The number of real roots of the equation √x2-4 x+3+√x2-9=√4 x2-14 x+6, is:

Question

The number of real roots of the equation √x2-4 x+3+√x2-9=√4 x2-14 x+6, is: (A) 2 (B) 3 (C) 1 (D) 0

Tardigrade Admin · Accepted Answer

√(x-1)(x-3)+√(x-3)(x+3) =√4(x-(12/4))(x-(2/4)) ⇒ √x-3=0 ⇒ x=3 text which is in domain text or √x-1+√x+3=√4 x-2 2 √(x-1)(x+3)=2 x-4 x2+2 x-3=x2-4 x+4 6 x=7 x=7 / 6(rejected)

Q. The number of real roots of the equation $x^{2} - 4 x + 3 + x^{2} - 9 = 4 x^{2} - 14 x + 6$ , is:

2

3

1

0

$(x - 1) (x - 3) + (x - 3) (x + 3)$
$= 4 (x - \frac{12}{4}) (x - \frac{2}{4})$
$\Rightarrow x - 3 = 0 \Rightarrow x = 3 which is in domain$
$or$
$x - 1 + x + 3 = 4 x - 2$
$2 (x - 1) (x + 3) = 2 x - 4$
$x^{2} + 2 x - 3 = x^{2} - 4 x + 4$
$6 x = 7$
$x = 7/6$ (rejected)

Q. The number of real roots of the equation x2−4x+3​+x2−9​=4x2−14x+6​, is:

2

3

1

0

(x−1)(x−3)​+(x−3)(x+3)​ =4(x−412​)(x−42​)​ ⇒x−3​=0⇒x=3 which is in domain or x−1​+x+3​=4x−2​ 2(x−1)(x+3)​=2x−4 x2+2x−3=x2−4x+4 6x=7 x=7/6(rejected)

Q. The number of real roots of the equation $x^{2} - 4 x + 3 + x^{2} - 9 = 4 x^{2} - 14 x + 6$ , is:

$(x - 1) (x - 3) + (x - 3) (x + 3)$
$= 4 (x - \frac{12}{4}) (x - \frac{2}{4})$
$\Rightarrow x - 3 = 0 \Rightarrow x = 3 which is in domain$
$or$
$x - 1 + x + 3 = 4 x - 2$
$2 (x - 1) (x + 3) = 2 x - 4$
$x^{2} + 2 x - 3 = x^{2} - 4 x + 4$
$6 x = 7$
$x = 7/6$ (rejected)