3 unit maths ur 11 Cambridge questions help (1 Viewer)

drsabz101 · Jan 21, 2016

Ex 1A: 18 .
It says for what values of c is it true that :
B) x * x* x is less than or equal to x +x+x

The answer was x is less an than negative square root of 3 , or 0 is less than or equal to x which is less than and equal to square root of 3

Can someone please explain how they obtained this answer. Thanks

Paradoxica · Jan 21, 2016

Martin_SSEDU · Jan 21, 2016

They solved x^3-3x=0 and found all the x-intercepts needed and sketched it.

After doing that they returned back to the inequality x^3-3x<=0 and just deduced what values of x was the curve below the y-axis (y=0).

drsabz101 · Jan 21, 2016

Oh okay . Thanks !

drsabz101 · Jan 21, 2016

Another question in exercise 1C question 8d.
I don't get the answer

InteGrand · Jan 21, 2016

drsabz101 said:
Another question in exercise 1C question 8d.
I don't get the answer

What is the question?

drsabz101 · Jan 21, 2016

To fully factorise
4x^4 -2x^3y-3xy^3-9y^4

I got ( 2x^2+3y^2)(2x^2-3y^2) -Xy

The answer is : (2x^2+3y^2)(2x-3y)(x+y)

InteGrand · Jan 21, 2016

drsabz101 said:
To fully factorise
4x^4 -2x^3y-3xy^3-9y^4

I got ( 2x^2+3y^2)(2x^2-3y^2) -Xy

The answer is : (2x^2+3y^2)(2x-3y)(x+y)

$\noindent Your answer is correct except that you should have a bracket around the $-xy$ instead. I.e. it should be $\left( 2x^2+3y^2\right)\left(2x^2-3y^2 -xy\right)$, so maybe you just made a silly mistake somewhere? Once you have this, you just need to factorise the $\left(2x^2-3y^2 -xy\right)=\left(2x^2-xy-3y^2\right)$, which is relatively simple to do.$

InteGrand · Jan 21, 2016

$\noindent So here's how to do it. We have$

$\begin{align*}4x^4 - 2x^3 y -3xy^3 - 9y^4 &= \left(4x^4 - 9y^4\right) - \left(2x^3y+3xy^3 \right) \\ &= \left(2x^2 + 3y^2 \right) \left(2x^2 - 3y^2 \right) - xy \left(2x^2 + 3y^2 \right) \quad (\text{since }4x^4 - 9y^4= \left(2x^2 + 3y^2 \right) \left(2x^2 - 3y^2 \right) \text{ by difference of two squares and }2x^3y+3xy^3 = xy \left(2x^2 + 3y^2 \right) \text{ by basic factorisation}) \\ &= \left(2x^2 + 3y^2 \right) \left(2x^2 - 3y^2 - xy \right) \quad (\text{basic factorisation rules, namely }AB +AC = A(B+C))\\ &= \left(2x^2 + 3y^2 \right) \left(2x^2 -xy -3y^2 \right). \end{align*}$

$\noindent Now, to factorise the $\left(2x^2 -xy -3y^2 \right)$, we have$

$\begin{align*}2x^2 -xy -3y^2 &= \frac{\left(2x-3y \right)\left(2x+2y \right)}{2} \quad (\text{standard method for factorising a quadratic}) \\ &=\left(2x-3y \right)\left(x+y \right). \end{align*}$

$\noindent Thus $4x^4 - 2x^3 y -3xy^3 - 8y^4 =\left(2x^2 + 3y^2 \right)\left(2x-3y \right)\left(x+y \right)$.$

drsabz101 · Jan 21, 2016

Thanks ! I see what I did wrong, silly mistakes

drsabz101 · Jan 21, 2016

View attachment 32807

Also need help with factoring the 1st, 2nd, and last questions in the image above (it upside down).

thankyou

axwe7 · Jan 21, 2016

InteGrand said:
$\noindent So here's how to do it. We have$

$\begin{align*}4x^4 - 2x^3 y -3xy^3 - 8y^4 &= \left(4x^4 - 9y^4\right) - \left(2x^3y+3xy^3 \right) \\ &= \left(2x^2 + 3y^2 \right) \left(2x^2 - 3y^2 \right) - xy \left(2x^2 + 3y^2 \right) \quad (\text{since }4x^4 - 9y^4= \left(2x^2 + 3y^2 \right) \left(2x^2 - 3y^2 \right) \text{ by difference of two squares and }2x^3y+3xy^3 = xy \left(2x^2 + 3y^2 \right) \text{ by basic factorisation}) \\ &= \left(2x^2 + 3y^2 \right) \left(2x^2 - 3y^2 - xy \right) \quad (\text{basic factorisation rules, namely }AB +AC = A(B+C))\\ &= \left(2x^2 + 3y^2 \right) \left(2x^2 -xy -3y^2 \right). \end{align*}$

