MHB Expression containing coefficients of a quadratic equation

[Evgeny.Makarov]

Let $a$ be the smallest root of the equation $x^2-9x+10=0$. Find $a^4-549a$. Extra credit if the solution does not find the actual roots of the equation.

 

[Opalg]

Let $a$ be the smallest root of the equation $x^2-9x+10=0$. Find $a^4-549a$. Extra credit if the solution does not find the actual roots of the equation.

[sp]The factorisation $x^4 - 549x + 710 = (x^2-9x+10)(x^2+9x+71)$ shows that if $a$ is a root of $x^2-9x+10=0$ (it could be either of the roots, not necessarily the smaller one), then $a^4 - 549a + 710 = 0.$

Therefore $a^4 - 549a = -710.$[/sp]

 

[Evgeny.Makarov]

Correct.

 

[Greg]

Let $a$ be the smallest root of the equation $x^2-9x+10=0$. Find $a^4-549a$. Extra credit if the solution does not find the actual roots of the equation.

Spoiler

$$(a^2+9a-10)(a^2-9a+10)=0$$
$$a^4-9a^3+10a^2+9a^3-81a^2+90a-10a^2+90a-100=0$$
$$a^4-81a^2+180a-100=0$$
$$a^4+180a-729a+729a=81a^2+100$$
$$a^4-549a=81a^2-729a+100$$
$$a^4-549a=81\left(a^2-9a+\dfrac{100}{81}\right)$$
$$a^4-549a=81\left(-10+\dfrac{100}{81}\right)=-710$$

which holds for either root.

 

[MarkFL]

My solution:

Spoiler

We are given:

$$x^2=9x-10$$

Square:

$$x^4=81x^2-180x+100$$

Subtract through by $549x$:

$$x^4-549x=81x(x-9)+100$$

But, we are given $x(x-9)=-10$, hence:

$$x^4-549x=81(-10)+100=-710$$

Because of the squaring, this holds for all roots of:

$$\left(x^2-9x+10\right)\left(x^2+9x-10\right)=0$$

 

[kaliprasad]

Let $a$ be the smallest root of the equation $x^2-9x+10=0$. Find $a^4-549a$. Extra credit if the solution does not find the actual roots of the equation.

Spoiler

we have $a^2= 9a - 10$
Hence $a^4 = a(a(9a-10)) = a(9a^2- 10a) = a (9(9a-10)- 10a)= a(71a - 90)$
$= 71a^2 - 90 a = 71(9a-10) - 90 a = 549a - 710$
or $a^4-549a = - 710$