Why it can't have 4 real roots ?!!
( a − 1 ) ( x 2 + x + 1 ) 2 − ( a + 1 ) ( x 4 + x 2 + 1 ) = 0
Provided that two roots of the above equation are real and distinct for a ∈ R − A , find the set A .
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3 solutions
As This is Part Of JEE - Maths ! So I solved it in JEE style, I Know This is not mathematical But Still it has 4 options. ( Since Question Has Options )
So Check the equation for a = 2 & a = 0
@SandeepBhardwaj I tried solving it by dividing both sides by x^2 an got an expression (a+1)^2>=0 implying a belongs to R . Why is that wrong ?
I don't understand one thing: the problem says that the roots must be distinct, so (a+2)(a-2) should be just greater than zero, not greater or equal. This should lead to the answer (-2,2). Why is it [-2,2] ?
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SINCE a^2-4>0
THEREFORE (a+2)(a-2)>0
that is either a+2 and a-2 both are positive
that is a>-2 & a>+2
which reduces to a>2
case 2: a+2<0 that is a<-2 a-2<0 that is a<+2 which reduces to# a<-2 hence a>2 and a<-2
NOW A BELONGS TO SET OF INTEGERS MINUS [2 , -2]
HENCE ANSWER IS [2 , -2]
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Well the question clearly states that roots should be real and distinct. Therefore D>0.
your equation is − 2 ( x 2 + x + 1 ) ( x 2 − a ∗ x + 1 ) = 0
Could you please explain how did you reached the above simplification ?
I gone through some hard way but I think it's worth learning new methods to solve different problems and apply which you thing is the best at the time of exam.
Let's begin,
( a − 1 ) ( x 2 + x + 1 ) 2 − ( a + 1 ) ( x 4 + x 2 + 1 ) = 0
Now after opening the bracket and after simplification we get ,
x 4 − ( a − 1 ) x 3 − ( a − 2 ) x 2 − ( a − 1 ) x + 1 = 0
Now we observe that it is a symmetric quartic, so we divide the whole equation by x 2 .
We get after simplification,
( x + x 1 ) 2 − ( a − 1 ) ( x + x 1 ) − a = 0
Now substitute x + x 1 = t , we get a quadratic in t.
t 2 − ( a − 1 ) t − a = 0
( t − a ) ( t + 1 ) = 0
We came to this factorization by observing the product and sum of roots by using Sir Vieta's formula or we could also use Sir Shreedharachaya formula ,
We get two quadratic equation after back substituting. t = x + x 1
One quadratic x + x 1 = − 1 will have no real roots (check the discriminant).
Now so the other quadratic must have ∗ ∗ d i s t i n c t ∗ ∗ and real . I am emphasizing on distinct so as to put the discriminant greater than zero and not greater that or equal to zero. From there we get ,
a ∈ R − [ − 2 , 2 ] .
We can make use of the fact that x 4 + x 2 + 1 is a square "short" of being a perfect square to factor it. x 4 + x 2 + 1 = x 4 + 2 x 2 + 1 − x 2 = ( x 2 + 1 ) 2 − x 2 = ( x 2 + x + 1 ) ( x 2 − x + 1 ) This now allows us to factor the equation: ( a − 1 ) ( x 2 + x + 1 ) 2 − ( a − 1 ) ( x 4 + x 2 + 1 ) ( a − 1 ) ( x 2 + x + 1 ) 2 − ( a − 1 ) ( x 2 + x + 1 ) ( x 2 − x + 1 ) ( x 2 + x + 1 ) [ ( a − 1 ) ( x 2 + x + 1 ) − ( a − 1 ) ( x 2 − x + 1 ) ] = 0 = 0 = 0 and then expanding and simplifying the big bracket leads to − 2 ( x 2 + x + 1 ) ( x 2 − a x + 1 ) = 0 Since the first polynomial factor has a negative discriminant, any real solutions would have to come from the second polynomial factor. Since we want real and distinct solutions, we need its discriminant to be positive, i.e. a 2 − 4 > 0 a < − 2 or a > 2 a ∈ R − [ − 2 , 2 ]
This expression reduces to - 2 ( x 2 + x + 1 ) ( x 2 + a x + 1 ) = 0 .
Now, for this to be true, x = 2 a ± a 2 − 4 i.e, a 2 − 4 ≥ 0
or, ( a + 2 ) ( a − 2 ) ≥ 0 i.e,
a ∈ R ∼ [ 2 , − 2 ]
So A = [ 2 , − 2 ]