If alpha and beta are the two zeroes of the quadratic polynomial x^(2)

1.	If alpha and beta are the two zeroes of the quadratic polynomial x^(2)-3x+7 find a quadratic polynomial whose zeroes are (1)/(alpha) and (1)/(beta) .
Answer» $\bigstar$ Answer x² - 3/7 x + 1/7 $\bigstar$ Given If α and β are the TWO ZEROES of the quadratic POLYNOMIAL x²-3x+7 $\bigstar$ To Find A quadratic polynomial WHOSE zeroes are ¹/α & ¹/β $\bigstar$ Solution $\rm Given\ polynomial\ is\ x^2-3x+7\\\\\rm Compare\ this\ with\ ax^2+bx+c,we\ get,\\\\\rm \bullet \;\; a=1,b=-3,c=7\\\\\rm Sum\ of\ zroes\ , \alpha + \beta =\dfrac{-b}{a}\\\\\implies \rm \alpha + \beta =\dfrac{-(-3)}{1}\\\\\implies \rm \alpha + \beta =3...(1)\\\\\rm Product\ of\ zeroes\ ,\alpha .\beta =\dfrac{c}{a}\\\\\implies \rm \alpha .\beta =\dfrac{7}{1}\\\\\implies \rm \alpha .\beta =7 ...(2)$ $\rm Now,If\ \dfrac{1}{\alpha },\dfrac{1}{\beta } \; are\ zeroes\ of\ the\ polynomial\ ,\\\\\rm Sum\ of\ zeroes\ =\dfrac{1}{\alpha }+\dfrac{1}{\beta }\\\\\implies \rm \dfrac{\alpha + \beta}{\alpha . \beta}\\\\\implies \rm \dfrac{3}{7}\;\;[\;From\ (1) \& (2)\;]\\\\\rm Product\ of\ zeroes=\dfrac{1}{\alpha}.\dfrac{1}{\beta}\\\\\implies \rm \dfrac{1}{\alpha . \beta}\\\\\implies \rm \dfrac{1}{7}\;\; [\; From\ (2) \;]\\\\\rm General\ form\ of\ polynomial\ is ,\\\\$ $\rm x^2-(Sum\ of\ zeroes)x+(Product\ of\ zeroes)\\\\\implies \rm x^2-\dfrac{3}{7}x+\dfrac{1}{7}\\\\$

If alpha and beta are the two zeroes of the quadratic polynomial x^(2)-3x+7 find a quadratic polynomial whose zeroes are (1)/(alpha) and (1)/(beta) .

Answer»

$\bigstar$ Answer

x² - 3/7 x + 1/7

$\bigstar$ Given

If α and β are the TWO ZEROES of the quadratic POLYNOMIAL x²-3x+7

$\bigstar$ To Find

A quadratic polynomial WHOSE zeroes are ¹/α & ¹/β

$\bigstar$ Solution

$\rm Given\ polynomial\ is\ x^2-3x+7\\\\\rm Compare\ this\ with\ ax^2+bx+c,we\ get,\\\\\rm \bullet \;\; a=1,b=-3,c=7\\\\\rm Sum\ of\ zroes\ , \alpha + \beta =\dfrac{-b}{a}\\\\\implies \rm \alpha + \beta =\dfrac{-(-3)}{1}\\\\\implies \rm \alpha + \beta =3...(1)\\\\\rm Product\ of\ zeroes\ ,\alpha .\beta =\dfrac{c}{a}\\\\\implies \rm \alpha .\beta =\dfrac{7}{1}\\\\\implies \rm \alpha .\beta =7 ...(2)$

$\rm Now,If\ \dfrac{1}{\alpha },\dfrac{1}{\beta } \; are\ zeroes\ of\ the\ polynomial\ ,\\\\\rm Sum\ of\ zeroes\ =\dfrac{1}{\alpha }+\dfrac{1}{\beta }\\\\\implies \rm \dfrac{\alpha + \beta}{\alpha . \beta}\\\\\implies \rm \dfrac{3}{7}\;\;[\;From\ (1) \& (2)\;]\\\\\rm Product\ of\ zeroes=\dfrac{1}{\alpha}.\dfrac{1}{\beta}\\\\\implies \rm \dfrac{1}{\alpha . \beta}\\\\\implies \rm \dfrac{1}{7}\;\; [\; From\ (2) \;]\\\\\rm General\ form\ of\ polynomial\ is ,\\\\$

$\rm x^2-(Sum\ of\ zeroes)x+(Product\ of\ zeroes)\\\\\implies \rm x^2-\dfrac{3}{7}x+\dfrac{1}{7}\\\\$

If alpha and beta are the two zeroes of the quadratic polynomial x^(2)-3x+7 find a quadratic polynomial whose zeroes are (1)/(alpha) and (1)/(beta) .

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If alpha and beta are the two zeroes of the quadratic polynomial x^(2)-3x+7 find a quadratic polynomial whose zeroes are (1)/(alpha) and (1)/(beta) .​

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If alpha and beta are the two zeroes of the quadratic polynomial x^(2)-3x+7 find a quadratic polynomial whose zeroes are (1)/(alpha) and (1)/(beta) .