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Algebra Level 4

( a − 1 ) ( x 4 + x 2 + 1 ) + ( a + 1 ) ( x 2 + x + 1 ) 2 = 0 \large (a-1)(x^4 + x^2 +1 ) + (a+1)(x^2+x+1)^2 =0

If two roots of the equation above (in x x ) are real , distinct and non zero , then what is the set of values of a a ?

( − 1 2 , 1 2 ) \displaystyle \left ( \frac{-1}{2} , \frac{1}{2}\right) ( − 1 2 , 0 ) \left( \frac{-1}{2} ,0\right) ( − ∞ , − 2 ) ∪ ( 2 , ∞ ) \displaystyle ( -\infty,-2) \cup (2,\infty) None of these

Sumanth R Hegde by Sumanth R Hegde , 21, India

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1 solution

Sumanth R Hegde
Feb 15, 2017

The range of values of a a is ( − 1 2 , 1 2 ) − { 0 } \displaystyle (\dfrac{-1}{2}, \dfrac{1}{2} ) - \{ 0\}

IMG<em>20170215</em>230644.jpg IMG 20170215 230644.jpg

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Oh my god! . Missed the word non zero and became over excited on seeing matching option. Clever problem!!! :)

Prakhar Bindal - 4 years, 3 months ago

You can make life easier by noting that x^4 + x^2 +1 = (x^2+x+1)(x^2-x+1) and note that first factor cant have a real root, thus divide by it and then an easy quadratic equation will come that's easy to handle :)

Harsh Shrivastava - 4 years, 1 month ago

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