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Quadratic identies problem, that i dont understand (1 Viewer)

wolf7

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Oct 10, 2005
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HSC
2006
i need help with 3 question that i dont understand?

first question

if x² = ax(ax-1)+bx for more than two values of xd, find a and b

second question

Find a, b and c if:

x² = a(x+1)²+bx+c

the equal(=) sign suppose to mean similar

third question
Find the values for A and B if:
4x²-12x + 9 = (Ax+B)²

in the thrid question, the equal sign also means similar.

Can some one help me with these question?
 
I

icycloud

Guest
wolf7 said:
first question

if x² = ax(ax-1)+bx for more than two values of xd, find a and b
It can be proved that two quadratic equations which are equal for three or more distinct values of x (i.e. more than two values of x) agree for all values of x, and thus their coefficients are congruent (three horizontal lines). Thus,

LHS = x^2
RHS = ax^2 - ax + bx
= ax^2 + (b-a)x

Equating coefficients of LHS and RHS,

a = 1

b-a = 0
b-1=0
b = 1

Thus, a = 1, b = 1 #

wolf7 said:
second question

Find a, b and c if:

x² = a(x+1)²+bx+c

the equal(=) sign suppose to mean similar
Since the two equations are congruent (three horizontal lines), we can simply equate coefficients. There are two ways of doing this, one way is outlined below:

LHS = x^2
RHS = a[x^2 + 2x + 1] + bx + c
= ax^2 + 2ax + a + bx + c
= ax^2 + (2a+b)x + (a+c)

Thus, equating coefficients of LHS and RHS, we have:

a = 1

2a + b = 0,
b = -2(1)
= - 2

a+c = 0
c = -1

Thus, a = 1, b = -2, c = -1 #

wolf7 said:
third question
Find the values for A and B if:
4x²-12x + 9 = (Ax+B)²

in the thrid question, the equal sign also means similar.
As above, expand and equate coefficients. (The other way is to substitute various values of x to work out the coefficients. Sometimes this way is faster, it depends on the question.)

LHS = 4x^2 - 12x + 9
RHS = (Ax+B)^2
=(Ax)^2 + 2ABx + B^2
=A^2 x^2 + 2ABx + B^2

Equating coefficients,

A^2 = 4
A = +2 or -2

B^2 = 9
B = +3 or -3

2AB = -12
AB = -6
A = -6 / B

Thus,
A = 2 and B = -3 or
A = -2 and B = 3 #
 

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