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 August 29th, 2019, 04:51 AM #1 Newbie   Joined: Aug 2019 From: India Posts: 15 Thanks: 0 How to prove that a rational algebraic expression is capable of all values? Well, I want to express myself using an example from Higher Algebra by Hall and Knight . Find the limits between which a must lie in order that $\displaystyle \frac{ax^2 - 7x +5} {5x^2-7x+a}$ may be capable of all values, x being any real quantity. Solution:- Put $\displaystyle \frac{ax^2-7x+5} {5x^2-7x+a} = y$ then $\displaystyle (a-5y)x^2 - 7x(1-y)+(5-ay) = 0$ In order that the values of x found from this quadratic may be real the expression $\displaystyle 49(1-y)^2 - 4(a-5y)(5-ay)$ must be positive, that is, $\displaystyle (49-20a)y^2 + 2(2a^2+1)y +(49-20a)$ must be positive; hence $\displaystyle (2a^2 +1)^2 - (49-20a)^2$must be negative or zero, and $\displaystyle 49-20a$ must be be positive. Now $\displaystyle (2a^2 +1)^2 - (49-20a)^2$ is negative or zero, according as $\displaystyle 2(a^2-10a+25)\times 2(a^2+10a-24)$ is negative or zero that is, according as $\displaystyle 4(a-5)^2(a+12)(a-2)$ is negative or zero. This expression is negative as long as a lies between $\displaystyle 2$ and $\displaystyle -12$ and for such values $\displaystyle 49-20a$ is positive; the expression is zero when $\displaystyle a=5 , -12 , or 2$ but $\displaystyle 49-20a$ is negative when $\displaystyle a=5$. Hence the limiting values are 2 and -12 , and a may have any intermediate value. I have lot of problems in this, so first question is :- How does that prove that $\displaystyle \frac{ax^2 -7x +5} {5x^2-7x+a}$ is capable of all values? We just have proved that x will be real between those limiting values? Now, when we assumed $\displaystyle y = \frac{ax^2 - 7x + 5} {5x^2 - 7x +a}$ and we multiplied y with $\displaystyle 5x^2 - 7x + a$ it is pretty obvious thing that when you multiply a fraction by its denominator you will certainly get numerator, and then we subtracted $\displaystyle y\times (5x^2-7x+a)$ from $\displaystyle ax^2 - 7x +5$ which must be equal to zero as we are subtracting not two equal things but two same things. Then, we are treating the expression $\displaystyle (a-5y)x^2 - 7x(1-y) + (5-ay) = 0$ a quadratic but I want say that y is concealing an expression and when you put it in you will get zero so this is not a equation involving two unknowns. How can you go on solving it? I hope that experienced people over here will understand my problem and will help me. Thank You
August 29th, 2019, 07:09 AM   #2
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 Originally Posted by Adesh Mishra We just have proved that x will be real between those limiting values?[
No. The limiting values were for $a$, not $x$.

 August 29th, 2019, 07:11 AM #3 Newbie   Joined: Aug 2019 From: India Posts: 15 Thanks: 0 I was saying that x will be real between those limiting values of a. Last edited by skipjack; August 29th, 2019 at 07:56 AM.
 August 29th, 2019, 08:27 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,972 Thanks: 2222 The aim was to show that the rational expression, i.e. $y$, which is a function of $x$ and $a$, where $x$ is real, can have any real value if $a$ satisfies certain conditions (to be found). That means that if $a$ has some value that satisfies those conditions, you could choose any real value, and it would then be possible (by solving a quadratic equation) to find a real value of $x$ such that $y$ equals the value you chose. Once you've chosen values for $a$ and $y$, the quadratic equation for $x$ will have real coefficients that you can find easily, as they're already known in terms of $a$ and $y$. Thanks from Adesh Mishra
 August 29th, 2019, 08:55 AM #5 Newbie   Joined: Aug 2019 From: India Posts: 15 Thanks: 0 Okay. Means for any value of y I can find a real x. Wow! Thank you Sir your understanding is great because it was very hard to understand what was going on in the mind of Mr. Hall and Mr. Knight. Thank you.

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