If two roots of the equation (a-1)(x2+x+1)2-(a+1)(x4+x2+1)=0 are real and distinct, then 'a' lies in the interval

Question

If two roots of the equation (a-1)(x2+x+1)2-(a+1)(x4+x2+1)=0 are real and distinct, then 'a' lies in the interval (A) (-2,2) (B) (-∞,-2) ∪(2, ∞) (C) (2, ∞) (D) (-3,3)

Tardigrade Admin · Accepted Answer

(a-1)(x2+x+1)2-(a+1)(x4+x2+1)=0 ..........(1) ∵ x4+x2+1=(x2+x+1)(x2-x+1) ∴ (1) text becomes ⇒ (x2+x+1)[(x2+x+1)(a-1)-(a+1)(x2-x+1)]=0 ⇒ (x2+x+1)(x2-a x+1)=0 Here two roots are imaginary and for other two roots to be real D>0 ⇒ a2-4>0 ⇒ a ∈(-∞,-2) ∪(2, ∞)

Q. If two roots of the equation $(a - 1) (x^{2} + x + 1)^{2} - (a + 1) (x^{4} + x^{2} + 1) = 0$ are real and distinct, then 'a' lies in the interval

$(- 2, 2)$

$(- \infty, - 2) \cup (2, \infty)$

$(2, \infty)$

$(- 3, 3)$

Q. If two roots of the equation (a−1)(x2+x+1)2−(a+1)(x4+x2+1)=0 are real and distinct, then 'a' lies in the interval

(−2,2)

(−∞,−2)∪(2,∞)

(2,∞)

(−3,3)

(a−1)(x2+x+1)2−(a+1)(x4+x2+1)=0..........(1) ∵x4+x2+1=(x2+x+1)(x2−x+1) ∴(1) becomes ⇒(x2+x+1)[(x2+x+1)(a−1)−(a+1)(x2−x+1)]=0 ⇒(x2+x+1)(x2−ax+1)=0 Here two roots are imaginary and for other two roots to be real D>0 ⇒a2−4>0⇒a∈(−∞,−2)∪(2,∞)

Q. If two roots of the equation $(a - 1) (x^{2} + x + 1)^{2} - (a + 1) (x^{4} + x^{2} + 1) = 0$ are real and distinct, then 'a' lies in the interval

$(- 2, 2)$

$(- \infty, - 2) \cup (2, \infty)$

$(2, \infty)$

$(- 3, 3)$