another qn (1 Viewer)

[bos1234]

If one root of the eqn x^3 - bx^2 + cx - d = 0 is equal to the product of the other two, show that (c+d)^2 = d(b+1)^2

Let the roots be a,B,aB

a+B + aB = b

aB+ a^2.B + aB^2 = c

a^2B^2 = d

 

[jyu]

If one root of the eqn x^3 - bx^2 + cx - d = 0 is equal to the product of the other two, show that (c+d)^2 = d(b+1)^2

Let the roots be a,B,aB

a+B + aB = b

aB+ a^2.B + aB^2 = c

a^2B^2 = d

:wave:

 

[bos1234]

mm kkkk.. how did u know to make a + B the subject in the first one?

alot of manipulation :mad1:

thanks and nice work as usual

 

[pLuvia]

Simply by looking at the sum of paired roots and then you see you can factorise it and so there's a a+b in it, and so just make a+b the subject.

There are a lot of algebra bashing in HSC polynomials

 

[AlvinCY]

Despite the complexity of the question by first glance, once you sit down with a piece of paper, you'll actually kick yourself, watch:

If one root of the eqn x³ - bx² + cx - d = 0 is equal to the product of the other two, show that (c+d)² = d(b+1)²

Let the roots be A, B, AB; you know then by sum and product of roots, that:

A + B + AB = b

AB+ A² B + AB² = c

A²B² = d
<!--[if !supportEmptyParas]--> <!--[endif]-->
then
(c+d)²
= (AB + A²B + AB² + A²B²)²
= (AB)² (1 + A + B + AB)²
= d(b + 1)²


See!


Enjoy,
Alvin
B Education / B Maths @ Usyd (3rd year)[FONT=&quot][/FONT]

 

[bos1234]

bloody easy