$\noindent Now, to factorise the $\left(2x^2 -xy -3y^2 \right)$, we have$

$\begin{align*}2x^2 -xy -3y^2 &= \frac{\left(2x-3y \right)\left(2x+2y \right)}{2} \quad (\text{standard method for factorising a quadratic}) \\ &=\left(2x-3y \right)\left(x+y \right). \end{align*}$

$\noindent Thus $4x^4 - 2x^3 y -3xy^3 - 8y^4 =\left(2x^2 + 3y^2 \right)\left(2x-3y \right)\left(x+y \right)$.$

Question says that it's a 9y^4 not a 8y^4?

axwe7 · Jan 21, 2016

drsabz101 said:
View attachment 32807

Also need help with factoring the 1st, 2nd, and last questions in the image above (it upside down).

thankyou

Attachment is invalid XD

InteGrand · Jan 21, 2016

axwe7 said:
Question says that it's a 9y^4 not a 8y^4?

Oh yeah, sorry, that was just a typo by me. It doesn't affect the rest of the working.

drsabz101 · Jan 22, 2016

I realised that, but I knew it was just a mistake.

I'll re-upload the photo again

drsabz101 · Jan 22, 2016

Okay so I may have to type these questions out again since the photo just won't open.
Factorise fully"

Q.1. (a^2-b^2-c^2)^2 - 4b^2c^2

ANS: (a-b-c)(a+b+c)(a-b+c)(a+b-c)

Q.2. (ax+by)^2+(ay-bx)^2+c^2(x^2+y^2)

ANS: (X^2+Y^2)(a^2+b^2+c^2)

Q3. ( a^4+a^2+b^2+b^4)

ANS: (a^2 - ab+b^2)(a^2+ab+b^2)

Q4. a^4+ ab^4

ANS: (a^2-2ab+2b^2)(a^2+2a^2+b^2)

InteGrand · Jan 22, 2016

drsabz101 said:
Okay so I may have to type these questions out again since the photo just won't open.
Factorise fully"

Q.1. (a^2-b^2-c^2)^2 - 4b^2c^2

ANS: (a-b-c)(a+b+c)(a-b+c)(a+b-c)

Q.2. (ax+by)^2+(ay-bx)^2+c^2(x^2+y^2)

ANS: (X^2+Y^2)(a^2+b^2+c^2)

Q3. ( a^4+a^2+b^2+b^4)

ANS: (a^2 - ab+b^2)(a^2+ab+b^2)

Q4. a^4+ ab^4

ANS: (a^2-2ab+2b^2)(a^2+2a^2+b^2)

$\underline{QUESTION 1}$

$We have$

$\begin{align*}\left(a^2-b^2-c^2\right)^2 - 4b^2c^2 &=\left(a^2-b^2-c^2\right)^2 - \left(2bc\right)^2 \\ &= \left(a^2-b^2-c^2 + 2bc \right)\left(a^2-b^2-c^2 - 2bc \right) \quad (\text{difference of two squares}) \\ &= \left ( a^2-\left ( b^2-2bc+c^2 \right ) \right )\left ( a^2-\left ( b^2+2bc+c^2 \right ) \right ) \\ &=\left ( a^2-\left ( b-c \right )^2 \right )\left ( a^2-\left ( b+c\right )^2 \right ) \\ &= \left ( a+b-c \right )\left ( a-b+c \right )\left ( a+b+c \right )\left ( a-b-c \right ) \quad (\text{difference of two squares}). \end{align*}$

InteGrand · Jan 22, 2016

drsabz101 said:
Okay so I may have to type these questions out again since the photo just won't open.
Factorise fully"

Q.1. (a^2-b^2-c^2)^2 - 4b^2c^2

ANS: (a-b-c)(a+b+c)(a-b+c)(a+b-c)

Q.2. (ax+by)^2+(ay-bx)^2+c^2(x^2+y^2)

ANS: (X^2+Y^2)(a^2+b^2+c^2)

