Question

If x=1/√3 is root of the equation px^2+(√3-√2)x-1=0, then the value of p^2+1 is.. A)√6 B)6 C)7 D)8

Answer 1

Answer:

option C is correct.  $p^2+1 = 7$

Step-by-step explanation:

the given equation is :

$px^2+(\sqrt{3} -\sqrt{2}) x - 1 =0$

it is also given that :

$x = 1/\sqrt{3}$ is  the root of above equation.

so , substitute the value of 'x' in given equation:

$px^2+(\sqrt{3} -\sqrt{2}) x - 1 =0$

$p(1/\sqrt{3} )^2+(\sqrt{3} -\sqrt{2} )(1/\sqrt{3} ) - 1 =0$

we know that ;

$(1/\sqrt{3} )^2=(1/\sqrt{3} )(1/\sqrt{3} )$

= $1/(\sqrt{3} )^2$

= 1/3

the equation becomes:

$(p)(1/3)+(\sqrt{3} -\sqrt{2} )(1/\sqrt{3} ) - 1 =0$

p/3 + $(\sqrt{3} )(1/\sqrt{3} ) - (\sqrt{2} )(1/\sqrt{3} )-1=0$

p/3 + 1 -1 - $\sqrt{(2/3)}$=0

p/3 = $\sqrt{(2/3)}$

p = (3)( $\sqrt{(2/3)}$)

now

$p^2=(p)(p)$

substitute value of p;

$p^2=(3)^2(\sqrt{(2/3)})^2$

$p^2 = (9)(2/3)\\p^2=(3)(2)=6$

so the value of $p^2+1 = 6 +1 = 7$

$p^2+1 = 7$

option C is correct.

Answer 2

Answer:

Option (c). The value of p²+1  = 7

Step-by-step explanation:

• In context to the given question we have to find the value of $p^{2} +1$

Given equation,

px²+(√3-√2) x-1=0

by putting the given value of x=1/√3

p(1/√3)²+(√3-√2) (1/√3) - 1=0

p/3 + [(1.732 - 1.414) / 1.732] = 1   [√3= 1.732 and √2= 1.414]

p/3 + 0.318/ 1.732 = 1

p/3 + 0.183 = 1

by transposing method,

p/3 = 1 - 0.183

p/3 = 0.817

p = 0.817 x 3

p = 2.451

Further, by putting the value of p in p²+1

we get,

p²+1  = (2.45)²+1

p²+1  = 6.00 + 1

p²+1  = 7

Option (c). The value of p²+1  = 7