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February 22nd, 2014, 02:11 AM   #1
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quadratic equation roots

The roots of a quadratic equation h(x)=0 are given by

x = -5+-sqr root 25 - 3k / 6
For which value of k will one of the roots be 0? Now write down the value of the other root
According to marking memoranda, the answer is sqr 25 - 3k =5
25 - 3k = 5
k = 0
I don't understand why ws sqr root of 25 - 3k was set equal to b i.e 5
after finding value of k I can substitute to get the other value.
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February 22nd, 2014, 02:20 AM   #2
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Re: quadratic equation roots

If k = 0, what are the two roots?
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February 22nd, 2014, 04:14 AM   #3
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Re: quadratic equation roots

Quote:
Originally Posted by bongantedd
The roots of a quadratic equation h(x)=0 are given by

x = -5+-sqr root 25 - 3k / 6
It would help if you wrote this clearly! What you wrote is -5+ 5- k2= -k/2 and -5- 5- k/2= -10- k/2.
But I am sure what you meant was (-5+- sqr root(25- 3k))/6. What a difference parentheses make!

Quote:
For which value of k will one of the roots be 0? Now write down the value of the other root
According to marking memoranda, the answer is sqr 25 - 3k =5
25 - 3k = 5
This does NOT follow from the above! if sqrt(25- 3k)= 5 then 25- 3k= 25.

Quote:
k = 0
I don't understand why ws sqr root of 25 - 3k was set equal to b i.e 5
after finding value of k I can substitute to get the other value.
Well, set the root equal to 0: so


Now square both sides: and solve for k.
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February 22nd, 2014, 07:10 AM   #4
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Assuming x = (-5 ?(25 - 3k))/6 was meant, the equation was equivalent to 3x + 5x + k/4 = 0.
If k = 0, that equation becomes 3x + 5x = 0, which factorizes as x(3x + 5) = 0, and so has roots 0 and -5/3.
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