Is it a 5th degree Polynomial?

Algebra Level 4

$\large \begin{aligned} a + 4b + 9c + 16d + 25e & = 1 \\ 4a + 9b + 16c + 25d + 36e & = 8 \\ 9a + 16b + 25c+ 36d + 49e & = 23 \end{aligned}$

Real numbers $a$ , $b$ , $c$ , $d$ , and $e$ satisfy the system of equations above. Find $a + b + c+ d + e$ .

The answer is 4.

2 solutions

Wei Chen
Aug 29, 2016

Add the first and third equation, subtract twice the second equation, we get

$2a+2b+2c+2d+2e=8$

Divide by 2, leads to

$a+b+c+d+e=\boxed{4}$

Exact (+1), or in other words, $x(1,4,9,16,25) + y(4,9,16,25,36) + z(9,16,25,36,49) = (1,1,1,1,1)\Rightarrow$ $x=1/2, y = -1, z = 1/2 \Rightarrow a + b +c + d+ e = (1/2)\cdot 1 - 1\cdot 8 + (1/2)\cdot 23 = 4$

Guillermo Templado - 4 years, 9 months ago

Good point, instead of solving a,b,c,d and e individually, the idea is to express (1,1,1,1,1) as linear combination of the given vectors.

Wei Chen - 4 years, 9 months ago

In this case, if the vector $(1,1,1,1,1)$ could not be expressed as a linear combination of the other vectors, this question would have infinite solutions for $a + b + c + d + e$ due to Rouche-Frobenius theorem. In other cases, it's possible that $a + b + c + d + e$ even can't be calculated,i.e, it doesn't have any solution

Guillermo Templado - 4 years, 9 months ago

@Guillermo Templado – Yeah, luckily for the 3x5 coefficient matrix here, the rank is 3 so we know there is at least one solution for $a+b+c+d+e$ . As whether there is one solution or infinite number of solutions, luckily again for the 5X3 coefficient matrix of linear combination equations you wrote, the rank is also 3, so we get an unique solution.

Btw, thanks for asking Pi to give me the Latex guide, think I'm getting hang of it now.

Wei Chen - 4 years, 9 months ago

@Wei Chen – yes,it's like this. I hope not to be wrong this time... and you're welcome ,very soon you'll drive Latex very good. I'm sure.

Guillermo Templado - 4 years, 9 months ago

Its really a brilliant solution.Thanks for the same.

Priyanshu Mishra - 4 years, 9 months ago

maybe this years amti question

Aditya Narayan Sharma - 4 years, 9 months ago

Correct. Have you appeared in it?

Priyanshu Mishra - 4 years, 9 months ago

Aditya Dhawan
Sep 17, 2016

$Consider\quad the\quad function\quad f(x)=\quad a{ x }^{ 2 }+b({ x+1) }^{ 2 }+c({ x+2) }^{ 2 }+d({ x+3) }^{ 2 }+e{ (x+4) }^{ 2 }\quad \\ \\ =\quad (a+b+c+d+e){ x }^{ 2 }+2x(b+2c+3d+4e)+(b+4c+9d+16e)\\ =\quad p{ x }^{ 2 }+qx+r\\ \\ Now,\\ \begin{cases} f(1)=p+q+r=1\quad \\ f(2)=4p+2q+r=8 \\ f(3)=9p+3q+r=23 \end{cases}\\ \Rightarrow \quad p=\quad a+b+c+d+e\quad =\quad \boxed { 4 } \\ \\$

Nice solution. How you generated $f(x)$ ?

Priyanshu Mishra - 4 years, 8 months ago

Thanks! I merely looked at the coefficients of the given system of equations and generated the function accordingly.

Aditya Dhawan - 4 years, 8 months ago

Is it a 5th degree Polynomial?

The answer is 4.

2 solutions

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