solution
Given that x² - 3x + 2 =0
⇒x² - 3x + 2 =0
⇒x² - (2+1)x + 2 =0
⇒x² - (2x+x) + 2 =0
⇒x² - 2x - x + 2 =0
⇒x(x-2)-1(x-2)=0 { take common (x-2) }
⇒(x-2)(x-1)=0
⇒(x-2)=0 or (x-1)=0
⇒x=0+2 or x=0+1
⇒x=2 or x=1
Hence the value of x=1,2
x²+3x+2 factorise
x² + 3x + 2
= x² + (2+1)x + 2
= x² + (2x+x) + 2
= x² + 2x + x + 2
= x(x+2)+1(x+2) { take common (x+2) }
= (x+2)(x+1)
x^2-3x-2=0 completing the square
Solution
given that x² - 3x - 2 =0
(x)² - 2.x.3/2+ (3/2)² -(3/2)²-2 =0
(x-3/2)² -(3/2)²-2 =0
(x-3/2)² =0+(3/2)²+2
(x-3/2)² =(3/2)²+2
(x-3/2)² =9/4+2
(x-3/2)² =17/4
(x-3/2) = ±√(17/4)
x = 3/2 ± √17/2
x = 3/2 ± √17/2
here x=3/2+√17/2
x=3/2 - √17/2
x^2-3x+2=0 quadratic formula
solution
given that x² - 3x + 2 =0
here a=1 ,b=-3, c=2
x ={-b ± √(b^2-4ac)}/2a
put value a,b and c
x ={-(-3) ± √((-3)^2-4x1x2)}/2x1
x ={3± √(9-8)}/2x1
x ={3 ± √(1)}/2
for positive sign,
x=(3+1)/2
x=4/2=2
for negative sign
x=(3-1)/2
x=2/2=1
hence x=2,1 answer.
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