Question

If α and β are zeros of the polynomial 6x^2-7x-3. Then form a quadratic polynomial whose zeros are 1/α and 1/β.

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Answer 1

Step-by-step explanation:

$6 {x}^{2} - 7x - 3 = 0$

let Roots are A and B

$\\ \frac{1}{a} + \frac{1}{b} = \frac{7}{ - 0.5} = -14$

$\frac{1}{ab} = - 2$

Quadratic Required

${x}^{2} + 14x - 2 = 0$

Answer 2

From the given quadratic equation $6x^2-7x-3=0,$ since the roots are $\alpha$ and $\beta,$

$\alpha+\beta=\dfrac {7}{6}\\\\\\\alpha\beta=-\dfrac {3}{6}=-\dfrac {1}{2}$

We're given to find a quadratic equation whose roots are $\dfrac {1}{\alpha}$ and $\dfrac {1}{\beta}.$ Then,

sum of the roots,

$\dfrac {1}{\alpha}+\dfrac{1}{\beta}=\dfrac {\alpha+\beta}{\alpha\beta}\\\\\\\dfrac {1}{\alpha}+\dfrac{1}{\beta}=\dfrac {\frac {7}{6}}{-\frac {1}{2}}\\\\\\\dfrac {1}{\alpha}+\dfrac{1}{\beta}=-\dfrac {7}{3}$

and, product of the roots,

$\dfrac {1}{\alpha}\cdot\dfrac {1}{\beta}=\dfrac {1}{\alpha\beta}\\\\\\\dfrac {1}{\alpha}\cdot\dfrac {1}{\beta}=\dfrac {1}{-\frac {1}{2}}\\\\\\\dfrac {1}{\alpha}\cdot\dfrac {1}{\beta}=-2$

Then the equation is,

$x^2+\dfrac {7}{3}x-2=0$

Well, we can multiply the equation by suitable integers to avoid the fractional coefficients. So, the equation is multiplied by 3, and it can be,

$\mathbf{3x^2+7x-6=0}$

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Suppose we're given the equation,

$ax^2+bx+c=0$

whose roots are $\alpha$ and $\beta$ and we're given to find a quadratic equation whose roots are $\dfrac {1}{\alpha}$ and $\dfrac {1}{\beta}.$ Then we have,

$\alpha+\beta=-\dfrac {b}{a}\\\\\\\alpha\beta=\dfrac {c}{a}$

Then, in the new equation, sum of the roots,

$\dfrac {1}{\alpha}+\dfrac {1}{\beta}=\dfrac {\alpha+\beta}{\alpha\beta}\\\\\\\dfrac {1}{\alpha}+\dfrac {1}{\beta}=\dfrac {-\frac{b}{a}}{\frac{c}{a}}\\\\\\\dfrac {1}{\alpha}+\dfrac {1}{\beta}=-\dfrac {b}{c}}$

and product of the roots,

$\dfrac {1}{\alpha}\cdot\dfrac {1}{\beta}=\dfrac {1}{\alpha\beta}\\\\\\\dfrac {1}{\alpha}\cdot\dfrac {1}{\beta}=\dfrac {1}{\frac{c}{a}}\\\\\\\dfrac {1}{\alpha}\cdot\dfrac {1}{\beta}=\dfrac {a}{c}$

Then the equation is,

$x^2+\dfrac {b}{c}x+\dfrac {a}{c}=0$

Multiplying the equation by $c$ to avoid the fractional coefficients, we get one of our answers (because there are infinitely many possible answers, according to the non - zero integer multiplied with the fraction to avoid the fractional coefficients),

$cx^2+bx+a=0$

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So we can directly reach to the conclusion from the given equation, that our answer can be,

$\mathbf{-3x^2-7x+6=0}$

Just multiply it by -1. We get what we've got earlier!

#answerwithquality

#BAL