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Re: What is the sum of roots of the equation x^2 – 40x + 399 = 0?
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06 Sep 2018, 02:57

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\(x^2-40x+399=0\) \(x^2-40x+400-1=0\) \(x^2-40x+400=1\) \((x-20)^2=1\) \((x-20)=\sqrt{1}\) \(x-20=1\) and \(x-20=-1\) \(x=21\) and \(x=19\) Sum of roots=21+19=40 Answer: D

Re: What is the sum of roots of the equation x^2 – 40x + 399 = 0?
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09 Sep 2018, 17:26

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Let \(p\) and \(q\) be the roots of the equation \(x^2 – 40x + 399 = 0.\) Then \(x^2 – 40x + 399 = (x-p)(x-q) = x^2 – (p+q)x + pq.\) Equating coefficients gives\(p + q = 40\) from the coefficient of \(x\).

Therefore, the answer is D. Answer: D
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Re: What is the sum of roots of the equation x^2 – 40x + 399 = 0?
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22 Sep 2018, 01:33

While you could find the prime factors \(399\) and determine that the two roots are \(19\) and \(21\), there is a quadratic pattern that we can use to answer this question.

The quadratic equation is given by the general form: \(ax^2\)+\(bx\)+\(c\)\(=0\), where:

\(a =\) coefficient of \(x^2\) \(b =\) coefficient of \(x\) \(c =\) constant

A quadratic equation can have two roots and the sum of its roots is given by \(-b\) / \(a\), therefore: