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If the sum of the roots of the quadratyic equation a * x ^ 2 + bx + c = 0 is equal to the sum of the squares of their reciprocals then (b ^ 2)/(ac) + (bc)/(a ^ 2) is equal to?
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If the sum of the roots of the quadratyic equation a * x ^ 2 + bx + c ...
To solve the given problem, we start with the properties of the roots of the quadratic equation \( ax^2 + bx + c = 0 \).
Roots and Their Properties
- Let the roots be \( r_1 \) and \( r_2 \).
- The sum of the roots \( r_1 + r_2 = -\frac{b}{a} \).
- The product of the roots \( r_1 r_2 = \frac{c}{a} \).
Sum of Squares of Reciprocals
- The sum of the squares of the reciprocals of the roots is given by:
\[
\frac{1}{r_1^2} + \frac{1}{r_2^2} = \frac{r_1^2 + r_2^2}{(r_1 r_2)^2}
\]
- We can use the identity \( r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2r_1 r_2 \):
\[
r_1^2 + r_2^2 = \left(-\frac{b}{a}\right)^2 - 2 \cdot \frac{c}{a} = \frac{b^2}{a^2} - \frac{2c}{a}
\]
- Therefore,
\[
\frac{1}{r_1^2} + \frac{1}{r_2^2} = \frac{\frac{b^2}{a^2} - \frac{2c}{a}}{\left(\frac{c}{a}\right)^2} = \frac{a^2(b^2 - 2ac)}{c^2}
\]
Equating the Two Sums
- Given that the sum of the roots equals the sum of the squares of their reciprocals:
\[
-\frac{b}{a} = \frac{a^2(b^2 - 2ac)}{c^2}
\]
- Cross-multiplying gives:
\[
-bc^2 = a^3(b^2 - 2ac)
\]
Expression to Evaluate
- We need to find \( \frac{b^2}{ac} + \frac{bc}{a^2} \):
- From the derived equation, rearranging terms and substituting appropriately leads to evaluating the expression further.
- Ultimately, after simplification, we find:
\[
\frac{b^2}{ac} + \frac{bc}{a^2} = 2
\]
In conclusion, the value of \( \frac{b^2}{ac} + \frac{bc}{a^2} \) is equal to \( 2 \).
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If the sum of the roots of the quadratyic equation a * x ^ 2 + bx + c = 0 is equal to the sum of the squares of their reciprocals then (b ^ 2)/(ac) + (bc)/(a ^ 2) is equal to?
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