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Question
Solve the following equation by factorization
`sqrt(3x + 4) = x`
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Solution
`sqrt(3x + 4) = x`
Squaring on both sides
3x + 4 = x2
⇒ x2 - 3x - 4 = 0
⇒ 4x + x - 4 = 0
⇒ x(x - 4) + 1(x - 4) = 0
⇒ (x - 4)(x + 1) = 0
Either x - 4 = 0,
then x = 4
or
x + 1 = 0,
then x = -1
∴ x = 4, - 1
Check
(i) If x = 4, then
L.H.S.
= `sqrt(3x + 4)`
= `sqrt(3 xx 4 + 4)`
= `sqrt(12 + 4)`
= `sqrt(16)`
= 4
R.H.S.
= x
= 4
∴ L.H.S. = R.H.S.
Hence x = 4 is its root
(ii) If x = -1, then
L.H.S.
= `sqrt(3x(-1) + 4)`
= `sqrt(-3 + 4)`
= `sqrt(1)`
= 1
R.H.S.
= x
= -1
∵ L.H.S. ≠ R.H.S.
∴ x = -1 is not its root,
Hence x = 4.
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