The number of roots of the equation √x2-4-(x-2)=√x2-5 x+6 is

Question

The number of roots of the equation √x2-4-(x-2)=√x2-5 x+6 is (A) 0 (B) 1 (C) 2 (D) 3

Tardigrade Admin · Accepted Answer

√x2-4-(x-2)=√x2-5 x+6 √x-2(√x+2-√x-2)=√x-2 √x-3 So x=2 or √x+2-√x-2=√x-3 squaring both side x+2+x-2-2 √x2-4=x-3 x+3=2 √x2-4 x2+6 x+9=4 x2-16 ⇒ 3 x2=6 x-25=0 D>0

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

0

1

2

3

$x^{2} - 4 - (x - 2) = x^{2} - 5 x + 6$
$x - 2 (x + 2 - x - 2) = x - 2 x - 3$
So $x = 2$ or $x + 2 - x - 2 = x - 3$
squaring both side
$x + 2 + x - 2 - 2 x^{2} - 4 = x - 3$
$x + 3 = 2 x^{2} - 4$
$x^{2} + 6 x + 9 = 4 x^{2} - 16$
$\Rightarrow 3 x^{2} = 6 x - 25 = 0$
$D > 0$

Q. The number of roots of the equation x2−4​−(x−2)=x2−5x+6​ is

0

1

2

3

x2−4​−(x−2)=x2−5x+6​ x−2​(x+2​−x−2​)=x−2​x−3​ So x=2 or x+2​−x−2​=x−3​ squaring both side x+2+x−2−2x2−4​=x−3 x+3=2x2−4​ x2+6x+9=4x2−16 ⇒3x2=6x−25=0 D>0

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

$x^{2} - 4 - (x - 2) = x^{2} - 5 x + 6$
$x - 2 (x + 2 - x - 2) = x - 2 x - 3$
So $x = 2$ or $x + 2 - x - 2 = x - 3$
squaring both side
$x + 2 + x - 2 - 2 x^{2} - 4 = x - 3$
$x + 3 = 2 x^{2} - 4$
$x^{2} + 6 x + 9 = 4 x^{2} - 16$
$\Rightarrow 3 x^{2} = 6 x - 25 = 0$
$D > 0$