Question:
The number of integers that satisfy the equality \((x^2-5x+7)^{x+1} = 1\) is
The number of integers that satisfy the equality \((x^2-5x+7)^{x+1} = 1\) is
Updated On: Sep 17, 2024
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The Correct Option is B
Solution and Explanation
The correct answer is (B): \(3\)
\((x^2-5x+7)^{x+1} = 1\)
We know, for \(a^b=1\), if
\(-a =-1\) then \(b\) is even.
\(-a = 1 \) then \(b\) is any number
\(-a>0\) then \(b=0\)
Case 1: \(x+1=0 ⇒ x = -1\)
Case 2: \(x^2-5x+7 = 1 ⇒ x^2-5x+6=0 ⇒ x = 2 \;or \;3\)
Case 3: \(x^2-5x+7 = -1 ⇒ x^2-5x+8=0\)
but \( x\) is not an integer
\(∴\) The number of integers satisfies the equation is \(3\)
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