Question

Find the value of p, when
px² + (√3 - √2)x - 1 = 0 and
x = - 1/√3 is one root of the equation​

Answer 1

Answer:

Hope this will help you!

Answer 2

The value of p is $\sqrt{6} (\sqrt{6}-1 )$

Step-by-step explanation:

To find: The value of p for the equation $px^2+(\sqrt{3} -\sqrt{2} )x-1=0$

When $x=\dfrac{-1}{\sqrt{3} }$ is one of its root

As given $x=\dfrac{-1}{\sqrt{3} }$ is one of the root

Therefore

Put the value of x in the given equation

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

$\Rightarrow \dfrac{p}{3} -1+\dfrac{\sqrt{2} }{\sqrt{3} } -1=0\\\\\\\Rightarrow \dfrac{p}{3} +\dfrac{\sqrt{2} }{\sqrt{3} } -2=0\\\\\Rightarrow \dfrac{p}{3} =2-\dfrac{\sqrt{2} }{\sqrt{3} }\\\\\Rightarrow p= 3(2-\dfrac{\sqrt{2} }{\sqrt{3} })\\\\\Rightarrow p= 6-3\dfrac{\sqrt{2} }{\sqrt{3} }$

$\Rightarrow p= 6- \sqrt{3} \times \sqrt{2}\\\\\Rightarrow p= 6-\sqrt{6} = \sqrt{6}(\sqrt{6}-1 )$

Hence, The value of p is $\sqrt{6} (\sqrt{6}-1 )$

#Learn more

Find the value of p, for which one root of the quadratic equation px²-14x+8=0 is 6 time the other.

brainly.in/question/1141570