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problem involving roots
- Thread starter shweta
- Start date
Something is wrong in your question .. the two roots of the given quadratic are NOT reciprocals of each other. Sometimes problems like this have an unknown parameter in the quadratic, which you are to find such that the statement about reciprocal roots would be true. For instance, the "1" might be a parameter "k" which is to be determined.how to solve this one
If one root of 5x2+ 13x +1=0, is reciprocal of other then what is x equal to??
We need to see your work. Please check that you have copied the problem correctly, and show us what steps you have considered.
This is kind of a neat problem about roots.
\(\displaystyle h\ and\ \dfrac{1}{h}\ are\ roots\ of\ ax^2 + bx + c \implies \ c = a,\ and\ b = \dfrac{-a(h^2 + 1)}{h}.\)
Proof
\(\displaystyle a(x - h)\left(x - \dfrac{1}{h}\right) = 0 \implies a\left(x^2 - hx - \dfrac{x}{h} + 1\right) = 0 \implies\)
\(\displaystyle a\left(x^2 - \dfrac{h^2x - x}{h} + 1\right) = 0 \implies ax^2 + \left(\dfrac{-a(h^2 + 1)}{h}\right)x + a = 0 \implies\)
\(\displaystyle b = \dfrac{-a(h^2 + 1)}{h}\ and\ c = a.\)
\(\displaystyle \ c = a,\ and\ b = \dfrac{-a(h^2 + 1)}{h} \implies h\ and\ \dfrac{1}{h}\ are\ roots\ of\ ax^2 + bx + c.\)
Proof
\(\displaystyle ax^2 + bx + c = 0 \implies ax^2 + \left(\dfrac{-a(h^2 + 1)}{h}\right)x + a = 0 \implies x^2 + \left(\dfrac{-(h^2 + 1)}{h}\right)x + 1 = 0 \implies\)
\(\displaystyle x = \dfrac{\dfrac{h^2 + 1}{h} \pm \sqrt{\dfrac{h^4 + 2h^2 + 1}{h^2} - (4 * 1 * 1)}}{2 * 1} = \dfrac{\dfrac{h^2 + 1}{h} \pm \sqrt{\dfrac{h^4 + 2h^2 + 1}{h^2} - \dfrac{4h^2}{h^2}}}{2} =\)
\(\displaystyle \dfrac{\dfrac{h^2 + 1}{h} \pm \sqrt{\dfrac{h^4 - 2h^2 + 1}{h^2}}}{2} = \dfrac{\dfrac{h^2 + 1}{h} \pm \sqrt{\left(\dfrac{h^2 - 1}{h}\right)^2}}{2} \implies\)
\(\displaystyle x = \dfrac{\dfrac{h^2 + 1}{h} + \dfrac{h^2 - 1}{h}}{2} = \dfrac{\dfrac{2h^2}{h}}{2} = h\ OR\)
\(\displaystyle x = \dfrac{\dfrac{h^2 + 1}{h} - \dfrac{h^2 - 1}{h}}{2} = \dfrac{\dfrac{2}{h}}{2} = \dfrac{1}{h}.\)
Needless to say, the original post will not work because \(\displaystyle 5 \ne 1.\)
\(\displaystyle h\ and\ \dfrac{1}{h}\ are\ roots\ of\ ax^2 + bx + c \implies \ c = a,\ and\ b = \dfrac{-a(h^2 + 1)}{h}.\)
Proof
\(\displaystyle a(x - h)\left(x - \dfrac{1}{h}\right) = 0 \implies a\left(x^2 - hx - \dfrac{x}{h} + 1\right) = 0 \implies\)
\(\displaystyle a\left(x^2 - \dfrac{h^2x - x}{h} + 1\right) = 0 \implies ax^2 + \left(\dfrac{-a(h^2 + 1)}{h}\right)x + a = 0 \implies\)
\(\displaystyle b = \dfrac{-a(h^2 + 1)}{h}\ and\ c = a.\)
\(\displaystyle \ c = a,\ and\ b = \dfrac{-a(h^2 + 1)}{h} \implies h\ and\ \dfrac{1}{h}\ are\ roots\ of\ ax^2 + bx + c.\)
Proof
\(\displaystyle ax^2 + bx + c = 0 \implies ax^2 + \left(\dfrac{-a(h^2 + 1)}{h}\right)x + a = 0 \implies x^2 + \left(\dfrac{-(h^2 + 1)}{h}\right)x + 1 = 0 \implies\)
\(\displaystyle x = \dfrac{\dfrac{h^2 + 1}{h} \pm \sqrt{\dfrac{h^4 + 2h^2 + 1}{h^2} - (4 * 1 * 1)}}{2 * 1} = \dfrac{\dfrac{h^2 + 1}{h} \pm \sqrt{\dfrac{h^4 + 2h^2 + 1}{h^2} - \dfrac{4h^2}{h^2}}}{2} =\)
\(\displaystyle \dfrac{\dfrac{h^2 + 1}{h} \pm \sqrt{\dfrac{h^4 - 2h^2 + 1}{h^2}}}{2} = \dfrac{\dfrac{h^2 + 1}{h} \pm \sqrt{\left(\dfrac{h^2 - 1}{h}\right)^2}}{2} \implies\)
\(\displaystyle x = \dfrac{\dfrac{h^2 + 1}{h} + \dfrac{h^2 - 1}{h}}{2} = \dfrac{\dfrac{2h^2}{h}}{2} = h\ OR\)
\(\displaystyle x = \dfrac{\dfrac{h^2 + 1}{h} - \dfrac{h^2 - 1}{h}}{2} = \dfrac{\dfrac{2}{h}}{2} = \dfrac{1}{h}.\)
Needless to say, the original post will not work because \(\displaystyle 5 \ne 1.\)
HallsofIvy
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