Question

completing the square

Answer 1

Hey buddy!



Here is yr answer
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Method l:


=> x² - (√3 + 1 )x +√3 = 0

=> x² - (√3 + 1)x + √3 × 1 = 0


It is in the form of-------


x² + (a+b)x + ab = (x+a) (x+b)

x = x, a = √3 , b = 1

bt there is minus (-)...

so,


=> (x-√3) (x-1) = 0

x = √3

x = 1


Method ll:


=> x² - (√3+1)x + √3 = 0

=> x² -√3x -1x + √3 = 0

=> x (x-√3) -1 (x-√3) = 0

=> (x-√3) (x-1)


x = √3

x = 1



Hope it hlpz....

Answer 2

4x^2 + 4√3x +3 =0

(2x)^2 + 2X2xX√3 +(√3)^2-(√3)^2+3=0

(2x)^2 +2X2xX√3 + 3-3=0

[2x +√3]^2 = 0

(2x+√3)=0

x= -√3/2 &-√3/2

Use the same procedure for any other relatives equation