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Question
Solve the following equation by factorisation :
`sqrt(3x^2 - 2x - 1) = 2x - 2`
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Solution
`sqrt(3x^2 - 2x - 1) = 2x - 2`
Squaring both sides
3x2 - 2x – 1 = (2x - 2)2
⇒ 3x2 – 2x – 1 = 4x2 – 8x + 4
⇒ 4x2 – 8x + 4 – 3x2 + 2x + 1 = 0
⇒ x2 – 6x + 5 = 0
⇒ x2 – 5x – x + 5 = 0
⇒ x(x – 5) –1(x – 5) = 0
⇒ (x – 5)(x – 1) = 0
Either x – 5 = 0,
then x = 5
or
x – 1 = 0,
then x = 1
Check :
(i) If x = 5, then
L.H.S. = `sqrt(3x^2 - 2x - 1)`
= `sqrt(3 xx (5)^2 - 2 xx 5 - 1)`
= `sqrt(3 xx 25 - 10 - 1)`
= `sqrt(75 - 10 - 1)`
= `sqrt(64)`
= 8
R.H.S. = 2x – 2
= 2 x 5 – 2
= 10 – 2
= 8
∵ L.H.S. = R.H.S.
∴ x = 5 is a root
(ii) If x = 1, then
L.H.S. = `sqrt(3x^2 - 2x - 1)`
= `sqrt(3(1)^2 - 2(1) - 1)`
= `sqrt(3 xx 1 - 2 - 1)`
= `sqrt(3 - 2 - 1)`
= 0
R.H.S. = 2x – 2
= 2 x 1 – 2
= 2 – 2
= 0
∵ L.H.S. = R.H.S.
∴ x = 1 is also its root
Hence x = 5, 1.
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