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February 22nd, 2014, 02:11 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, 02:20 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, 04:14 AM  #3  
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 5  Re: quadratic equation roots Quote:
But I am sure what you meant was (5+ sqr root(25 3k))/6. What a difference parentheses make! Quote:
Quote:
Now square both sides: and solve for k.  
February 22nd, 2014, 07:10 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 17,729 Thanks: 1360 
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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