Please solve the sum number 8. Find the value of qx2 -2px + q
Attachments:
Answers
Answered by
1
x = { √(P + q) + √(P- q)}/{√(P+q) - √(P-q)}
use, componendo or dividendo rule
x/1 ={√(P + q) + √(P- q)}/{√(P+q) - √(P-q)}
(x + 1)/(x -1) = 2√(P + q)/2√(P - q)
(x + 1)/(x -1) = √(P + q)/√(P -q)
take square both sides ,
(x + 1)²/(x -1)² = (P +q)/(P -q)
(x +1)² (P - q) = (x -1)²(P+ q)
{(x+1)² - (x -1)² }P = q{ (x -1)² +(x+1)² }
2Px = q{x² + 1 }
qx² + q = 2Px
qx² - 2Px + q = 0 ( answer )
use, componendo or dividendo rule
x/1 ={√(P + q) + √(P- q)}/{√(P+q) - √(P-q)}
(x + 1)/(x -1) = 2√(P + q)/2√(P - q)
(x + 1)/(x -1) = √(P + q)/√(P -q)
take square both sides ,
(x + 1)²/(x -1)² = (P +q)/(P -q)
(x +1)² (P - q) = (x -1)²(P+ q)
{(x+1)² - (x -1)² }P = q{ (x -1)² +(x+1)² }
2Px = q{x² + 1 }
qx² + q = 2Px
qx² - 2Px + q = 0 ( answer )
sukh6:
Thanks
Similar questions