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This an example problem in Hall&Knight from the chapter the theory of Quadratic equation Under Art 121.

The question is,

Find the Limits between which 'a' must lie in order that
$$\dfrac{ax^2-7x+5}{5x^2-7x+a}$$ may be capable of all values,x being real quantity.

A) I am not able to understand the meaning of Equations "may be of all values",
I assume the meaning as
"what is the limit of 'a' so that the two equations can have real and positive roots ?"

Answer (given in the book):
Put,
$$\frac{ax^2-7x+5} {5x^2-7x+a}=y;$$ $$(a-5y)x^2-7x(1-y)+(5-ay)=0$$ using $b^2-4ac$ to check the nature of the roots we get, $$49(1-y)^2-4(a-5y)(5-ay)$$ must be positive,(makes sense), On simplifying we get,
$$(49-20a)y^2 +2(2a^2+1)y+(49-20a)$$ must be positive, (for this quadratic equation we again check the nature of the roots) We get,
$$(2a^2+1)^2 - (49-20a)^2$$must be negative or zero ,(B) this part I don't get,Why must this equation be negative or zero?),

Now , $(2a^2+1)^2 - (49-20a)^2$ is negative or zero,
according as $2(a^2-10a+25)*2(a^2+10a-24)$ is negative or zero:(C) I don't get this part also)

That is according as $4*(a-5)^2(a+12)(a-2)$ is negative or zero .

Finally, This expression is negative as long as a lies between 2 and -12 and for such values $49-2a$ is positive; the expression is zero when a=5,-12 or 2, but $49-20a$ is negative when a = 5 . Hence the limiting values are 2 and -12 and a may have any intermediate values.

I have three doubts A, B, C, If possible please explain the problem on the whole and be patient with me if the questions are silly, I am an amateur Learner.

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  • $\begingroup$ The wording is very odd, but I would guess, and this is a very different thing from what you guessed at (A), that "capable of all values" means that, together, both quadratics have $\;\Bbb R\;$ as all their image. This already would mean that it must be $\;a<0\;$, for example...But this is only if my guess is correct. It is, imo, a very poor, lousy way to word things in that book. $\endgroup$
    – DonAntonio
    Commented Aug 31, 2018 at 21:08
  • $\begingroup$ @DravidHemanth I fixed the question so that it shows the fraction, rather than two separate quadratics. Also, (B) and (C) are answered in the text right before the Example. $\endgroup$
    – dxiv
    Commented Aug 31, 2018 at 21:33
  • $\begingroup$ @DonAntonio The wording is very odd The book was first published in $\,1887\,$ ;-) $\endgroup$
    – dxiv
    Commented Aug 31, 2018 at 21:37
  • $\begingroup$ @DonAntonio Thank you ! $\endgroup$
    – Dravid Hemanth
    Commented Sep 1, 2018 at 6:58
  • $\begingroup$ @dxiv Yes I got that ,so it is that any quadratic equation will be positive if $b^2-4ac$ is negative or zero .That is why $(2a^2+1)2−(49−20a)2$ should be negative or zero so that $(49−20a)y^2+2(2a^2+1)y+(49−20a)$ will be positive.And thanks for the editing ,it was actually confusing as to why those two separate equations were put as a fraction and were equated with y $\endgroup$
    – Dravid Hemanth
    Commented Sep 1, 2018 at 7:03

1 Answer 1

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$A)$ They mean the fraction $f(x)=\dfrac{ax^2-7x+5}{5x^2-7x+a}$ can attain any real value.

$B)$ A quadratic polynomial has a constant sign or is $0$ if and only if its discriminant is non-positive.

$C)$ They use the factorisation $\; X^2-Y^2=(X-Y)(X+Y)$.

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