relationship between coefficients of a quadratic equations

**swordfish774** · June 29th 2012, 12:33 PM

The condition that a root of the equation

may be reciprocal to a root of

is

Possible Answers

	Possible Answer

**Soroban** · June 29th 2012, 04:24 PM

Hello, swordfish774!

$\text{The condition that a root of the equation }ax^2+bx+c \:=\:0$
$\text{may be reciprocal to a root of }a_1x^2 + b_1x + c_1 \:=\:0\,\text{ is:}$

. . $\begin{array}{|c|c|}\hline 1 & (b_1c-a_1b)^2 \:=\:(ac_1-b_1c)(ab_1 - bc_1) \\ \hline 2 & (bb_1-aa_1)^2 \:=\:(ab_1-bc_1)(a_1b-b_1c) \\ \hline 3 & (cc_1 - aa_1)^2 \:=\:(ab_1 - bc_1)(a_1b - b_1c) \\ \hline 4 & a + b + c \:=\:0 \\ \hline \end{array}$ . . Are they kidding?

$\text{If }r\text{ is a root of }\,ax^2 + bx+c,\,\text{ then: }\:r \;=\;\frac{-b \pm\sqrt{b^2-4ac}}{2a}$ .[1]

$\text{If }\tfrac{1}{r}\text{ is a root of }\,a_1x^2 + b_1x + c_1 \:=\:0,\,\text{ we have: }\:\frac{a_1}{r^2} + \frac{b_1}{r} + c_1 \:=\:0$

. . $c_1r^2 + b_1r + a_1 \:=\:0 \quad\Rightarrow\quad r \;=\;\frac{-b_1 \pm\sqrt{b_1^2 - 4a_1c _1}}{2c_1}$ .[2]

Equating [1] and [2], we have: . $\begin{Bmatrix} a \:=\:c_1 & (a) \\ b \:=\:b_1 & (b) \\ c \:=\:a_1 & (c) \end{Bmatrix}$

Multiply (a) and (b): . $\begin{Bmatrix}a &=& c_1\\ b_1 &=& b \end{Bmatrix} \quad\Rightarrow\quad ab_1 \:=\:bc_1 \quad\Rightarrow\quad ab_1 - bc_1 \:=\:0 \;\;(d)$

Multiply (b) and (c): . $\begin{Bmatrix}b &=& b_1 \\ a_1 &=& c \end{Bmatrix} \quad\Rightarrow\quad a_1b \:=\:b_1c \quad\Rightarrow\quad a_1b - b_1c \:=\:0 \;\;(e)$

Multiply (a) and (c): . $\begin{Bmatrix}a &=& c_1 \\ a_1 &=& c \end{Bmatrix} \quad\Rightarrow\quad aa_1 \:=\:cc_1 \quad\Rightarrow\quad cc_1 - aa_1 \:=\:0 \;\;(f)$

Square (f): . $(cc_1 - aa_1)^2 \;=\;0\;\;(g)$

Multiply (e) and (d):. . $(a_1b-b_1c)(ab_1-bc_1) \:=\:0 \;\;(h)$

Therefore, (g) = (h): . $(cc_1-aa_1)^2 \;=\;(a_1b-b_1c)(ab_1-bc_1) \quad\hdots\quad \text{Answer 3}$

**swordfish774** · June 29th 2012, 06:20 PM

What are the curly brackets? How did you know that a=c1, b= b1, c=a1 for sure?

**biffboy** · June 29th 2012, 10:55 PM

This reminds me of a similar question: If p and q are rhe roots of ax^2+bx+c=0 find the equation whose roots are 1/p and 1/q
We have p+q=-b/a and pq=c/a So 1/p+1/q=(q+p)/pq=(-b/a)/(c/a)=-b/c and (1/p)*(1/q)=1/pq=a/c
So required equation is x^2+(b/c)x+(a/c)=0 That is cx^2+bx+a=0

**swordfish774** · July 1st 2012, 04:01 AM

Originally Posted by Soroban

Hello, swordfish774!

Multiply (a) and (b): . $\begin{Bmatrix}a &=& c_1\\ b_1 &=& b \end{Bmatrix} \quad\Rightarrow\quad ab_1 \:=\:bc_1 \quad\Rightarrow\quad ab_1 - bc_1 \:=\:0 \;\;(d)$

Multiply (b) and (c): . $\begin{Bmatrix}b &=& b_1 \\ a_1 &=& c \end{Bmatrix} \quad\Rightarrow\quad a_1b \:=\:b_1c \quad\Rightarrow\quad a_1b - b_1c \:=\:0 \;\;(e)$

Multiply (a) and (c): . $\begin{Bmatrix}a &=& c_1 \\ a_1 &=& c \end{Bmatrix} \quad\Rightarrow\quad aa_1 \:=\:cc_1 \quad\Rightarrow\quad cc_1 - aa_1 \:=\:0 \;\;(f)$

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consider (x-1)(x-3)=0 and (x-1)(x-4)= 0 1 is reciprocal of 1

(d), (c) and (f) are not =0 in the above example. I understood till [2]. How did you jump to the next step. It seems incorrect to me. Please explain.

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