桃子

Consider the quadratic equation x^2+kix+(k+1)(k-1)=0.
(a) Find the value or range of k such that the equation has repeated root.
(b) Find the value or range of k^2 such that the equation has two real roots.

A student gave the following answers.
(a) discriminant=0
(ki)^2-4(k+1)(k-1)=0
-5k^2+4=0
k^2=4/5
k=2/sqrt(5)

(b) discriminant>0
(ki)^2-4(k+1)(k-1)>0
-5k^2+4>0
k^2<4/5

Point out the student's mistake in each part.

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[ 本帖最後由 桃子 於 2014-3-7 11:47 PM 編輯 ]

ckh822

This is not a equation
Why there is a "+" sign in-front of "0"？

Simon

haha, this should be a typo!

桃子

引用:

原帖由 ckh822 於 2014-3-7 11:39 PM 發表
This is not a equation
Why there is a "+" sign in-front of "0"？

typo
should be =0

thomas3253

引用:

原帖由 桃子 於 2014-3-7 23:29 發表
Consider the quadratic equation x^2+kix+(k+1)(k-1)=0.
(a) Find the value or range of k such that the equation has repeated root.
(b) Find the value or range of k^2 such that the equation has two rea ...

I wonder the discrimination formula can be used when there exists non-real coefficient in the quadratic equation

桃子

引用:

原帖由 thomas3253 於 2014-3-7 11:54 PM 發表

I wonder the discrimination formula can be used when there exists non-real coefficient in the quadratic equation

Yes for part (a) but no for part (b). Think about why..

萬陽

真係唔知ｗｏ，求解！

smartboypm

留名睇答案，真係諗唔到

thomas3253

引用:

原帖由 桃子 於 2014-3-8 00:03 發表

Yes for part (a) but no for part (b). Think about why..

for quadratic equation y=ax^2+bx+c
the root is x=[-b+sqrt(b^2-4ac)]/2a or x=[-b-sqrt(b^2-4ac)]/2a

Now, in this question
x^2+kix+(k+1)(k-1)=0
x=[-(ki)+sqrt((ki)^2-4(k^2-1))]/2 or x=[-b-sqrt((ki)^2-4(k^2-1))]/2
x=[-(ki)+sqrt(-5k^2+4)]/2 or x=[-(ki)-sqrt(-5k^2+4)]/2

If both root are real,
we first consider -5k^2+4 is positive and ki is real=> k^2<4/5 and (k is pure imaginary or k=0)=>k^2<4/5 and k^2<=0 => k^2<=0

Then, if k^2>=4/5 , both ki and sqrt(-5k^2+4) will be imaginary.
The sum of two imaginary number equal to real number if they sum to 0.
So, k+sqrt(4-5k^2)=0 => k=-sqrt(4-5k^2) => k^2=4-5k^2 => 6k^2-4=0 =>k=sqrt(2/3) => k^2=2/3 < 4/5
Hence, the case do not satisfy the assumption.

=>k^2<=0

oh!

真係諗唔到 求解！

桃子

引用:

原帖由 桃子 於 2014-3-8 12:03 AM 發表

Yes for part (a) but no for part (b). Think about why..

To elaborate that, the following properties works for all quadratic equations.
1. If discriminant=0, then the equation has a repeated root.
2. If discriminant is non-zero, then the equation has two distinct roots.

The following properties works for quadratic equations with real coefficients.
1. If discriminant>0, then the equation has two real roots.
2. If discriminant<0, then the equation has two distinct non-real roots. (no real roots)

These properties can be easily derived by noting the roots are [-b+sqrt(delta)]/(2a) and [-b-sqrt(delta)]/(2a).

directioner

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jwyl

see see

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