Quadratic Equation problems

As per the given above equations , we can say that
?.
?500x + ?402 = 0
? ?500x = - ?402
? x = - ?( 402/500 ) = - ?( 400/500 ) = - 0.9
?. ?360y + ( 200 )^1/2 = 0
? ?360y = - ?200
?
y = - ?( 200 / 360 ) = - 0.74
Clearly, we can see that x < y is required answer .

As per the given above question , we have
Let, x = 6 + ?6 + ?6 + ?6 +........
On squaring both sides, we have
x² = 6 + ?6 + ?6 + ?6 +........
? x² = 6 + x
? x² - x - 6 = 0
? x² - 3x + 2x - 6 = 0
? x( x - 3 ) + 2( x - 3 ) = 0
? ( x - 3 ) ( x + 2 ) = 0
? x = 3 and x ? ?2 because numbers are positive.
Thus , correct answer will be 3 .

Given : - Sum of the roots ( k₁ + k₂ ) = 6
One root k₁ = 3 - ?5
? the other root k₂ = 6 - ( 3 - ?5) = 3 + ?5
? Product of roots k₁k₂ = ( 3 - ?5 ) ( 3 + ?5 )
product of roots = 9 - 5 = 4
Hence , the required equation is x² - (sum of the roots)x + (product of roots) = 0
? x² - 6x + 4 = 0.

Let k₁ and k₂ be the roots of quadratic equation .
Given :- k₁ = ?2 and k₂ = 2 ?2
Sum of the roots = ?2 + 2 ?2 = 3 ?2
Product of the roots = ( ?2 ) ( 2?2 ) = 4
Hence the required quadratic equation is x ² = (sum of the roots) x + (Product of two roots) = 0
? x ² - 3 ?2x + 4 = 0.