O-Level Additional Math

[iFruit]

Let's say √(x² + 2x +3) = y


Then we can write the original equation as

x² + 2x +3 - 110 = √(x² + 2x + 3)

which is

y² - 110 = y ----> y² - 110 -y =0 ---> (y+10)(y-11)=0

so, y=11 or -10 -----> √(x² + 2x +3) = 11 or -10

Taking the postive value y = 11,

(x² + 2x + 3) = 121 ---> x² + 2x -118 = 0

considering the roots α and β for the equation,

α + β = -2, α.β = -118 ---> (α² + β²) = 4+236 = 240, α².β² = 118²

Now We need an equation with roots α² and β² or of the form

(x-α²)(x - β²) = x² -(α² + β²)x + α².β² = 0

or

x² -240x+ 118² = 0


You can do the same exercise for the negative value y = -10

SKT:

Please help:

It is given that α and β are roots of the following equation:

x² + 2x - 107 = √(x² + 2x + 3)

Form an equation with roots α² and β².

[SKT]

Hi iFruit, tks for the solution.

I think y = -10 is rejected.

Need help for another Q:
Obtain the range of values of x for which │2x² - 7│ > 1/2 (x² + 1), for -2 ≤ x ≤ 5.

TIA.

[iFruit]

Hi SKT,


1) y = -10 is valid but it just means α, β are complex numbers. For real numbers it can be ignored.

2)

│2x² - 7│ > 1/2 (x² + 1) ----> │4x² - 14│ > (x² + 1)


i.e. 4x² - 14 > (x² + 1) or -(4x² - 14) > (x² + 1)

For 4x² - 14 > (x² + 1),


4x² - x² > 15 —> x² > 5----> x < -√5 or x > √5


For -(4x² - 14) > (x² + 1),

-(4x² - 14) > (x² + 1) —> 4x² - 14 < - (x² + 1) —> 5x² < 13

so x² < 13/5 –> x < √(13/5) or x > -√(13/5)

so we have,

x < -√5 or x > √5; x < √(13/5) or x > -√(13/5)

and

-2 ≤ x ≤ 5.


Combining all we get,

-√(13/5) < x < √(13/5) and √5 < x ≤ 5

[SKT]

Can anyone help to clear my DD’s doubt for the following:


(x+1)/(x-1) = (-x-1)/(1-x), but the remainder of (x+1)/(x-1) is 2, and the remainder of (-x-1)/(1-x) is -2. Why are they not the same?

TIA.

[iFruit]

(x+1)/(x-1) is of the form


D = d.q + r where D = x+1, q =1, d = x-1, r =2

whereas (-x-1)/(1-x) is of the form

-D = -d.q -r ==> (-D) = (-d).q + (-r) where (-D) = (-x-1), q = 1, (-d) = (1-x) and -r = -2


Because the divisor is changed to -ve sign, the remainder also has to change to -ve sign so that dividend also changes sign

HTH.

SKT:

Can anyone help to clear my DD's doubt for the following:

(x+1)/(x-1) = (-x-1)/(1-x), but the remainder of (x+1)/(x-1) is 2, and the remainder of (-x-1)/(1-x) is -2. Why are they not the same?

TIA.

[benorito]

Hi,


Can you kindly help with this question :
Given that 280 and a number y have a LCM of 6160 and a HCF of 40, find the number y.

Thanks in advance!

[iFruit]

benorito:

Hi,


Can you kindly help with this question :
Given that 280 and a number y have a LCM of 6160 and a HCF of 40, find the number y.

Thanks in advance!

You need to know that product of two numbers = HCF x LCM of those two numbers


280 * y = 6160 * 40

y = (6160 * 40)/280 = 880

[benorito]

iFruit:

You need to know that product of two numbers = HCF x LCM of those two numbers


280 * y = 6160 * 40

y = (6160 * 40)/280 = 880

Thank you !!

[SKT]

Hi,

Given that x and y satisfy the simultaneous equations
mx + (m-1)y = 10,
(m-2)x + 3my = 20.

(a) If the equations have no unique solution, find the values of m.
(b) If the equations have no solutions, find the value of m.

TIA.

[iFruit]

SKT:

Hi,

Given that x and y satisfy the simultaneous equations
mx + (m-1)y = 10,
(m-2)x + 3my = 20.

(a) If the equations have no unique solution, find the values of m.
(b) If the equations have no solutions, find the value of m.

TIA.

Using Cramer’s rule

x = [30m - 20(m-1)]/ [3m² – (m² -3m +2)] = (10m+20)/ (2m² + 3m -2)

y = [20m -10(m-2)] / (2m² + 3m -2) = (10m+20)/ (2m² + 3m -2)


Recall that equation has
1) infinite solutions when (10m+20) = 0 and (2m²+ 3m -2)=0
2) No solutions when (10m+20) # 0 and (2m² + 3m -2)=0

(2m² + 3m -2)= (2m-1)(m+2) =0----> m=-2 or m=1/2


When m=-2, 10m+20 = 0,
when m=1/2, 10m+20 = 25


So infinite solutions when m=-2
No solutions when m=1/2