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# Maths question watch

Find the value of k such that the equation 2x^2−kx+2=0
(The x is a letter not multiplication)
2. Have you tried completing the square?
3. Can you do that with the k?
4. Anything else that goes with the question?
5. (Original post by akay8)
Anything else that goes with the question?
6. I got x=(k/4)+/-sqrt((k^2/16)-1)
7. 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)
8. What level is this, GCSE Higher?
9. (Original post by MWills99)
I got x=(k/4)+/-sqrt((k^2/16)-1)
How did you get that ?
10. (Original post by peeked)
What level is this, GCSE Higher?
It’s national 5, I think it’s the same as GCSE though
11. 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.
12. (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
13. (Original post by Bananasplitxxx)
Ohh is it this formula
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.
14. (Original post by Bananasplitxxx)
Ohh is it this formula
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.
15. (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!
16. (Original post by Bananasplitxxx)
How did you get that ?
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)
17. (Original post by Mehru1214)
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
Ohh thanks I finally get it now
18. (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)
19. Is it k=+-4 then?
20. (Original post by peeked)
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)

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.

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Updated: February 27, 2018
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