Help!- Quadratic Equations

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 xx,

x2+2x+px2+4x+3\dfrac{x^2+2x+p}{x^2+4x+3} can take real value if:

(A)0<p<2\text{(A)} 0<p<2

(B)0p1\text{(B)} 0\leq p\leq 1

(C)1<p<1\text{(C)} -1<p<1

(D)3p1\text{(D)} -3\leq p\leq 1

Answer given in answer keys- They don't give solutions:(

(D)3p1\text{(D)} -3\leq p\leq 1

Note by Vinayak Srivastava
4 months, 3 weeks ago

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Let y=x2+2x+px2+4x+3y=\dfrac{x^2+2x+p}{x^2+4x+3} Then yx2+4xy+3y=x2+2x+pyx^2+4xy+3y=x^2+2x+p yx2x2+4xy2x+3yp=0yx^2-x^2+4xy-2x+3y-p=0     (y1)x2+(4y2)x+3yp=0\implies (y-1)x^2+ (4y-2)x +3y-p=0 Since xx is real, D0D\geq0 (2y1)2(3yp)(y1)0(2y-1)^2-(3y-p)(y-1) \geq 0 Solving, y2+(p1)yp+10y^2+(p-1)y-p+1 \geq 0 Since yy is also real, here's DD is also 0\geq 0 (p1)2+4p40(p-1)^2 +4p-4 \geq 0 Solving, (p1)(p+3)0(p-1)(p+3)\geq 0 p(,3)(1,) \therefore \boxed{p \in (-\infty,-3) \cup (1,\infty)}

Vinayak Srivastava - 4 months, 3 weeks ago

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This is confusing! You should start with x2+4x+30x^2+4x+3\neq 0

Páll Márton (no activity) - 4 months, 3 weeks ago

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But the task says all values!!! So the task is wrong

Páll Márton (no activity) - 4 months, 3 weeks ago

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The first to second step needs to include the condition “for x3x \neq -3 and x1x \neq -1”. This condition is never dealt with.

David Vreken - 4 months, 3 weeks ago

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I agree. If x=1x=-1 or x=3x=-3, then you're basically multiplying the equation by 0, which makes the entire working pointless.

Pi Han Goh - 4 months, 3 weeks ago

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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(p1)(p+3)isthedisrcrimnantforanyvaluebetween3:(3+3)(31)=0(AsyousaidwhenD>0orD=0Therearerealsolutions)orbetween1:(11)(1+3)=0Youwillseethattheybothworkandcanconcludethatitworksforthatrange.If\quad (p-1)(p+3)\quad is\quad the\quad disrcrimnant\quad for\quad any\quad value\quad between\quad -3:\quad (-3+3)(-3-1)=0\quad (As\quad you\quad said\quad when\quad D>0\quad or\quad D=0\quad There\quad are\quad real\quad solutions)\quad or\quad between\quad 1:\quad (1-1)(1+3)=0\\ You\quad will\quad see\quad that\quad they\quad both\quad work\quad and\quad can\quad conclude\quad that\quad it\quad works\quad for\quad that\quad range.

Joshua Olayanju - 4 months, 3 weeks ago

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Ok??? Use \ ( \ ) :)

Páll Márton (no activity) - 4 months, 3 weeks ago

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@Páll Márton (no activity) :) is only a happppppy bracket

Páll Márton (no activity) - 4 months, 3 weeks ago

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@Páll Márton (no activity) Thanks

Joshua Olayanju - 4 months, 3 weeks ago

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It looks like something is wrong with either the question or answer. Let's say that D is the answer. That means pp could be 00, and according to the question, f(x)=x2+2x+0x2+4x+3=x(x+2)(x+1)(x+3)f(x) = \frac{x^2 + 2x + 0}{x^2 + 4x + 3} = \frac{x(x + 2)}{(x + 1)(x + 3)} is a real value for all real values xx. However, this is not true for x=1x = -1 and x=3x = -3 because f(1)=10f(-1) = \frac{-1}{0} and f(3)=30f(-3) = \frac{3}{0}, neither of which are real numbers.

David Vreken - 4 months, 3 weeks ago

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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,)p \in (-\infty,-3) \cup (1,\infty)

Vinayak Srivastava - 4 months, 3 weeks ago

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Hmmm... Your code is wrong :)

Páll Márton (no activity) - 4 months, 3 weeks ago

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@Páll Márton (no activity) Was wrong :)

Páll Márton (no activity) - 4 months, 3 weeks ago

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David Vreken is right. Test by plugging in some arbitrary values for pp, say p=524364536543543543p=524364536543543543. You will still realize that the fraction x2+2x+px2+4x+3\frac{x^2+2x+p}{x^2+4x+3} does not always take real values. So the interval you've found/given/told is still wrong.

