Question

If alpha and beta are zeroes of 6y^2 -7y +2 . Find a quadratic polynomial whose zeroes are 1/alpha and 1/beta.
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Answer 1

Answer:

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Answer 2

Answer:

The required quadratic equation is

$\sf\:2y^{2}\:-\:7y\:+\:6\:=\:0$

Step-by-step-explanation:

NOTE : Kindly refer to the attachment first.

The given quadratic equation is

$\sf\:6y^{2}\:-\:7y\:+\:2\:=\:0$

$\sf\alpha\:and\:\beta$ are zeroes of the given quadratic equation.

Now, comparing $\sf\:6y^{2}\:-\:7y\:+\:2\:=\:0$ with $\sf\:ax^{2}\:+\:bx\:+\:c\:=\:0$, we get,

$\sf\:a\:=\:6,\:b\:=\:-\:7,\:c\:=\:2$

Now, we know that,

$\boxed{\red{\sf\:\alpha\:+\:\beta\:=\:-\:\frac{b}{a}}}\\\\\implies\sf\:\alpha\:+\:\beta\:=\:-\:\frac{(-\:7\:)}{6}\\\\\implies\boxed{\sf\:\alpha\:+\:\beta\:=\:\frac{7}{6}}\:\:-\:-\:-\:(\:1\:)</p><p>$

Now,

$\boxed{\red{\sf\:\alpha\:.\:\beta\:=\:\frac{c}{a}}}\\\\\implies\sf\:\alpha\:.\:\beta\:=\:\frac{\cancel2}{\cancel6}\\\\\implies\boxed{\sf\:\alpha\:.\:\beta\:=\:\frac{1}{3}}\:\:-\:-\:(\:2\:)</p><p>$

Now, the required quadratic equation is in the form

$\boxed{\pink{\sf\:x^{2}\:-\:(\:\alpha\:+\:\beta\:)\:x\:+\:(\:\alpha\:.\:\beta\:)\:=\:0}}$

But, here the roots of required quadratic equation are

$\sf\:\frac{1}{\alpha}\:and\:\frac{1}{\beta}$

$\therefore$ The required quadratic equation is in form

$\sf\:y^{2}\:-\:(\:\frac{1}{\alpha}\:+\:\frac{1}{\beta}\:)\:y\:+\:(\:\frac{1}{\alpha}\:.\:\frac{1}{\beta}\:)\:=\:0\\\\\implies\sf\:y^{2}\:-\:(\:\frac{\beta\:+\:\alpha}{\alpha\:.\:\beta}\:)\:y\:+\:(\:\frac{1}{\alpha\:.\:\beta}\:)\:=\:0\\\\\implies\sf\:y^{2}\:-\:(\:\frac{\alpha\:+\:\beta}{\alpha\:.\:\beta}\:)\:y\:+\:(\:\frac{1}{\alpha\:.\:\beta}\:)\:=\:0$

$\implies\sf\:y^{2}\:-\:(\:\dfrac{\frac{7}{6}}{\frac{1}{3}}\:) \: y\:+\:(\:\dfrac{\frac{1}{1}}{\frac{1}{3}}\:)\:=\:0\:\:\:-\:-\:[\:From\:(\:1\:)\:\&\:(\:2\:)\:]\\\\\implies\sf\:y^{2}\:-\:(\:\frac{7}{\cancel6}\:\times\:\frac{\cancel3}{1}\:)\:y\:+\:(\:\frac{\cancel1}{1}\:\times\:\frac{3}{\cancel1}\:)\:=\:0\\\\\implies\sf\:y^{2}\:-\:(\:\frac{7}{2}\:)\:y\:+\:3\:=\:0\\\\\implies\sf\:y^{2}\:-\:\frac{7}{2}\:y\:+\:3\:=\:0\\\\\implies\boxed{\red{\sf\:2y^{2}\:-\:7y\:+\:6\:=\:0}}\:\:[\sf\:Multiplying\:both\:sides\:by\:2\:]</p><p>$