Question

Determine. if 3 is a root of the equation given below: $\sqrt{x^{2}-4x+3}+ \sqrt{x^{2}-9}= \sqrt{4x^{2}-14x+16}$

Answer 1

SOLUTION :  

Given : √(x² - 4x + 3) + √(x² - 9) = √(4x² -14x +16)

3 is a root of the equation.

Therefore, x = 3  

On putting x = 3 in LHS & RHS of the given equation.

LHS = √(x² - 4x + 3) + √(x² - 9)

=  √(3² - 4×3 + 3) + √(3² - 9)

= √(9 - 12 +3) + √(9 - 9)

= √( -3 +3) + √0

= √0 + √0 = 0

LHS = 0

RHS = √(4x² -14x +16)

= √(4 × 3² - 14 × 3 + 16)

= √ (4 × 9 - 42 +16)

= √(36 - 26)

=√10

RHS = √10

LHS ≠ RHS  

Here, LHS &  RHS are not equal.  

Hence, x = 3 is not the solution of the given quadratic equation.  

★★ ROOTS OF A QUADRATIC EQUATION :  

All the values of variable which satisfy the given quadratic equation, are called roots or solutions of given quadratic equation.

**Any quadratic equation can have atmost two roots.

HOPE THIS ANSWER WILL HELP YOU….

Answer 2

$\huge\bold{AYY}$

==>>$\sqrt{x^{2}-4x+3}+ \sqrt{x^{2}-9}= \sqrt{4x^{2}-14x+16}$

whether to find out the given value is root of the equations,,, let substitute x = 3... IF both LHS and RHS are equal then the given value is said to be the root of the equations..but the given equation or any quad equations always contain atleast two roots that's one of the hint

$\sqrt{3^{2}-4×3+3}+ \sqrt{3^{2}-9}= \sqrt{4×3^{2}-14×3+16}$

squaring on both side..

we get,,,

LHS==>>

9 - 12 + 3 + 9 - 9 = 4 × 9 - 42 + 16

12 - 12 = 36 - 42 + 16

0 = 10...i.e,.

0 =/= 10...

hence 3 is not the root of the equation...

$\huge{RGRDSM\:@\:STYLGG}$