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प्रश्न
In the following, determine whether the given values are solutions of the given equation or not:
`sqrt(x^2 - 4x + 3) + sqrt(x^2 - 9) = sqrt(4x^2 - 14x + 16), x = 3`
बेरीज
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उत्तर
Given: `sqrt(x^2 - 4x + 3) + sqrt(x^2 - 9) = sqrt(4x^2 - 14x + 16), x = 3`
x2 – 4x + 3 = 9 – 12 + 3 = 0
⇒ `sqrt(0) = 0`
x2 – 9 = 9 – 9 = 0
⇒ `sqrt(0) = 0`
LHS = 0 + 0 = 0
4x2 – 14x + 16 = 36 – 42 + 16 = 10
⇒ RHS = `sqrt(10) (≈ 3.1623)`
`0 ≠ sqrt(10)`; also the radicands are defined at x = 3 (x2 – 9 = 0), so substitution is valid.
x = 3 is not a solution of the equation.
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