Pi Han Goh - 4 months, 3 weeks ago

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Yeah! I can't understand the 0 denominator too! The problem is wrong. I think the solution is (;3)and(3;1)and(1;)(-\infty;-3)and(-3;-1)and(-1;\infty)

Páll Márton (no activity) - 4 months, 3 weeks ago

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I agree that answer given is incorrect. Plus, I don't think so 0 is only one possible value of pp. But, If I am not mistaken, you took p=0p = 0 to prove something is flawed, right?

Mahdi Raza - 4 months, 3 weeks ago

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But, what I did not understand was when he graphed the equation, D came out to be the answer! I don't know how??!!

Vinayak Srivastava - 4 months, 3 weeks ago

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Can you take a photo about the solution? Or the solution isn't in english?

Páll Márton (no activity) - 4 months, 3 weeks ago

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"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=1,3x=-1, -3.

Pi Han Goh - 4 months, 3 weeks ago

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When can a fraction give a non-real number, if the numerator and the denominator are real numbers?

Páll Márton (no activity) - 4 months, 3 weeks ago

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00 \frac 00 .

Pi Han Goh - 4 months, 3 weeks ago

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00\cfrac{0}{0} is undefinied. This can be 1 or 0.

Páll Márton (no activity) - 4 months, 3 weeks ago

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@Páll Márton (no activity) First, i ask what is x0\frac{x}{0}? if you say a value kk for instance either 1 or 0, then for every xx, the values are equal. For example:

40=k=90    4=9?\dfrac{4}{0} = k = \dfrac{9}{0} \implies \boxed{4 = 9}?

Mahdi Raza - 4 months, 3 weeks ago

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@Mahdi Raza I learned like this at school: n0=;n+0=+\cfrac{n}{-0}=-\infty;\cfrac{n}{+0}=+\infty link to 00\cfrac{0}{0}

Páll Márton (no activity) - 4 months, 3 weeks ago

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@Páll Márton (no activity) That involves limits I guess. But there isn't a perfect definition for it, that's why it is undefined. You can get a very close approximation but the value of it might not be exact, you place a hollow circle on the graph for it. I think this where calculus comes in with determinate and indeterminate forms.

Mahdi Raza - 4 months, 3 weeks ago

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For @David Vreken, @Mahdi Raza, @Páll Márton

I am writing what he wrote in my notebook, I don't fully understand, please wait! Thanks!

Vinayak Srivastava - 4 months, 3 weeks ago

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However, the graph shows:

The equation really has all values of yy in the range given in option DD!!!!!!

Vinayak Srivastava - 4 months, 3 weeks ago

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That's not correct. When x=1x = -1 or when x=3x=-3, the value of yy is undefined.

Pi Han Goh - 4 months, 3 weeks ago

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Even at p=0, y has all real values!

Vinayak Srivastava - 4 months, 3 weeks ago

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The question is not asking if the equation has all values of yy for a given range of pp, it’s asking if there is a real value of yy for all values of xx for a given range of pp. Unfortunately, for any value of pp, there is at least one vertical asymptote (either at x=3x = -3 or x=1x = -1), so it’s not true for any given range or value of pp.

David Vreken - 4 months, 3 weeks ago

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Also, something which I was thinking before I gave up:

x2+2x+p,x^2+2x+p, needs to be real, so its discriminant is 0\geq 0

(2)24p0(2)^2-4p \geq 0     p1\implies p\leq1

Vinayak Srivastava - 4 months, 3 weeks ago

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Yes, this is the basic solution. But you should give the intervals, where the fraction isn't real

Páll Márton (no activity) - 4 months, 3 weeks ago

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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!

Vinayak Srivastava - 4 months, 3 weeks ago

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This doesn't make sense. If x2+2x+px^2 + 2x + p is real and xx is real, then trivially, pp must be real as well. We can't have real number+another real number+some non-real number=some real number\text{real number} + \text{another real number} + \text{some non-real number} = \text{some real number} .

Pi Han Goh - 4 months, 3 weeks ago

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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!

Vinayak Srivastava - 4 months, 3 weeks ago

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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 xx. Which means x2+4x+30x^2 + 4x + 3 \ne 0 for all real xx. But this is obviously false because x=1,3x=-1,-3 makes the expression (the ratio of the two polynomials) undefined.

Pi Han Goh - 4 months, 3 weeks ago

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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!

Vinayak Srivastava - 4 months, 3 weeks ago

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No problem! =D

Pi Han Goh - 4 months, 3 weeks ago

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@Pi Han Goh @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.

A Former Brilliant Member - 4 months, 3 weeks ago

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