pls solve 13th question
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Answers
qx² - 2px + q = 0
Step-by-step explanation:
If, x = [root(p+q) +root(p-q)] /[root(p+q) - root(p-q)] then find the value of qx2 – 2px + q .
x = (√(p + q) + √(p-q))/(√(p + q) - √(p-q))
Multiplying & diving by √(p + q) + √(p-q)
=> x = (p + q + p - q + 2√(p² - q²) )/ ( (p + q) - (p - q))
=> x = 2 ( p + √(p² - q²) ) / 2q
=> x = ( p + √(p² - q²) ) / q
=> qx = p + √(p² - q²)
=> qx - p = √(p² - q²)
Squaring both sides
=> q²x² + p² - 2pqx = p² - q²
=> q²x² - 2pqx = - q²
diving by q both sides
=> qx² - 2px = -q
=> qx² - 2px + q = 0
I hope it will help you siso
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Concept Used :-
Componendo and Dividendo :-
Given that,
can be rewritten as
On applying Componendo and Dividendo, we get
On squaring both sides, we get
On applying Componendo and Dividendo, we get
We know,
and
So, using these Identities, we get
Hence, Value of
Additional Information :-
1. Invertendo :-
2. Alternendo
3. Componendo
4. Dividendo