I was solving some questions related to Quadratic Equations from my book, but I was unable to solve this question. Any help will be greatly appreciated!
Question
For all real values of x,
x2+4x+3x2+2x+p can take real value if:
(A)0<p<2
(B)0≤p≤1
(C)−1<p<1
(D)−3≤p≤1
Answer given in answer keys- They don't give solutions:(
(D)−3≤p≤1
#Algebra
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Comments
@Finnley Paolella, @Mahdi Raza, @David Vreken, @Pi Han Goh, @Páll Márton, please help me!
It looks like something is wrong with either the question or answer. Let's say that D is the answer. That means p could be 0, and according to the question, f(x)=x2+4x+3x2+2x+0=(x+1)(x+3)x(x+2) is a real value for all real values x. However, this is not true for x=−1 and x=−3 because f(−1)=0−1 and f(−3)=03, neither of which are real numbers.
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I was also feeling like something is wrong. I asked someone, and they calculated that none of the options is true, instead, the answer is p∈(−∞,−3)∪(1,∞)
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Hmmm... Your code is wrong :)
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David Vreken is right. Test by plugging in some arbitrary values for p, say p=524364536543543543. You will still realize that the fraction x2+4x+3x2+2x+p does not always take real values. So the interval you've found/given/told is still wrong.
Yeah! I can't understand the 0 denominator too! The problem is wrong. I think the solution is (−∞;−3)and(−3;−1)and(−1;∞)
I agree that answer given is incorrect. Plus, I don't think so 0 is only one possible value of p. But, If I am not mistaken, you took p=0 to prove something is flawed, right?
But, what I did not understand was when he graphed the equation, D came out to be the answer! I don't know how??!!
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Can you take a photo about the solution? Or the solution isn't in english?
"D" does not came out to be the answer. If it does, you should expect to see that it's continuous everywhere, which is obviously not true at the points x=−1,−3.
Let y=x2+4x+3x2+2x+p Then yx2+4xy+3y=x2+2x+p yx2−x2+4xy−2x+3y−p=0 ⟹(y−1)x2+(4y−2)x+3y−p=0 Since x is real, D≥0 (2y−1)2−(3y−p)(y−1)≥0 Solving, y2+(p−1)y−p+1≥0 Since y is also real, here's D is also ≥0 (p−1)2+4p−4≥0 Solving, (p−1)(p+3)≥0 ∴p∈(−∞,−3)∪(1,∞)
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This is confusing! You should start with x2+4x+3=0
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But the task says all values!!! So the task is wrong
The first to second step needs to include the condition “for x=−3 and x=−1”. This condition is never dealt with.
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I agree. If x=−1 or x=−3, then you're basically multiplying the equation by 0, which makes the entire working pointless.
The problem asks for what values will the solutions be real. You did the right thing by finding the discrimnant but all you have to do is notice that
If(p−1)(p+3)isthedisrcrimnantforanyvaluebetween−3:(−3+3)(−3−1)=0(AsyousaidwhenD>0orD=0Therearerealsolutions)orbetween1:(1−1)(1+3)=0Youwillseethattheybothworkandcanconcludethatitworksforthatrange.
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Ok??? Use \ ( \ ) :)
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These kind of problems are in H.S.Hall and R.S.Knight's book and i dont like this method employed:/
When can a fraction give a non-real number, if the numerator and the denominator are real numbers?
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00.
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00 is undefinied. This can be 1 or 0.
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0x? if you say a value k for instance either 1 or 0, then for every x, the values are equal. For example:
First, i ask what is04=k=09⟹4=9?
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−0n=−∞;+0n=+∞ link to 00
I learned like this at school:Log in to reply
For @David Vreken, @Mahdi Raza, @Páll Márton
I am writing what he wrote in my notebook, I don't fully understand, please wait! Thanks!
However, the graph shows:
The equation really has all values of y in the range given in option D!!!!!!
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That's not correct. When x=−1 or when x=−3, the value of y is undefined.
Even at p=0, y has all real values!
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The question is not asking if the equation has all values of y for a given range of p, it’s asking if there is a real value of y for all values of x for a given range of p. Unfortunately, for any value of p, there is at least one vertical asymptote (either at x=−3 or x=−1), so it’s not true for any given range or value of p.
Also, something which I was thinking before I gave up:
x2+2x+p, needs to be real, so its discriminant is ≥0
(2)2−4p≥0 ⟹p≤1
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Yes, this is the basic solution. But you should give the intervals, where the fraction isn't real
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I have written the exact same words as in my book, I haven't changed even a single word of the question or options. So, if it is wrong, the book is wrong!
This doesn't make sense. If x2+2x+p is real and x is real, then trivially, p must be real as well. We can't have real number+another real number+some non-real number=some real number.
See my problem here!
So, from all the comments, I conclude that my book is wrong and the question was flawed. Thanks a lot @David Vreken, @Mahdi Raza, @Páll Márton for helping me!
I don't see how this is possible. If the ratio of two polynomials with real coefficients is always real, then the denominator of this fraction must be non-zero for all x. Which means x2+4x+3=0 for all real x. But this is obviously false because x=−1,−3 makes the expression (the ratio of the two polynomials) undefined.
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Yes, thanks for replying. I am thinking of contacting the book authors to know what they thought while making the question, to know if there is some misprint. Thanks a lot @Pi Han Goh!
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No problem! =D
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@Pi Han Goh Now please post the solution of this Problem
I think may be, now you have deleted your solution, so wouldn't get another chance to upload solution.