Question

x^4-9x^2+c=0
a) Find the value of c such that the equation has four distinct real roots
b) Find the value of c such that the equation has two distinct roots
c) Find the value of c such that the equation has no roots

Answer 1

how do I begin?

Answer 2

HINT: replace \(x^2\) with \(y\)

Answer 3

ok.then y^2-9y+c=0. then..

Answer 4

do you know about the "discriminant" of a quadratic equation?

Answer 5

No.

Answer 6

let x^2 = y
the equation will be :
y^2 - 9y + c = 0

then take D > 0

Answer 7

this might help you learn about this property: http://www.mathwarehouse.com/quadratic/discriminant-in-quadratic-equation.php

Answer 8

is it use b^2-4ac.

Answer 9

yes

Answer 10

the value of the discriminant determines the number of real roots that a quadratic equation has

Answer 11

if we call the discriminant D, then:

D = 0 => two equal real roots
D > 0 => two distinct real roots
D < 0 => no real roots

Answer 12

C > 81/4.but...how to get 4 roots

Answer 13

remember you substituted \(y=x^2\) to get:\[y^2-9y+c=0\]so if you get two real roots for y, then that means 4 real roots for x since \(x=\pm\sqrt{y}\)

Answer 14

I still didn't understand. is my ans correct?

Answer 15

do you agree that C > 81/4 will give you two rel roots for y?

Answer 16

No. how to solve....pls step by step

Answer 17

you need to understand the significance of discriminant fully in order to understand how to solve this. I suggest you first spend some time understanding that and then come back to this problem. Use the web link I gave above or just search for "quadratic discriminant" in google.

Answer 18

if you are still stuck afterwards then give me a ping and I will try to explain further.

Answer 19

ok - so do you understand the discriminant fully now?

Answer 20

yes.

Answer 21

good, so what would be the discriminant for:\[y^2-9y+c=0\]

Answer 22

D>0

Answer 23

no - what is the expression for the discriminant for that equation?

Answer 24

e.g. if we have: \(ax^2+bx+c=0\) then the discriminant would be equal to \(b^2-4ac\)

Answer 25

81-4c>0

Answer 26

there is no ">0" here. the discriminant is equal to 81-4c

agreed?

Answer 27

agree

Answer 28

good - so if now call the discriminant D, then we can write:\[D=81-4c\]now what are the 3 rules for the value of the discriminant?

Answer 29

as u ve mentioned bf,
D = 0 => two equal real roots
D > 0 => two distinct real roots
D < 0 => no real roots

Answer 30

perfect - so now find what value(s) of 'c' will lead to each of the 3 above cases. what do you get?

Answer 31

81-4c>0
81-4c<0
81-4c=0

Answer 32

is that so?

Answer 33

yes, and now solve each of these to find 'c' in each case

Answer 34

c>81/4
c<81/4
c=81/4

Answer 35

1st two are incorrect

Answer 36

how do I solve then?

Answer 37

please list the steps you took to solve the 1st one and I will try to spot where you may have made a mistake

Answer 38

81-4c>0
81-4c<0
is it c < 81/4
c> 81/4......bit confused

Answer 39

>=?

Answer 40

where did I go?

Answer 41

take a simpler case:

81-c>0

what does this mean for 'c'?

Answer 42

c<81

Answer 43

exactly - no retry the case we had above please

Answer 44

*now

Answer 45

c<81/4
c>81/4
c=81/4

Answer 46

perfect, so now you know that:

c = 81/4 => two equal real roots for y
c < 81/4 => two distinct real roots for y
c > 81/4 => no real roots for y

agreed?

Answer 47

after solving for c, I ve to subst back to the eqn. as \[y^2-9y+c=0\]

Answer 48

you don't need to substitute back into the equation

Answer 49

\[y^2-9y+81/4=0\]

Answer 50

you are not asked to find the roots - you are asked to find the value for 'c' to give you a certain number of real roots

Answer 51

Sry, hw to determine the roots?

Answer 52

do you agree that you have been asked to find the values for 'c' that give you:
a) four distinct real roots
b) two distinct real roots
c) no real roots

for the equation: \(x^4-9x^2+c=0\)

Answer 53

Yes. surely

Answer 54

so, we first substituted \(y=x^2\) to get: \(y^2-9y+c=0\) and found that:

a) c < 81/4 => two distinct real roots for y
b) c = 81/4 => two equal real roots for y
c) c > 81/4 => no real roots for y

agreed?

Answer 55

agree.

Answer 56

so, finally, we know that \(x=\pm\sqrt{y}\) which means:

a) two distinct real roots for y => four distinct real roots for x
b) two equal real roots for y => two distinct real roots for x
c) no real roots for y => no real roots for x

Answer 57

remember, for each root of y we get two roots for x since \(x=\pm\sqrt{y}\)

Answer 58

we can therefore say that:

a) c < 81/4 => four distinct real roots for x
b) c = 81/4 => two distinct real roots for x
c) c > 81/4 => no real roots for x

Answer 59

OK. thanks a lot!!!!! Clearly explained.
Cn I ask a qn on inequality?

Answer 60

<<<---- sure - just post it as a new question on the left