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 February 22nd, 2014, 03:11 AM #1 Member   Joined: Sep 2013 Posts: 75 Thanks: 0 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.
 February 22nd, 2014, 03:20 AM #2 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,312 Thanks: 115 Re: quadratic equation roots If k = 0, what are the two roots?
February 22nd, 2014, 05:14 AM   #3
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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: $\frac{5\pm\sqrt{25- 3k}}{6}= 0$ so $5\pm\sqrt{25- 3k}= 0$
$5= \pm\sqrt{25- 3k}$

Now square both sides: $25= 25- 3k$ and solve for k.

 February 22nd, 2014, 08:10 AM #4 Global Moderator   Joined: Dec 2006 Posts: 18,241 Thanks: 1438 Assuming x = (-5 ± ?(25 - 3k))/6 was meant, the equation was equivalent to 3x² + 5x + k/4 = 0. If k = 0, that equation becomes 3x² + 5x = 0, which factorizes as x(3x + 5) = 0, and so has roots 0 and -5/3.

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