# Urgent maths help!!!!

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#1
Guys plz help with math question:

f(x)=x^2-(k+8)x+(8k+1)

b) If the equation f(x) = 0 has two equal roots, find possible values of k
c)Show that k=8 , f(x) >0 for all values of x
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3 years ago
#2
which part are you having trouble with
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#3
(Original post by mullac)
which part are you having trouble with
for part a i got kx^2+64x^2-32k-4 but it seems a bit off to find values of k since it has 2 unknown values
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3 years ago
#4
(Original post by Yr_11_MATHS)
for part a i got kx^2+64x^2-32k-4 but it seems a bit off to find values of k since it has 2 unknown values
the discriminant involves only the coeficients not x itself. E.g for ax^2 + bx + c the discriminant is b^2 - 4ac. In your answer you have included both the coefficients and x so you have 2 unkowns.
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#5
(Original post by mullac)
the discriminant involves only the coeficients not x itself. E.g for ax^2 + bx + c the discriminant is b^2 - 4ac. In your answer you have included both the coefficients and x so you have 2 unkowns.
for part b i need to find K how would this be done?????
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3 years ago
#6
(Original post by Yr_11_MATHS)
for part b i need to find K how would this be done?????
the discriminant comes from the quadratic formula which im guessing you know - x = (-b+/-(b2-4ac)1/2)/2 (excuse the formatting) - for the quadratic to have two roots b2-4ac >0 so that the plus or minus has an effect. So you need to take the discriminant you found in part a and set it to >0 and then solve the inequality for a range of values for k
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#7
Yes but for part b of the question k needs to be found wtf do I do?
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3 years ago
#8
(Original post by mullac)
the discriminant comes from the quadratic formula which im guessing you know - x = (-b+/-(b2-4ac)1/2)/2 (excuse the formatting) - for the quadratic to have two roots b2-4ac >0 so that the plus or minus has an effect. So you need to take the discriminant you found in part a and set it to >0 and then solve the inequality for a range of values for k
Yes, but for two equal roots the discriminant = 0 and you'll get a single value, not a range? If its a quadratic in k, you may get two values.
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3 years ago
#9
(Original post by Yr_11_MATHS)
Yes but for part b of the question k needs to be found wtf do I do?
i forgot how to do part c but for b you expand and simplify the factorise (I think)
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3 years ago
#10
(Original post by mqb2766)
Yes, but for two equal roots the discriminant = 0 and you'll get a single value, not a range? If its a quadratic in k, you may get two values.
you are right I misread the question
1
3 years ago
#11
(Original post by Yr_11_MATHS)
Yes but for part b of the question k needs to be found wtf do I do?
as mbq said for two equal roots the discriminant (the expression you derived in terms of k for part a) is equal to zero. so set your answer for part a equal to 0 and solve the quadratic for k
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#12
(Original post by mullac)
as mbq said for two equal roots the discriminant (the expression you derived in terms of k for part a) is equal to zero. so set your answer for part a equal to 0 and solve the quadratic for k
oh f*ck im so stupid... why did i include the Xs in my a b and c... I get what im doing now... ill give you rep
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3 years ago
#13
(Original post by mullac)
you are right I misread the question
I got k=10 or k=6?
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3 years ago
#14
Deleted
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#15
(Original post by BrandonS03)
I got k=10 or k=6?
You're right what about c tho
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3 years ago
#16
(Original post by Yr_11_MATHS)
You're right what about c tho
substitute k = 8 and complete the square.
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3 years ago
#17
(Original post by mqb2766)
substitute k = 8 and complete the square.
u sure since u are meant to show that k is 8
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#18
(Original post by mqb2766)
substitute k = 8 and complete the square.
9 and 7?
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#19
(Original post by Rolls_Reus_0wner)
u sure since u are meant to show that k is 8
it says that in the question....
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3 years ago
#20
(Original post by Rolls_Reus_0wner)
u sure since u are meant to show that k is 8
(Original post by Yr_11_MATHS)
9 and 7?
Rolls, not 100% because the language isn't the original question but it is my best guess.

Y11, not sure what those numbers are. Completing the square is
https://www.mathsisfun.com/algebra/c...ng-square.html
so you end up with
(x+a)^2 + b
if b>0, the quadratic is always positive.
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