Question

if x*2-6x+8=0then find the value of alpha*2-beta*2
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Answer 1

x²-6x+8 = 0

x²-2x-4x+8 = 0

x(x-2) - 4(x-2) = 0

(x-4)(x-2) = 0

x = 4 or 2

∴ α = 4 or 2 and β = 2 or 4

α²-β² = 16-4 or 4-16

= ±12

Alternatively,

Let x²-6x+8 = ax²+bx+c

Then a = 1

b = -6

c = 8

Sum of roots (α+β) = -b/a

= -(-6)/1

= 6

Difference of roots (α-β) = |D|/2a

where D is the determinant and |D| is the modulus of determinant

D = b²-4ac

= 36 - 4×1×8

= 36 - 32

= 4

|D|/2a = 4/2×1

= 2

Removing the modulus, we get α-β = ±2

α²-β² = (α+β)(α-β)

= 6 × ±2

= ±12