Question

hey mate

plz help me na

stuck in this

Answer 1

Step-by-step explanation:

a/b = a^2/ab

So 1st term is sqrt(a^2/ab) = a/sqrt(ab)

b/a = b^2/ab (multiplying Nr & Dnr of b/a with b)

2nd term, therefore is sqrt(b^2/ab) = b/sqrt(ab)

So sum of first two terms = (a+b)/ sqrt(ab)

Now, a+b = -q/p ab q/p - substituting in (3), we get

= -(q/p) / sqrt (q/p) = - sqrt (q/p)

So LHS is - sqrt(q/p) + sqrt (q/p) = 0

Answer 2

Answer:

Since α and β are the roots of the equation px^2 + qx + q = 0,

α + β = -q / p ----- ( 1 )

α β = q / p ----- ( 2 )

Given : α / β = a / b ----- ( 3 )

Now, by ( 2 ) × ( 3 ),

α β × ( α / β ) = ( q / p ) × ( a / b )

α ^2 = q a / p b

α = √ ( q a / p b ) ----- ( 4 )

In ( 2 ),

α β = q / p

β = q / p α

β = q / [ p { √ ( q a / p b ) } ]

β = √ ( q b / p a ) ----- ( 5 )

Thus, in ( 1 ),

α + β = - q / p

√ ( q a / p b ) + √ ( q b / p a ) = -q / p

√ q / p [ √ ( a / b ) + √ b / a ) = -q / p

√ ( a / b ) + √ ( b / a ) = - √ ( q / p )

√ a / b ) + √ ( b / a ) + √ ( q / p ) = 0

Hence, proved.