algebra problem

**ahmsimi** · January 26th 2009, 05:54 AM

need ur help people on few questions

Q1>>>Show that the roots of the equation
(x – a) (x – b) + (x – b) (x – c) + (x – c) (x – a) = 0
are always real and they cannot be equal unless a = b = c

Q2>>If w is a complex cube root of unity, then show that
(1 + 5w2 + w4) (1 + 5w + w2) (5 + w + w2) =64

Q3>>>A question paper contains 12 questions divided into two sections. Section I contains
7 questions and section II contains 5 questions. In how many ways can a candidate
choose the questions if he has to select 8 questions in all with teh restriction of
atleast 3 questions from each section?

Q4>>Let A =
|1 -1 0 |
|2 2 4 |
|2 3 4 |

and B =

|4 2 4 |
|0 1 2 |
|2 1 5 |
Find AB.

Use this to solve the following system of equations:
x – y = 3
2x + 3y + 4z = 17
y + 2z = 7

thanku for ur help frndz

**red_dog** · January 26th 2009, 07:48 AM

Q1)
The equation can be written as $3x^2-2(a+b+c)x+ab+ac+bc=0$
The discriminant is $\Delta=4(a^2+b^2+c^2-ab-ac-bc)$

But it is well known that $a^2+b^2+c^2\geq ab+ac+bc \ \forall a,b,c\in\mathbf{R}$

so, $\Delta\geq 0$ and the roots are real.

The equality stands if and only if $a=b=c$

**red_dog** · January 26th 2009, 07:54 AM

Q2)

We have $w^3=1$ and $w^2+w+1=0$ .

Then $w^4=w^3\cdot w=w$

$1+5w^2+w^4=(1+w+w^2)+4w^2=4w^2$

$1+5w+w^2=(1+w+w^2)+4w=4w$

$5+w+w^2=(1+w+w^2)+4=4$

Then the product is equal to $4w^2\cdot 4w\cdot 4=64w^3=64$

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