Question

Find the value of P when p f x square + root 3 minus root 2 into x minus 1 is equal to zero and X = 1 by root 3 is one root of the equation

Answer 1

Given:

px²+(√3-√2)x-1=0

x=1/√3

To find:

The value of p

Solution:

The value of p is √6.

We can find the value by following the given steps-

We know that the root of an equation is the value of the variable which makes the equation equal to 0.

So, on putting x=1/√3, px²+(√3-√2)x-1 should be equal to 0.

The equation=px²+(√3-√2)x-1=0

The value of the root of the equation=x=1/√3

Now we will put the value of the root in the equation to find p.

On putting the value of x in the equation,

p(1/√3)²+(√3-√2)×1/√3-1=0

p/3+(√3-√2)/√3=1

(√3p+3(√3-√2))/3√3=1

√3p+3(√3-√2)=3√3

√3p+3√3-3√2=3√3

√3p-3√2=0

√3p=3√2

p=3√2/√3

p=√3×√2

p=√6

Therefore, the value of p is √6.

Answer 2

Answer:-

√6 is the final answer

hope you got the answer

and understood the sum (☞ ಠ_ಠ)☞

(firstly i was also in trouble (*_*) ಠ︵ಠ but than I tried again with different style (◍•ᴗ•◍))