^2-frac{a^2}{4}x^2+2x^2+bx+1=$$
$$=(x^2+frac{a}{2}x)^2-frac{a^2}{4}x^2+2x^2-frac{b^2}{4}x^2+(frac{b}{2}x+1)^2=$$
$$=(x^2+frac{a}{2}x)^2+x^2(2-frac{a^2+b^2}{4})+(frac{b}{2}x+1)^2$$
If $2-frac{a^2+b^2}{4}>0$, then we have a problem; the sum of three nonnegative terms equals zero (at our real root), which can only happen if $x=0$ to make the second term zero. But then $(frac{b}{2}x+1)^2>0$. So if $a^2+b^2<8$ there is no real root. Hence $a^2+b^2ge 8$.
Now, if $a^2+b^2=8$, then the middle term is zero. Can we make the other two terms zero as well? We need $x=-frac{a}{2}$ and $x=-frac{2}{b}$, to make the first and last terms zero. This can be achieved, with $a=b=2$ and $x=-1$.
Hence, the minimum possible value of $a^2+b^2$ is indeed $8$.
answered Jan 10 at 5:38
vadim123vadim123
75.9k897189
$endgroup$
2
$begingroup$
Whoa, that was quick..
$endgroup$
– Yellow
Jan 10 at 5:42
2
$begingroup$
@AnuRadha, I've seen similar problems before, involving completing the square repeatedly.
$endgroup$
– vadim123
Jan 10 at 5:49
add a comment |
$begingroup$
$$x^4+ax^3+2x^2+bx+1=$$
$$=(x^2+frac{a}{2}x)^2-frac{a^2}{4}x^2+2x^2+bx+1=$$
$$=(x^2+frac{a}{2}x)^2-frac{a^2}{4}x^2+2x^2-frac{b^2}{4}x^2+(frac{b}{2}x+1)^2=$$
$$=(x^2+frac{a}{2}x)^2+x^2(2-frac{a^2+b^2}{4})+(frac{b}{2}x+1)^2$$
If $2-frac{a^2+b^2}{4}>0$, then we have a problem; the sum of three nonnegative terms equals zero (at our real root), which can only happen if $x=0$ to make the second term zero. But then $(frac{b}{2}x+1)^2>0$. So if $a^2+b^2<8$ there is no real root. Hence $a^2+b^2ge 8$.
Now, if $a^2+b^2=8$, then the middle term is zero. Can we make the other two terms zero as well? We need $x=-frac{a}{2}$ and $x=-frac{2}{b}$, to make the first and last terms zero. This can be achieved, with $a=b=2$ and $x=-1$.
Hence, the minimum possible value of $a^2+b^2$ is indeed $8$.
answered Jan 10 at 5:38
vadim123vadim123
75.9k897189
$endgroup$
2
$begingroup$
Whoa, that was quick..
$endgroup$
– Yellow
Jan 10 at 5:42
2
$begingroup$
@AnuRadha, I've seen similar problems before, involving completing the square repeatedly.
$endgroup$
– vadim123
Jan 10 at 5:49
add a comment |
$begingroup$
$$x^4+ax^3+2x^2+bx+1=$$
$$=(x^2+frac{a}{2}x)^2-frac{a^2}{4}x^2+2x^2+bx+1=$$
$$=(x^2+frac{a}{2}x)^2-frac{a^2}{4}x^2+2x^2-frac{b^2}{4}x^2+(frac{b}{2}x+1)^2=$$
$$=(x^2+frac{a}{2}x)^2+x^2(2-frac{a^2+b^2}{4})+(frac{b}{2}x+1)^2$$
If $2-frac{a^2+b^2}{4}>0$, then we have a problem; the sum of three nonnegative terms equals zero (at our real root), which can only happen if $x=0$ to make the second term zero. But then $(frac{b}{2}x+1)^2>0$. So if $a^2+b^2<8$ there is no real root. Hence $a^2+b^2ge 8$.
Now, if $a^2+b^2=8$, then the middle term is zero. Can we make the other two terms zero as well? We need $x=-frac{a}{2}$ and $x=-frac{2}{b}$, to make the first and last terms zero. This can be achieved, with $a=b=2$ and $x=-1$.
