Equal Roots !

Level pending

If two of the roots of the equation ( a − 2 ) ( x 2 + x + 1 ) 2 − ( a + 2 ) ( x 4 + x 2 + 1 ) = 0 (a-2)(x^2+x+1)^2 - (a+2)(x^4+x^2+1) = 0 are equal, then the value(s) of a a are

− 2 , 2 { -2 , 2 } − 4 , 4 { -4 , 4 } 4 { 4 } 2 { 2 }

Nishant Rai by Nishant Rai , 25, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Tom Engelsman
Aug 29, 2020

The above polynomial factors according to:

( x 2 + x + 1 ) [ ( a − 2 ) ( x 2 + x + 1 ) − ( a + 2 ) ( x 2 − x + 1 ) ] = 0 (x^2 + x + 1)[(a-2)(x^2+x+1) - (a+2)(x^2 - x + 1)]=0 ;

or ( x 2 + x + 1 ) [ ( a − 2 − a − 2 ) x 2 + ( a − 2 + a + 2 ) x + ( a − 2 − a − 2 ) ] = 0 ; (x^2+x+1)[(a-2 - a -2)x^2 + (a-2+ a+2)x + (a-2-a-2)] = 0;

or ( x 2 + x + 1 ) ( − 4 x 2 + 2 a x − 4 ) = 0. (x^2+x+1)(-4x^2 + 2ax - 4) = 0.

Since the factor x 2 + x + 1 x^2+x+1 yields a complex-conjugate pair of roots, we require the factor − 4 x 2 + 2 a x − 4 -4x^2 + 2ax - 4 to yield one real double root. If we apply the Quadratic Equation, we obtain:

x = − 2 a ± 4 a 2 − 4 ( − 4 ) 2 − 8 = a ± a 2 − 16 4 x = \frac{-2a \pm \sqrt{4a^2 - 4(-4)^2}}{-8} = \frac{a \pm \sqrt{a^2 - 16}}{4}

of which the discriminant must equal zero, or a 2 − 16 = 0 ⇒ a = ± 4 . a^2 - 16 = 0 \Rightarrow \boxed{a = \pm 4}.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...