Maths question

Hi im stuck on a question could someone please help me out:

Find the value of k such that the equation 2x^2−kx+2=0
(The x is a letter not multiplication) :rofl3:

(edited 6 years ago)

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Reply 1

Radioactivedecay

Have you tried completing the square?

Reply 2

Bananasplitxxx

Can you do that with the k?

Reply 3

akay8

Anything else that goes with the question?

Reply 4

Bananasplitxxx

Original post by akay8

Anything else that goes with the question?

E6D3C3F2-B713-43A6-913A-51484F5E6749.jpg.jpeg

(edited 6 years ago)

Reply 5

MWills99

I got x=(k/4)+/-sqrt((k^2/16)-1)

Reply 6

thekidwhogames

2x^2 - kx + 2 = 0

Equal roots hence b^2-4ac=0

k^2 - 4(2)(2) = 0 --> k^2 - 16 = 0 --> what is k (finish it off)

Reply 7

peeked

What level is this, GCSE Higher?

Reply 8

Bananasplitxxx

Original post by MWills99

I got x=(k/4)+/-sqrt((k^2/16)-1)

How did you get that ?

Reply 9

Bananasplitxxx

Original post by peeked

What level is this, GCSE Higher?

It’s national 5, I think it’s the same as GCSE though

Reply 10

Wellfish

Giving x the value of 1 and k the value of 6 seems to work out pretty fine but i guess that would be too easy.

Reply 11

Bananasplitxxx

Original post by thekidwhogames

2x^2 - kx + 2 = 0

Equal roots hence b^2-4ac=0

k^2 - 4(2)(2) = 0 --> k^2 - 16 = 0 --> what is k (finish it off)

Ohh is it this formula
C0A7B18D-0763-4765-A87F-5345D6486C5C.jpg.jpeg

Reply 12

peeked

Original post by Bananasplitxxx

Ohh is it this formula
C0A7B18D-0763-4765-A87F-5345D6486C5C.jpg.jpeg

that formula is the quadratic formula, you use it to find 'x' if you already know a,b and c. In this case you don't know what b equals. I don't think you need the formula for this question.

Reply 13

thekidwhogames

Original post by Bananasplitxxx

Ohh is it this formula
C0A7B18D-0763-4765-A87F-5345D6486C5C.jpg.jpeg

Are you in GCSE?

That is the quadratic formula. To have equal root that means the thing under the root = 0 to cause one repeated solution e.g. b^2 - 4ac = 0.

Reply 14

Mehru1214

Original post by Bananasplitxxx

How did you get that ?

The discriminant is (b^2 - 4ac)

If this expression is equal to 0, then the solution only has one real root.

a = 2
b = -k
c = 2

So:

(-k)^2 -4 x 2 x 2 = 0
k^2 -16 = 0
k^2 = 16 [now square root both sides]
k = +-4

EDIT: k = +-4 due to the square root. Just remembered!

(edited 6 years ago)

Reply 15

MWills99

Original post by Bananasplitxxx

How did you get that ?

So you start with 2x^2-kx+2=0
Then you divide by 2: x^2-(k/2)x+1=0
The complete the square: (x-(k/4))^2-((K^2)/16)+1=0
Then solve for x: (x-(k/4))^2=((k^2)/16)-1
(x-(k/4))=+/-sqrt(((k^2)/16)-1)
x=(k/4)+/-sqrt(((k^2)/16)-1)

Reply 16

Bananasplitxxx

Original post by Mehru1214

Ohh thanks I finally get it now :wink:

Reply 17

peeked

Original post by thekidwhogames

2x^2 - kx + 2 = 0

Equal roots hence b^2-4ac=0

k^2 - 4(2)(2) = 0 --> k^2 - 16 = 0 --> what is k (finish it off)

Wait so with a quadratic with equal roots will the equation b^2-4ac=0 always be correct?
Because in this situation does it make x -4 or 4?
I never done 'equal roots' problems before but does 'equal roots' just mean that the two x values are the same number (so like -4 and 4)

Is it k=+-4 then?

Original post by peeked

If a quadratic has equal roots thn b^2-4ac=0 is always true e.g. (x-2)^2
If a quadratic has 2 real distinct roots then b^4-4ac>0 and no roots b^2-4ac<0

Think about it. If the thing under the root, the discriminant, is 0 then x=-b+-0/2a --> x=-b/2a (one equal root).

If discriminant > 0 then you have 2 values for the root and hence 2 solutions. If <0, no real root hence no real solutoin.