Question

Solve this please
Determine if 3 is the root of the equation given below

Answer 1

Answer:-

Yes , 3 is a root of these equations.

Explanation:-

Given :-

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

Also given that 3 is a root of these equation.

Solution:-

Firstly take left hand side ,

$\bf{\sqrt{ {x}^{2} - 4x + 3} + \sqrt{ {x}^{2} - 9 } } \\ \\ \bf{ \: if \: 3 \: is \: a \: root \: of \: these \: rquations \: } \\ \bf{ hence \: put \: x = 3 }\\ \\ \\ \implies \: \bf{ \sqrt{ {(3)}^{2} - 4 \times 3 + 3} + \sqrt{ {(3)}^{2} - 9} } \\ \\ \implies \: \bf{ \sqrt{9 - 12 + 3} + \sqrt{9 - 9} } \\ \\ \implies \: 0$

Now taking right hand side,

$\implies \: \bf{ \sqrt{4 {x}^{2} - 14x + 16} } \\ \bf{ put \: x = 3} \\ \ \\ \implies \: \bf{ \sqrt{ 4 \times 9 - 14 \times 3 + 16} } \\ \\ \implies \bf{\sqrt{36 - 42 + 16} } \\ \\ \implies \: 0$

Because LHS = RHS

Now ,we can say that 3 is the root of these equations.

Answer 2

Answer:

Step-by-step explanation:

For 3 to be root of the equation, the result of the equation should be zero when x is substituted with 3 and solved.

f(x) =  √(x² - 4x + 3) + √(x² - 9) - [√4x² - 14x + 16] = 0

f(3) = 0

= √(3² - 4*3 + 3) + √(3² - 9) - [√4(3)² - 14*3 + 16]

= √ 9 + 3 - 12 + √9 - 9 - [√(36+16 - 42)]

= √0 + √√0 - √0

= 0.

Hence 3 is the root of the equation.