Question

a²-ma+1 is a equation
1.a-(1÷a)=?

Answer 1

Answer:

We know that while finding the root of a quadratic equation ax

2

+bx+c=0 by quadratic formula x=

2a

−b±

b

2

−4ac

,

if b

2

−4ac>0, then the roots are real and distinct

if b

2

−4ac=0, then the roots are real and equal and

if b

2

−4ac<0, then the roots are imaginary.

Here, the given quadratic equation a

2

−ma+1=0 is in the form ax

2

+bx+c=0 where a=1,b=−m and c=1.

(i) If the roots are equal then b

2

−4ac=0, therefore,

b

2

−4ac=0

⇒(−m)

2

−(4×1×1)=0

⇒m

2

−4=0

⇒m

2

=4

⇒m=±

4

⇒m=±2

(ii) If the roots are distinct then b

2

−4ac>0, therefore,

b

2

−4ac>0

⇒(−m)

2

−(4×1×1)>0

⇒m

2

−4>0

⇒m

2

>4

⇒m>±

4

⇒m>±2

(iii) If the roots are imaginary then b

2

−4ac<0, therefore,

b

2

−4ac<0

⇒(−m)

2

−(4×1×1)<0

⇒m

2

−4<0

⇒m

2

<4

⇒m<±

4

⇒m<±2

Hence m=±2 if the roots are equal, m>±2 if the roots are distinct and m<±2 if the roots are imaginary.

Answer 2

$\sqrt{ {m}^{2} - 4 }$

Step-by-step explanation:

$Given, \: \: {a}^{2} - ma + 1 = 0 \\ \: \: \: \: \: \: \: \: {a}^{2} + 1 = ma \\ On \: \: dividing \: \: by \: \: a \: \: on \: \: both \: \: sides \\ \frac{ {a}^{2} + 1 }{a} = \frac{ma}{a} \\ a + \frac{1}{a} = m \\ we \: \: know \: \: \boxed{(a - \frac{1}{a} )^{2} = (a + \frac{1}{a} )^{2} -4 \times a\times \frac{1}{a}} \\putting \: \: value \\ (a - \frac{1}{a} )^{2} = {m}^{2} - 4 \\ then, \: \: a - \frac{1}{a} = \sqrt{ {m}^{2} - 4 }$