Question

√(112) - √(80) / √(20) - √(28) = ?​

Answer 1

`\large\underline{\sf{Solution-}}`

We need to solve,

`(√(112)-√(80))/(√(20)-√(28))`

We can write the above mentioned expression as,

`\sf\longmapsto(√(2*2*2*2*7)-√(2*2*2*5))/(√(2*2*5)-√(2*2*7))`

So,

`\sf\longmapsto(√(4*4*7)-√(4*4*5))/(√(2^2*5)-√(2^2*7))`

So,

`\sf\longmapsto(√(4^2*7)-√(4^2*5))/(√(2^2*5)-√(2^2*7))`

Hence,

`\sf\longmapsto(4√(7)-4√(5))/(2√(5)-2√(7))`

Taking common in respective terms,

`\sf\longmapsto(4(√(7)-√(5)))/(2(√(5)-√(7)))`

On cancelling 4 with 2,

`\sf\longmapsto(4\!\!\!/^(\:2)(√(7)-√(5)))/(2\!\!\!/(√(5)-√(7)))`

`\sf\longmapsto(2(√(7)-√(5)))/((√(5)-√(7)))`

Taking (-) common,

`\sf\longmapsto(-2(√(5)-√(7)))/((√(5)-√(7)))`

So, (√5 - √7) gets cut,

Hence,

`\longmapsto\bf(√(112)-√(80))/(√(20)-√(28))=-2`