Question

Find the roots of the following equations
x-1/x =3,x≠0​

Answer 1

Answer:

hope this helps u

mark me as the brainliest if it helped u a lot

thanks:)

Step-by-step explanation:

x-1/x=3

=>x(x)-1/x=3

=>x²-1/x=3

=>x²-1=3x

=>x²-3x-1=0

comparing equation ax²+bx+c,

hence, a=1,b=-3 and c=-1

we know that,

D=b²-4ac

D=(-3)²-(4)(1)-1

D=9+4

D=13

hence, roots to equation are

x= -b±√D/2a

=>x= -(-3) ±√13/2*1

=x=3±√13/2

hence root to the equations are:

3+√13/2 and 3-√13/2

Answer 2

$\tt \implies\large\bold\green{x - \dfrac{1}{x} = 3} \\ \\ \tt \implies \: \dfrac{x {}^{2} - 1 }{x} = 3 \\ \\ \tt \implies \: x {}^{2} - 1 = 3 \times x \\ \\ \tt \implies \: x {}^{2} - 3x - 1 = 0$

By Comparing equation with ax² + bx + c = 0

• a = 1
• b = -3
• c = -1

We know that,

D = b² - 4ac

D = (-3)² - 4(1)(-1)

D = 9 + 4

D = 13

Roots of the equation will be,

$\bf \implies x = \dfrac{ - b \: \pm \: \sqrt{d} }{2a} \\ \\ \tt \implies \: x = \dfrac{ - ( - 3) \pm \: \sqrt{13} }{2 \times 1}$

$\large \implies \boxed {\boxed {\tt \blue {x = \dfrac{3 \: \pm \sqrt{13} }{2} }}}$

Hence, roots of the equation are:

$\tt x = \dfrac{3 \: + \sqrt{13} }{2} \: or \: x = \dfrac{3 \: - \sqrt{13} }{2}$