A quadratic equation x2+bx+c=0 has two real roots. It the difference between the reciprocals of the roots is $1 / 3$and the sum of the reciprocals of the squares of the roots is $5 / 9 , then the largest possible value of (b+c) is
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Official Correct Answer: 9. Okay, so I've got this quadratic equation problem from CAT 2023 Slot 3. It's about a quadratic equation \(x^2 + bx + c = 0\) with two real roots. The problem gives me two conditions: the difference between the reciprocals of the roots is \(1/3\), and the sum of the reciprocals of the squares of the roots is \(5/9\). We find the largest possible value of \(b + c\). Hmm, let's break this down step by step.
First, for a quadratic equation \(x^2 + bx + c = 0\), the sum of the roots, let's call them \(r_1\) and \(r_2\), is equal to \(-b\). That's from Vieta's formulas. So, \(r_1 + r_2 = -b\). Similarly, the product of the roots is \(c\), so \(r_1 r_2 = c\). Got that down.
Now, the problem mentions the difference between the reciprocals of the roots is \(1/3\). So, that would be \(\frac{1}{r_1} - \frac{1}{r_2} = \frac{1}{3}\). I can rewrite this as \(\frac{r_2 - r_1}{r_1 r_2} = \frac{1}{3}\). Since \(r_1 r_2 = c\), this becomes \(\frac{r_2 - r_1}{c} = \frac{1}{3}\), which implies \(r_2 - r_1 = \frac{c}{3}\).
But wait, I also know that the quadratic has real roots, so the discriminant must be non-negative. The discriminant is \(b^2 - 4c \geq 0\). That might come into play later.
Next, the sum of the reciprocals of the squares of the roots is \(5/9\). So, \(\frac{1}{r_1^2} + \frac{1}{r_2^2} = \frac{5}{9}\). I can express this in terms of \(r_1 + r_2\) and \(r_1 r_2\). Let me recall that \(\frac{1}{r_1^2} + \frac{1}{r_2^2} = \left(\frac{1}{r_1} + \frac{1}{r_2}\right)^2 - 2\left(\frac{1}{r_1 r_2}\right)\).
From earlier, I have \(\frac{1}{r_1} - \frac{1}{r_2} = \frac{1}{3}\). Let \(S = \frac{1}{r_1} + \frac{1}{r_2}\) and \(D = \frac{1}{r_1} - \frac{1}{r_2} = \frac{1}{3}\). Then, \(\frac{1}{r_1^2} + \frac{1}{r_2^2} = S^2 - 2\left(\frac{1}{r_1 r_2}\right)\).
I know that \(S^2 - 2\left(\frac{1}{r_1 r_2}\right) = \frac{5}{9}\). Also, \(S\) can be expressed as \(\frac{r_1 + r_2}{r_1 r_2}\). Since \(r_1 + r_2 = -b\) and \(r_1 r_2 = c\), this becomes \(S = \frac{-b}{c}\).
So, plugging back into the equation: \(\left(\frac{-b}{c}\right)^2 - 2\left(\frac{1}{c}\right) = \frac{5}{9}\). Simplifying, that's \(\frac{b^2}{c^2} - \frac{2}{c} = \frac{5}{9}\).
Now, I also have the equation from the difference of reciprocals: \(r_2 - r_1 = \frac{c}{3}\). But \(r_2 - r_1\) can be expressed in terms of the discriminant. The difference of roots is \(\sqrt{(r_1 + r_2)^2 - 4 r_1 r_2}\). So, \(r_2 - r_1 = \sqrt{(r_1 + r_2)^2 - 4 r_1 r_2}\).
Substituting the known values, \(r_2 - r_1 = \sqrt{(-b)^2 - 4c} = \sqrt{b^2 - 4c}\). But we also have \(r_2 - r_1 = \frac{c}{3}\), so \(\sqrt{b^2 - 4c} = \frac{c}{3}\). Squaring both sides, we get \(b^2 - 4c = \frac{c^2}{9}\).
So now, I have two equations:
1. \(\frac{b^2}{c^2} - \frac{2}{c} = \frac{5}{9}\)
2. \(b^2 - 4c = \frac{c^2}{9}\)
Let me try to solve these simultaneously. From equation 2, I can express \(b^2\) as \(b^2 = 4c + \frac{c^2}{9}\). Plugging this into equation 1:
\(\frac{4c + \frac{c^2}{9}}{c^2} - \frac{2}{c} = \frac{5}{9}\)
Simplify the first term: \(\frac{4c}{c^2} + \frac{c^2}{9c^2} = \frac{4}{c} + \frac{1}{9}\)
So, the equation becomes: \(\frac{4}{c} + \frac{1}{9} - \frac{2}{c} = \frac{5}{9}\)
Combine like terms: \(\frac{2}{c} + \frac{1}{9} = \frac{5}{9}\)
Subtract \(\frac{1}{9}\) from both sides: \(\frac{2}{c} = \frac{4}{9}\)
So, \(\frac{2}{c} = \frac{4}{9}\) implies \(c = \frac{2 \times 9}{4} = \frac{18}{4} = \frac{9}{2}\). So, \(c = 4.5\) or \(9/2\).
Now, plug \(c = 9/2\) back into equation 2