Q3. ( a^4+a^2+b^2+b^4)

ANS: (a^2 - ab+b^2)(a^2+ab+b^2)

Q4. a^4+ ab^4

ANS: (a^2-2ab+2b^2)(a^2+2a^2+b^2)

$\underline{QUESTION 2}$

$\noindent We have$

$\begin{align*}\left (ax+by \right )^2+\left ( ay-bx \right )^2+c^2 \left ( x^2 + y^2 \right ) &= a^2 x^2 \color{red}{+ 2axby}\color{black} + b^2y^2 + a^2 y^2 \color{red}- 2aybx \color{black}+ b^2 x^2 + c^2x^2 + c^2 y^2 \quad (\text{via expansion; note the red terms will cancel}) \\ &= a^2\left ( x^2+y^2 \right )+b^2 \left ( x^2+y^2 \right ) + c^2 \left ( x^2+y^2 \right ) \quad (\text{partially factorising the remaining terms})\\ &= \left ( a^2+b^2 + c^2 \right ) \left ( x^2+y^2 \right ) \quad (\text{fully factorising}).\end{align*}$

InteGrand · Jan 22, 2016

drsabz101 said:
Okay so I may have to type these questions out again since the photo just won't open.
Factorise fully"

Q.1. (a^2-b^2-c^2)^2 - 4b^2c^2

ANS: (a-b-c)(a+b+c)(a-b+c)(a+b-c)

Q.2. (ax+by)^2+(ay-bx)^2+c^2(x^2+y^2)

ANS: (X^2+Y^2)(a^2+b^2+c^2)

Q3. ( a^4+a^2+b^2+b^4)

ANS: (a^2 - ab+b^2)(a^2+ab+b^2)

Q4. a^4+ ab^4

ANS: (a^2-2ab+2b^2)(a^2+2a^2+b^2)

$\underline{QUESTION 3}$

$\noindent I think the expression for Question 3 is meant to be $a^4 + a^2b^2 + b^4$, rather than what you typed (I think you may have done a typo; there's no really nice way to factorise the one you typed). Assuming it's $a^4 + a^2b^2 + b^4$, we have$

$\begin{align*}a^4 + a^2b^2 + b^4 &= a^4 + 2a^2 b^2 + b^4 - a^2 b^2 \\ &= \left ( a^2+b^2 \right )^2 - \left ( ab \right )^2 \\ &= \left ( a^2+b^2+ab \right )\left ( a^2+b^2-ab \right ) \quad (\text{difference of two squares}).\end{align*}$

InteGrand · Jan 22, 2016

drsabz101 said:
Okay so I may have to type these questions out again since the photo just won't open.
Factorise fully"

Q.1. (a^2-b^2-c^2)^2 - 4b^2c^2

ANS: (a-b-c)(a+b+c)(a-b+c)(a+b-c)

Q.2. (ax+by)^2+(ay-bx)^2+c^2(x^2+y^2)

ANS: (X^2+Y^2)(a^2+b^2+c^2)

Q3. ( a^4+a^2+b^2+b^4)

ANS: (a^2 - ab+b^2)(a^2+ab+b^2)

Q4. a^4+ ab^4

ANS: (a^2-2ab+2b^2)(a^2+2a^2+b^2)

$\underline{QUESTION 4}$

$\noindent I think this was a typo too, since the answer written here doesn't agree with the question (just try some values for $a$ and $b$ to see this; also, the right bracket of the answer would have had $3a^2$ rather than $a^2 + 2a^2$, so I suppose there's a typo there).$

$\noindent My guess is the expression was meant to be $a^4 + 4b^4$. Assuming this, we have$

$\begin{align*}a^4 + 4b^4 &= \left ( a^2 + 2b^2 \right )^2 - 2a^2 \cdot 2b^2 \\ &=\left ( a^2 + 2b^2 \right )^2 - 4a^2 b^2 \\ &= \left ( a^2 + 2b^2 \right )^2 - \left(2ab\right)^2 \\ &= \left ( a^2 + 2b^2 + 2ab \right )\left ( a^2 + 2b^2 - 2ab \right ) \quad (\text{difference of two squares}). \end{align*}$

$\textbf{Remark.} This formula is known as Sophie Germain's Identity.$

Sophie Germain's Identity: http://www.cut-the-knot.org/blue/SophieGermainIdentity.shtml

3 unit maths ur 11 Cambridge questions help (1 Viewer)

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