Question

Please give solution of the question in attachment.....

Class 10th (CBSE)

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Answer 1

Thus so


. THE ANS IS IT S


Let a and b are the roots of the given equations.

Now, √(p/q) + √(q/p) + √(n/l)

= √p/√q + √q/√p + √n/√l

= (√p2 + √q2 )/(√p*√q) + √n/√l

= (p + q)/(√p*√q) + √n/√l ...................1

Again given, a : b = p : q

Let a = px, b = qx

Now, a + b = -n/l

=> px + qx = -n/l

=> (p + q)x = -n/l

=> p + q = -n/lx .............2

and a*b = n/l

=> px * qx = n/l

=> pq*x2 = n/l

=> √pq * x = √(n/l)

=> √pq = √(n/l)/x ..........3

From equation 1, we get

{-n/lx}/{√(n/l)/x} + √(n/l)

= {-n/l}/{√(n/l)} + √(n/l)

= -√(n/l) + √(n/l)

= 0

So, √(p/q) + √(q/p) + √(n/l) = 0



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Answer 2

HEYA !!!

REFER TO THE ATTACHMENT

# GUDIYA

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