Question

If the equation x² − ax + 1 = 0 has two distinct roots, then
(a)|a| = 2
(b)|a| < 2
(c)|a| > 2
(d)None of these

Answer 1

SOLUTION :  

Option (c) is correct : |a| > 2

Given : x² - ax + 1 = 0

On comparing the given equation with ax² + bx + c = 0  

Here, a = 1 , b = - a  , c = 1

D(discriminant) = b² – 4ac

D = (- a)² - 4 × 1 × 1

D = a² - 4

D > 0 (Given :  distinct roots)

a² - 4  > 0

a² > 4

a > √4

a > ±2

Hence, the value of |a| is > 2

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Answer 2

Option ( c ) is correct.

Explanation :

Compare x²-ax+1 = 0 with

Ax²+Bx+C=0 , we get

A = 1 , B = -a , C= 1

Now ,

Discreminant (D)>0

[ Since , given roots are

real and distinct ]

=> B² - 4AC > 0

=> (-a)² - 4×1×1 > 0

=> a² - 4 > 0

=> a² > 4

=> a > ± √4

=> a > ± 2

=> |a| > 2

Therefore ,

| a | > 2

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