Question

The number of roots of the equation$\sqrt{x^{2} -4} - (x-2) = \sqrt{x^{2} -5x+6}$

Answer 1

Given Expression,

$\sf \: \sqrt{x^{2} -4} - (x-2) = \sqrt{x^{2} -5x+6} \\ \\ \implies \sf \: \sqrt{(x - 2)(x + 2)} - (x - 2) = \sqrt{(x - 2)(x - 3)} \\ \\ \implies \sf \: \sqrt{x - 2} \bigg \{\sqrt{x + 2} - \sqrt{x - 2} \bigg\} = ( \sqrt{x - 2} )( \sqrt{x - 3} ) \\ \\ \implies \sf \: \sqrt{x - 2} \bigg \{\sqrt{x + 2} - \sqrt{x - 2} - \sqrt{x - 3}\bigg \} = 0 \\ \\ \implies \sf \sqrt{x - 2} = 0 \\ \\ \implies \boxed{\boxed{\sf x = 2}}$

Consider,

$\implies \sf \: \sqrt{x + 2} - \sqrt{x - 2} - \sqrt{x - 3} = 0 \\ \\ \implies \sf \sqrt{x + 2} - \sqrt{x - 2} = \sqrt{x - 3}$

Squaring on both sides,

$\implies \sf \: ( \sqrt{x + 2} - \sqrt{x - 2}) {}^{2} =( \sqrt{x - 3} ) {}^{2} \\ \\ \implies \sf \: x + 2 + x - 2 - 2 \sqrt{ {x}^{2} - 4} = x - 3 \\ \\ \implies \sf \: 2x - x + 3 = 2 \sqrt{ {x}^{2} - 4} \\ \\ \implies \sf \: x + 3 = 2 \sqrt{ {x}^{2} - 4 }$

Squaring on both sides,

$\implies \sf \: {x}^{2} + 6x + 9 = 4( {x}^{2} - 4) \\ \\ \implies \sf \: 3 {x}^{2} - 6x - 25 = 0 \\ \\ \implies \sf \: x = \dfrac{ - ( - 6) \pm \sqrt{( - 6) {}^{2} - 4(3)( - 25)} }{2(3)} \\ \\ \implies \sf \: x = \dfrac{ 6 \pm \sqrt{336} }{6} \\ \\ \implies \sf \: x = \dfrac{ 6 \pm 4\sqrt{21} }{6} \\ \\ \implies \boxed{ \boxed{ \sf \: x \: = \dfrac{3 + 2 \sqrt{21} }{3} \: or \: \dfrac{3 - 2 \sqrt{21} }{3} }}$

Answer 2

Step-by-step explanation:

given :

The number of roots of the equation$\sqrt{x^{2} -4} - (x-2) = \sqrt{x^{2} -5x+6}$

to find :

sqrt{x^{2} -4} - (x-2) = \sqrt{x^{2} -5x+6}[/tex]

solution :

• Comparing the equation with ax²+bx+c = 0 gives,

• a = 1, b = -5 and c = 6

• b²-4ac = (-5)2-4×1x6 = 1

• Roots of the equations are

• -b +√(b² - 4ac) /2a

• = 5 + 12 = 62

• = 3

• Another root

• -b-√(b² - 4ac) /2a

• = 5-12

• = 42

• = 2

• Answer

• Hence the roots of the equation are 3 and 2