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Let a and b be the roots of the equation x^2 - mx + 2 = 0.  Suppose that a + 1/b + ab and b + 1/a + ab are the roots of the equation x^2 - px + q = 0.  What is q?

May 25, 2021

#1
+11625
+1

Let a and b be the roots of the equation x^2 - mx + 2 = 0.  Suppose that a + 1/b + ab and b + 1/a + ab are the roots of the equation x^2 - px + q = 0.  What is q?

Hello Guest!

$$x^2 - mx + 2 = 0$$

p           q

$$x=-\frac{p}{2}\pm \sqrt{\frac{p^2}{4}-q}\\ a=\frac{m}{2}+ \sqrt{\frac{m^2}{4}-2}\\ b=\frac{m}{2}- \sqrt{\frac{m^2}{4}-2}$$

$${\color{blue}x^2 - px + q =}\{x-(\frac{m}{2}+ \sqrt{\frac{m^2}{4}-2}+\frac{1}{\frac{m}{2}- \sqrt{\frac{m^2}{4}-2}}\\ +(\frac{m}{2}+ \sqrt{\frac{m^2}{4}-2})(\frac{m}{2}- \sqrt{\frac{m^2}{4}-2}))\}$$

$$\times\{x-(\frac{m}{2}- \sqrt{\frac{m^2}{4}-2}+\frac{1}{\frac{m}{2}+ \sqrt{\frac{m^2}{4}-2} }+(\frac{m}{2}+ \sqrt{\frac{m^2}{4}-2})(\frac{m}{2}- \sqrt{\frac{m^2}{4}-2}))\}$$

$$\sqrt{\frac{m^2}{4}-2}=z$$

$${\color{blue}x^2 - px + q =}\{x-(\frac{m}{2}+ z+\frac{1}{\frac{m}{2}- z} +(\frac{m}{2}+ z )(\frac{m}{2}- z ))\}$$

$$\times\{x-(\frac{m}{2}-z +\frac{1}{\frac{m}{2}+ z }+(\frac{m}{2}+ z )\times (\frac{m}{2}-z ))\}$$

$${\color{blue}x^2 - px + q =} (x-(a + 1/b + ab))(x-( b + 1/a + ab))$$

!

May 25, 2021
edited by asinus  May 25, 2021
edited by asinus  May 25, 2021
edited by asinus  May 26, 2021
#2
+113617
+1

Let a and b be the roots of the equation x^2 - mx + 2 = 0.

Suppose that a + 1/b + ab and b + 1/a + ab are the roots of the equation x^2 - px + q = 0.

What is q?

Let a and b be the roots of the equation x^2 - mx + 2 = 0.

a+b=m

ab=2

Suppose that a + 1/b + ab and b + 1/a + ab are the roots of the equation x^2 - px + q = 0

(a + 1/b + ab)   +   (b + 1/a + ab) = p      and     (a + 1/b + ab)   *   (b + 1/a + ab) = q

(a + 1/b + ab)   = (ab+1)/b + ab = (2+1)/b + 2 = 3/b + 2

likewise

(b + 1/a + ab) = 3/a + 2

so

$$q= (\frac{3}{a}+2) (\frac{3}{b}+2)\\ abq= (3+2a) (3+2b)\\ 2q=9+6b+6a+4ab\\ 2q=9+6(a+b)+4ab\\ 2q=9+6(m)+8\\ 2q=17+6m\\ q=3m+8.5$$

Maybe you can get rid of the m if you go further with the working.

But the question does not state whether you must get q as a number or as a function of m.

May 25, 2021