Question

find the quadratic equation whose zeroes are 2 and 1/3
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Answer 1

Given:-

• α = 2

• β = $\sf\dfrac{1}{3}$

Solution:-

• sum of roots (α+β) = $\sf 2+\dfrac{1}{3} = \dfrac{7}{3}$

• product of roots (αβ) = $\sf 2\times\dfrac{1}{3} = \dfrac{2}{3}$

Formula:-

• quadratic equation = x²- (α+β)x + (αβ)

substituting the above values in the formula:-

• quadratic equation = $\sf x^{2} -\bigg(\dfrac{7}{3}\bigg) x +\dfrac{2}{3}$

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ADDITIONAL INFORMATION:-

Discriminant of a quadratic equation is calculated by the following formula:-

• D = b² - 4ac

where, D,b, a, and c represents discriminant, coefficient of x, coefficient of x², and constant respectively.

Nature of roots with the help of discriminant:-

• when D > 0, we have real and distinct roots
• when D = 0, we have real and equal roots.  
• when D < 0, we have imaginary roots.

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Hope it's helpful !!

Answer 2

Given:-

$\sf 2 \: and \: \dfrac{1}{3}$ are the roots of a quadratic equation

To Find :-

The Quadratic equation

Solution :-

${ \sf { Let \: us \: assume \: that \: , } }$

$\sf \leadsto \alpha = 2$

$\sf \leadsto \beta = \dfrac{1}{3}$

${ \sf { Now \: Sum \: of \: roots \: (S) \: , } }$

${ \sf { : \longmapsto { S = \alpha + \beta } } }$

${ \sf { : \longmapsto { S = 2 + \dfrac{1}{3} } } }$

${ \sf { : \longmapsto { S = \dfrac{7}{3} } } }$

${ \sf { Now \: Product \: of \: roots \: (P) \: } }$

${ \sf { : \longmapsto { P = 2 × \dfrac{1}{3} } } }$

${ \sf { : \longmapsto { P = \dfrac{2}{3} } } }$

${ \sf { As \: we \: knows \: that , \: a \: Quadratic \: Equation \: is , } }$

${ \sf { x² - Sx + P = 0 } }$

${ \sf { : \longmapsto { x² - \dfrac{7}{3}x + \dfrac{2}{3} = 0 } } }$

${ \sf { Multiplying \: both \: sides \: by \: 3 } }$

${ \sf { : \longmapsto { 3 × x² - \dfrac{7}{3} × 3 × x + \dfrac{2}{3} × 3 } } }$

${ \sf { : \longmapsto { 3x² - 7x + 2 = 0 } } }$

Henceforth , The Required equation is " 3x² - 7x + 2 = 0 " .