Question

if one root of the equation px^(√3-√2)x-1=0 is x=1÷√3 then the value of p

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Answer 1

$\huge\red{A}\pink{N}\orange{S} \green{W}\blue{E}\gray{R} =$

option C is correct. p²+1=7

Step-by-step explanation:

the given equation is:

px² + (√3-√2)x-1=0 it is also given that : x=1/√3 is the root of above

equation.

so, substitute the value of 'x'

in given equation:

p(1/√3)²+(√3-√2)(1/√3)

1=0 we know that ;

(1/√3)² = (1/√3)(1/√3)

= 1/(√3)²

= 1/3

the equation becomes:

(p) (1/3)+(√3-√2)(1/√3) - 1=0

p/3 + (√3)(1/√3) -

(√2)(1/√3)-1=0

p/3 +1-1-(2/3)=0

p/3 = √(2/3) p = (3)(√(2/3))

now

p² = (p) (p)

substitute value of p; p² = (3)²(√(2/3))²

p² = (9)(2/3) p² = (3)(2)=6

so the value of p²+1=6+1=7

p² + 1 = 7