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01 Oct 2024, 11:11
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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If a and b are the two solutions of the equation \(x^2 - 5x + 4 = 0\), what is the value of \((\frac{1 + a}{a})(\frac{1 + b}{b})\)?
Give your answer as a fraction.
Give your answer as a fraction.
09 Feb 2025, 02:37
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Factoring the quadratic equation \(x^2 - 5x + 4 = 0\), you get \((x - 1)(x - 4) = 0\), so the two solutions are either a = 1 and b = 4 or a = 4 and b = 1. Note that a and b are interchangeable in the expression \((\frac{1+a}{a})(\frac{1 + b}{b})\) so the value of the expression will be the same regardless of the choices of a and b. Thus \((\frac{1 + a}{a})(\frac{1 + b}{b}) = (\frac{1 + 1}{1})(\frac{1 + 4}{4}) = (2)(\frac{5}{4}) = \frac{5}{2}\), and the correct answer is \(\frac{5}{2}.\)
Factoring the quadratic equation \(x^2 - 5x + 4 = 0\), you get \((x - 1)(x - 4) = 0\), so the two solutions are either a = 1 and b = 4 or a = 4 and b = 1. Note that a and b are interchangeable in the expression \((\frac{1+a}{a})(\frac{1 + b}{b})\) so the value of the expression will be the same regardless of the choices of a and b. Thus \((\frac{1 + a}{a})(\frac{1 + b}{b}) = (\frac{1 + 1}{1})(\frac{1 + 4}{4}) = (2)(\frac{5}{4}) = \frac{5}{2}\), and the correct answer is \(\frac{5}{2}.\)
02 Jun 2025, 19:26
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\((\frac{1 + a}{a})(\frac{1 + b}{b})\)
=> \(\frac{(1 + a) * (1 + b)}{ab} \)
=> \(\frac{1 + a + b + ab}{ab}\)
a and b are the two solutions of the equation \(x^2 - 5x + 4 = 0\)
We know that for a generic equation \(Ax^2 + Bx + C = 0\)
Sum of roots = \(\frac{-B}{A}\)
Product of roots = \(\frac{C}{A}\)
Comparing \(x^2 - 5x + 4 = 0\) with \(Ax^2 + Bx + C = 0\) we have
A = 1, B = -5, C = 4
=> Sum of Roots a and b = a + b = \(\frac{-B}{A}\) = \(\frac{-(-5)}{1}\) = 5
=> Product of roots a and b = a * b = \(\frac{C}{A}\) = \(\frac{4}{1}\) = 4
=> \(\frac{1 + a + b + ab}{ab}\) = \(\frac{1 + 5 + 4}{4}\) = \(\frac{10}{4}\) = \(\frac{5}{2}\)
So, Answer will be \(\frac{5}{2}\)
Hope it helps!
=> \(\frac{(1 + a) * (1 + b)}{ab} \)
=> \(\frac{1 + a + b + ab}{ab}\)
a and b are the two solutions of the equation \(x^2 - 5x + 4 = 0\)
We know that for a generic equation \(Ax^2 + Bx + C = 0\)
Sum of roots = \(\frac{-B}{A}\)
Product of roots = \(\frac{C}{A}\)
Comparing \(x^2 - 5x + 4 = 0\) with \(Ax^2 + Bx + C = 0\) we have
A = 1, B = -5, C = 4
=> Sum of Roots a and b = a + b = \(\frac{-B}{A}\) = \(\frac{-(-5)}{1}\) = 5
=> Product of roots a and b = a * b = \(\frac{C}{A}\) = \(\frac{4}{1}\) = 4
=> \(\frac{1 + a + b + ab}{ab}\) = \(\frac{1 + 5 + 4}{4}\) = \(\frac{10}{4}\) = \(\frac{5}{2}\)
So, Answer will be \(\frac{5}{2}\)
Hope it helps!
