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When \(a≠0\), how many solutions does the equation \(a(x+b)^2+c=0\) have? 1) \(bc=0\) 2) \(b+c=0\)
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Re M6012
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11 Jun 2018, 06:12
Official Solution:Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution. The first step of the VA (Variable Approach) method is to modify the original condition and the question, and then recheck the question. There are three cases to consider: Case 1: \(a > 0\), \(c > 0\) or \(a < 0\), \(c < 0\) The equation has no roots. Case 2: \(c = 0\). The equation has only one root. Case 3: \(a > 0\), \(c < 0\) or \(a < 0\), \(c > 0\) The equation has two roots. Condition 1): If \(bc = 0\), then when \(a = 1\), \(b = 0\), \(c = 1\), the equation has two roots, and when \(a = 1\), \(b = 0\), \(c = 0\), the equation has one root. As the question does not have a unique answer, condition 1) is not sufficient. Condition 2) \(b + c = 0\) ⇔ \(b = c = 0\). Since \(c = 0\), the equation has only one root. Condition 2) is sufficient. Therefore, the answer is B. Answer: B
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Re: M6012
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27 Jul 2018, 07:29
Not sure if I get the break down of the question. Can you please explain how you broke it down in a different way?



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Re M6012
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03 Oct 2018, 05:24
I think this the explanation isn't clear enough, please elaborate. Can some one please explain how is statement 1 manipulated into the 3 cases. I am not able to follow how the three cases have been formed and why they have been formed?
Thanks in advance!



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souvik101990 wrote: Official Solution:
Question prompt :
When a≠0, how many solutions does the equation \(a(x+b)^2+c=0\) have?
1) bc=0
2) b+c=0
> Solving \(a(x+b)^2+c=0\) gives
\((x+b)^2 = (c/a)\)
Now lets discuss each of the 3 cases one by one :
There are three cases to consider:
Case 1: \(a > 0\), \(c > 0\) or \(a < 0\), \(c < 0\)
The equation has no roots.
\((x+b)^2 = (c/a)\) > LHS is positive number.
So we need RHS also positive.
\(a > 0\), \(c > 0\) > (c/a) negative number.
\(a < 0\), \(c < 0\) > (c/a) negative number.
So LHS can not be equal to RHS. No roots.
Case 2: \(c = 0\).
The equation has only one root.
\((x+b)^2 = (c/a)\)
c= 0 then \((x+b)^2 = 0\)
This gives one root x = b.
Case 3: \(a > 0\), \(c < 0\) or \(a < 0\), \(c > 0\)
The equation has two roots.
\((x+b)^2 = (c/a)\) > LHS is positive number.
So we need RHS also positive.
\(a > 0\), \(c < 0\) > (c/a) positive number. or
\(a < 0\), \(c > 0\) > (c/a) positive number.
\((x+b)^2 = (c/a) = K\)
So roots will be \(x = b +/ \sqrt{k}\)
Condition 1):
If \(bc = 0\), then when
\(a = 1\), \(b = 0\), \(c = 1\), the equation has two roots, and when
\(a = 1\), \(b = 0\), \(c = 0\), the equation has one root.
As the question does not have a unique answer, condition 1) is not sufficient.
Condition 2)
\(b + c = 0\) ⇔ \(b = c = 0\).
Since \(c = 0\), the equation has only one root.
Condition 2) is sufficient.
Therefore, the answer is B.
Answer: B ridhi9jainI have tried to breakdown the OE 3 cases... Hope that helps!!
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Re: M6012
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03 Oct 2018, 09:43
ridhi9jain wrote: I think this the explanation isn't clear enough, please elaborate. Can some one please explain how is statement 1 manipulated into the 3 cases. I am not able to follow how the three cases have been formed and why they have been formed?
Thanks in advance! souvik101990 wrote: Official Solution:
Question prompt :
When a≠0, how many solutions does the equation \(a(x+b)^2+c=0\) have?
1) bc=0
2) b+c=0
> Solving \(a(x+b)^2+c=0\) gives
\((x+b)^2 = (c/a)\)
Now lets discuss each of the 3 cases one by one :
There are three cases to consider:
Case 1: \(a > 0\), \(c > 0\) or \(a < 0\), \(c < 0\)
The equation has no roots.
\((x+b)^2 = (c/a)\) > LHS is positive number.
So we need RHS also positive.
\(a > 0\), \(c > 0\) > (c/a) negative number.
\(a < 0\), \(c < 0\) > (c/a) negative number.
So LHS can not be equal to RHS. No roots.
Case 2: \(c = 0\).
The equation has only one root.
\((x+b)^2 = (c/a)\)
c= 0 then \((x+b)^2 = 0\)
This gives one root x = b.
Case 3: \(a > 0\), \(c < 0\) or \(a < 0\), \(c > 0\)
The equation has two roots.
\((x+b)^2 = (c/a)\) > LHS is positive number.
So we need RHS also positive.
\(a > 0\), \(c < 0\) > (c/a) positive number. or
\(a < 0\), \(c > 0\) > (c/a) positive number.
\((x+b)^2 = (c/a) = K\)
So roots will be \(x = b +/ \sqrt{k}\)
Condition 1):
If \(bc = 0\), then when
\(a = 1\), \(b = 0\), \(c = 1\), the equation has two roots, and when
\(a = 1\), \(b = 0\), \(c = 0\), the equation has one root.
As the question does not have a unique answer, condition 1) is not sufficient.
Condition 2)
\(b + c = 0\) ⇔ \(b = c = 0\).
Since \(c = 0\), the equation has only one root.
Condition 2) is sufficient.
Therefore, the answer is B.
Answer: B ridhi9jainI have tried to breakdown the OE 3 cases... Hope that helps!!
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Re: M6012
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03 Oct 2018, 09:51
Paugustin90 wrote: Not sure if I get the break down of the question. Can you please explain how you broke it down in a different way? Paugustin90Question prompt : When a≠0, how many solutions does the equation \(a(x+b)^2+c=0\) have? 1) bc=0 2) b+c=0 St 1 : bc = 0 so b= 0 or c= 0 or both b and c = 0.
c=0 gives \(a(x+b)^2+0=0\) > x = b
Both b= 0 and c= 0 gives \(a(x+0)^2+0=0\) > x =0
2 different answers. We can eliminate st 1. Option A and D are gone.
For sack of completeness of discussion when b =0, it gives\(x^2 = (c/a)\)  So 2 roots or no roots. (As discussed in my earlier post above).
St 2: b+c=0
This gives b = 0 and c = 0
So \(a(x+0)^2+0=0\) > x =0
Only one possible values for x.
Sufficient.
Ans : B Hope this helps!!
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Re: M6012
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26 Mar 2019, 18:51
the equation is ax^2 + 2abx +(ab^2+c) the roots are [(b + sqrt(b^24ac)]/2a
solving this we get the two roots as c1 (c+1)
so if c =0 then we can get one root.










