Polynomial Q (1 Viewer)

[untouchablecuz]

CSSA 1997, Question 4 a)

One of the roots of the equation x^3 + ax^2 +1 = 0 is equal to the sum of the other roots.

i) Show that x = (-a/2) is a root of the equation.

ii) Find a.

i) Fine, no troubles here.

ii) To find a, I subbed x = (-a/2) into the equation and solved for a, getting a = -2

Worked solutions say that the answer is a = 2 (after using a long winded process of sum of roots, product of roots etc)

After graphing x^3 - 2x^2 +1 = y (using graphmatica) the roots were 1, -0.6, 1.6 (being in the form a, b, a - b)

After graphing x^3 + 2x^2 +1 = y (the supposed "answer") there was only one root and not three (intersection with x axis), as stated by the question

Is the question flawed or am I doing something wrong?

 

[gurmies]

Ignore that^

I used the same method as you which I believe is far more efficient than using coefficient relations, and ended up with a = -2 as well. Looks like an error in the answers...

 

[bored of sc]

<TABLE cellSpacing=0 cellPadding=3 width="100%" border=0><TBODY><TR><TD class=alt2 style="BORDER-RIGHT: 1px inset; BORDER-TOP: 1px inset; BORDER-LEFT: 1px inset; BORDER-BOTTOM: 1px inset">CSSA 1997, Question 4 a)

One of the roots of the equation x^3 + ax^2 +1 = 0 is equal to the sum of the other roots.

i) Show that x = (-a/2) is a root of the equation.

ii) Find a. </TD></TR></TBODY></TABLE>

(ii) x³+ax²+1 = 0
(-a/2)³+a(-a/2)²+1 = 0
-a³/8 +a³/4 +1 = 0
(-a³+2a³)/8 +1 = 0
a³/8 +1 = 0
(a/2)³+1 = 0
(a/2+1)(a²/4-a/2+1) = 0
a/2+1 = 0
a/2 = -1
a = -2

Same as you guys.