H2 Math Sat 130pm

This page contains all questions and answers asked by students from this class. The most recent questions will be at the top.

MF26


Hence part

Hence part

We have that (2-3i)^2 = -5-12i
(z-i+1)^2 = 5+12i
(z-i+1)^2 = -(-5-12i)
(z-i+1)^2 = -(2-3i)^2
(z-i+1)^2 = i^2 (2-3i)^2
(z-i+1)^2 = [i(2-3i)]^2
(z-i+1)^2 = (2i+3)^2
(z-i+1)^2 = (2i+3)^2
(z-i+1) = \pm(2i+3)
z = \pm(2i+3) - 1 + i
z = 2i+3 - 1 + i or z = -(2i+3) - 1 + i
z = 3i+2 or z = -i+2


Question 3

Question 3

This is a worksheet from class.
There is a slight typo, we need to show that x^2 (x+2)^2 - 1 = (x+1)^2(x+1+\sqrt{2})(x+1-\sqrt{2})
So for all proof/ show questions, it is important that we can identify whether we should prove from LHS to RHS or RHS to LHS.
For this question, we observe that if we start from LHS, we are trying to expand our expression. But if we start from RHS, we are trying to simplify our expression. It will be generally easier to simplify.

RHS
= (x+1)^2(x+1+\sqrt{2})(x+1-\sqrt{2})
= (x+1)^2[(x+1)^2 - (\sqrt{2})^2]
= (x+1)^2[(x+1)^2 - 2]
= (x+1)^2(x^2 + 2x + 1 - 2]
= (x^2 + 2x + 1)(x^2 + 2x -1); we can also expand it fully
= [x(x+2)+1][x(x+2)-1]
= x^2 (x+2)^2 -1

Here’s a much faster but less obvious method!
LHS
= x^2 (x+2)^2 -1
= [x(x+2)+1][x(x+2)-1]
= (x^2 + 2x + 1)(x^2 + 2x -1); using quadratic formula
= (x+1)^2 (x - \frac{-2-\sqrt{8}}{2})x - \frac{-2+\sqrt{8}}{2})
= (x+)^2(x+1+1+\sqrt{2})(x+1-\sqrt{2})

Comments
    pingbacks / trackbacks
    • […] H2 Math Sat 130pm H2 Math Sat 330pm H2 Math Sun 930am H2 Math Sun 1130am H2 Math Sun 2pm H2 Math Mon 2pm H2 Math Mon 430pm H2 Math Mon 730pm H2 Math Tue 5pm H2 Math Tue 7pm H2 Math Thur 5pm […]

    Leave a Comment

    Contact Us

    CONTACT US We would love to hear from you. Contact us, or simply hit our personal page for more contact information

    Not readable? Change text. captcha txt
    0

    Start typing and press Enter to search