Q. 1 E3.9( 21 Votes )

# <span lang="EN-US

(2x - 1) (x - 3) = (x + 5)(x - 1)

Here LHS = (2x – 1)(x - 3) = 2x2 –x - 6x + 3 = 2x2 - 7x + 3

RHS = (x + 5)(x - 1) = x2 + 5x – x - 5 = x2 + 4x - 5

Equating LHS and RHS

2x2 - 7x + 3 = x2 + 4x - 5

x2 – 11x + 8 = 0

which is of the form ax2 + bx + c = 0

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