1. ## Quadratic Equation Part 4

1. Solve $(\frac{3-i}{2+i})z=\frac{5+5i}{-1+2i}$. Give your answer in form a+bi
a. Show z on an Argand diagram, find the modulus and argument of z

2. Find the set of values of x such that $\mid \frac{2x+3}{x-1}\mid<1$

3. Given $ax^2+bx+c=0$. Deduce that the equation has two real roots if $b^2-4ac\geq0$ and it has two complex roots if $b^2-4ac\leq0$
a. Determine all values of k so that $x^2-(k-3)x+k^2+2k+5=0$ has real roots
b. If $\alpha$ and $\beta$ are two real roots of this equation, show that $
\alpha^2+\beta^2=-(k+5)^2+24$
. Hence find the maximum value of $\alpha^2+\beta^2$

4. What is the difference between $\log$ and $\lg$?

2. Originally Posted by cloud5
1. Solve $(\frac{3-i}{2+i})z=\frac{5+5i}{-1+2i}$. Give your answer in form a+bi
a. Show z on an Argand diagram, find the modulus and argument of z
(1) $z=\frac{5+5i}{-1+2i}\cdot \frac{2+i}{3-i}$

$=\frac{10+5i+10i+5i^2}{-3+i+6i-2i^2}$

$=\frac{5+15i}{-1+7i}$

Then multiply by its conjugate .

$=\frac{5+15i}{-1+7i}\cdot \frac{-1-7i}{-1-7i}$

$=\frac{100-50i}{50}$

Therefore , $z=2-i$

$|z|=\sqrt{2^2+(-1)^2}$

$\tan \theta=\frac{1}{2}$

$\theta=26.57$

$
\arg z=-26.57
$

3. Originally Posted by cloud5
1. Solve $(\frac{3-i}{2+i})z=\frac{5+5i}{-1+2i}$. Give your answer in form a+bi
2. Find the set of values of x such that $\mid \frac{2x+3}{x-1}\mid<1$

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4. What is the difference between $\log$ and $\lg$?
(2) $\frac{|2x+3|}{|x-1|}<1$

$|2x+3|<|x-1|$ , square both sides

$4x^2+12x+9

$3x^2+14x+8<0$

$(3x+2)(x+4)<0$

Carry on ..

(4) lg is another way of writing $log_{10}$

In other words , lg refers to only log to the base 10

log can be log to the base 2 or 3 or 4 ..

4. Originally Posted by cloud5

3. Given $ax^2+bx+c=0$. Deduce that the equation has two real roots if $b^2-4ac\geq0$ and it has two complex roots if $b^2-4ac\leq0$
a. Determine all values of k so that $x^2-(k-3)x+k^2+2k+5=0$ has real roots
b. If $\alpha$ and $\beta$ are two real roots of this equation, show that $
\alpha^2+\beta^2=-(k+5)^2+24$
. Hence find the maximum value of $\alpha^2+\beta^2$
$ax^2+bx+c=0$

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

When $b^2-4ac\geq 0$

What can you deduce and when its \leq what can you deduce ?

(a) Solve $[-(k-3)]^2-4[k^2+2k+5]\geq 0$

(b) Let the 2 roots be p and q .

$p+q=k-3$ , $pq=k^2+2k+5$

$(p+q)^2-2pq=(k-3)^2-2[k^2+2k+5]$

$=-k^2-10k-1$

$=-(k^2+10k+1)$

$=-[(k+5)^2-24]$

$=-(k+5)^2+24$

For the max , observe the range of the real roots , sketch the graph , and take the one nearest to 24 .