Hence, the minimum possible value of $a^2+b^2$ is indeed $8$.
answered Jan 10 at 5:38
vadim123vadim123
75.9k897189
$endgroup$
$$x^4+ax^3+2x^2+bx+1=$$
$$=(x^2+frac{a}{2}x)^2-frac{a^2}{4}x^2+2x^2+bx+1=$$
$$=(x^2+frac{a}{2}x)^2-frac{a^2}{4}x^2+2x^2-frac{b^2}{4}x^2+(frac{b}{2}x+1)^2=$$
$$=(x^2+frac{a}{2}x)^2+x^2(2-frac{a^2+b^2}{4})+(frac{b}{2}x+1)^2$$
If $2-frac{a^2+b^2}{4}>0$, then we have a problem; the sum of three nonnegative terms equals zero (at our real root), which can only happen if $x=0$ to make the second term zero. But then $(frac{b}{2}x+1)^2>0$. So if $a^2+b^2<8$ there is no real root. Hence $a^2+b^2ge 8$.
Now, if $a^2+b^2=8$, then the middle term is zero. Can we make the other two terms zero as well? We need $x=-frac{a}{2}$ and $x=-frac{2}{b}$, to make the first and last terms zero. This can be achieved, with $a=b=2$ and $x=-1$.
Hence, the minimum possible value of $a^2+b^2$ is indeed $8$.
answered Jan 10 at 5:38
vadim123vadim123
75.9k897189
answered Jan 10 at 5:38
vadim123vadim123
75.9k897189
answered Jan 10 at 5:38
vadim123vadim123
75.9k897189
answered Jan 10 at 5:38
vadim123vadim123
75.9k897189
75.9k897189
2
$begingroup$
Whoa, that was quick..
$endgroup$
– Yellow
Jan 10 at 5:42
2
$begingroup$
@AnuRadha, I've seen similar problems before, involving completing the square repeatedly.
$endgroup$
– vadim123
Jan 10 at 5:49
add a comment |
2
$begingroup$
Whoa, that was quick..
$endgroup$
– Yellow
Jan 10 at 5:42
2
$begingroup$
@AnuRadha, I've seen similar problems before, involving completing the square repeatedly.
$endgroup$
– vadim123
Jan 10 at 5:49
2
2
$begingroup$
Whoa, that was quick..
$endgroup$
– Yellow
Jan 10 at 5:42
$begingroup$
Whoa, that was quick..
$endgroup$
– Yellow
Jan 10 at 5:42
2
2
$begingroup$
@AnuRadha, I've seen similar problems before, involving completing the square repeatedly.
$endgroup$
– vadim123
Jan 10 at 5:49
$begingroup$
@AnuRadha, I've seen similar problems before, involving completing the square repeatedly.
$endgroup$
– vadim123
Jan 10 at 5:49
add a comment |
$begingroup$
I would write first $(x^2+cx+d) (x^2+ex+f) $ and then find $c, d, e, f$ in terms of $a, b$.
answered Jan 10 at 5:53
dmtridmtri
1,4542521
$endgroup$
add a comment |
$begingroup$
I would write first $(x^2+cx+d) (x^2+ex+f) $ and then find $c, d, e, f$ in terms of $a, b$.
answered Jan 10 at 5:53
dmtridmtri
1,4542521
$endgroup$
add a comment |
$begingroup$
I would write first $(x^2+cx+d) (x^2+ex+f) $ and then find $c, d, e, f$ in terms of $a, b$.
answered Jan 10 at 5:53
dmtridmtri
1,4542521
$endgroup$
I would write first $(x^2+cx+d) (x^2+ex+f) $ and then find $c, d, e, f$ in terms of $a, b$.
answered Jan 10 at 5:53
dmtridmtri
1,4542521
answered Jan 10 at 5:53
dmtridmtri
1,4542521
answered Jan 10 at 5:53
dmtridmtri
1,4542521
answered Jan 10 at 5:53
dmtridmtri
1,4542521
1,4542521
add a comment |
add a comment |
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2
$begingroup$
How do you know the polynomial has such a factorisation?
$endgroup$
– Lord Shark the Unknown
Jan 10 at 5:27
$begingroup$
Any fourth degree polynomial can be factorised into product of two quadratic expressions, no matter what the nature of its roots are, right?
$endgroup$
– Yellow
Jan 10 at 5:32
$begingroup$
@Any Radha, yes but not in that way you write...
$endgroup$
– dmtri
Jan 10 at 5:38
$begingroup$
I can't find anything wrong in your solution.
$endgroup$
– Anik Bhowmick
Jan 10 at 5:43
1
$begingroup$
Wouldn't this imply that $alpha + beta = a = b$ which need not be the case? If $x^4 + ax^3 + 2x^2 +bx +1 = (m x^2 + nx + p)(jx^2 + kx + q)$ we know $j = frac 1m$ and wolog it factors to $(x^2 + alpha x + gamma)(x^2 + beta x + delta)$ but we can't assume $gamma = delta = 1$
$endgroup$
– fleablood
Jan 10 at 6:24