Question

if the distance of the roots of the quadratic equation is 3 and the difference between their cubes is 189 find the quadratic equation

Answer 1

$\underline{\underline{\mathfrak{\Large{Solution : }}}}$



$\sf Let \: the \: zeroes \: are \: \alpha \: and \: \beta .$


$\underline{\textsf{Given,}} \\ \\ \sf \implies \alpha \: - \: \beta \: = \: 3 \quad...(1) \\ \\ \sf \implies (\alpha)^3 \: - \: (\beta)^3 \: = \: 189 \quad...(2)$


$\textsf{Now,} \\ \\ \sf \implies ( \alpha \: - \: \beta ) \: = \: 3 \\ \\ \textsf{Squaring both sides , } \\ \\ \sf \implies ( \alpha \: - \: \beta)^2 \: = \: 3^2 \\ \\ \sf \implies (\alpha)^2 \: + \: (\beta)^2 \: - \: 2\alpha \beta \: = \: 9 \\ \\ \sf \: \: \therefore \: \: (\alpha)^2 \: + \: ( \beta )^2 \: = \: 9 \: + \: 2 \alpha \beta \quad...( 3)$




$\textsf{Again,} \\ \\ \sf \implies (\alpha^3 \: - \: \beta^3 ) \: = \: 189 \\ \\ \sf \implies ( \alpha \: - \: \beta)(\alpha^2 \: + \: \beta^2 \: + \: \alpha \beta ) \: = \: 189 \\ \\ \textsf{Plug the value of (1) and (3) , } \\ \\ \sf \implies 3( 9 \: + \: 2 \alpha \beta \: + \: \alpha \beta ) \: = \: 189 \\ \\ \sf \implies 3( 9 \: + \: 3 \alpha \beta ) \: = \: 189 \\ \\ \sf \implies 3 \: \times \: 3( 3 \: + \: \alpha \beta ) \: = \: 189 \\ \\ \sf \implies 9( 3 \: + \: \alpha \beta ) \: = \: 189\\ \\ \sf \implies 3 \: + \: \alpha \beta \: = \: 189\: \div \: 9 \\ \\ \sf \implies 3 \: + \: \alpha \beta \: = \: 21$



$\sf \implies \alpha \beta \: = \: 21 \: - \: 3 \\ \\ \sf \: \: \therefore \: \: \alpha \beta \: = \: 18 \quad...(4)$





$\textsf{Again from (3), } \\ \\ \sf \implies (\alpha)^2 \: + \: (\beta )^2 \: = \: 9 \: + \: 2 \alpha \beta \\ \\ \textsf{Add 2 \alpha \beta to both sides , } \\ \\ \sf \implies (\alpha)^2 \: + \: (\beta )^2 \: + \: 2 \alpha \beta \: = \: 9 \: + \: 2 \alpha \beta \: + \: 2 \alpha \beta \\ \\ \sf \implies ( \alpha \: + \: \beta )^2 \: = \: 9 \: + \: 4 \alpha \beta \\ \\ \sf \implies ( \alpha \: + \: \beta )^2 \: = \: 9 \: + \: 4 \: \times \: 18 \\ \\ \sf \implies (\alpha \: + \: \beta )^2 \: = \: 9 \: + \: 72 \\ \\ \sf \implies (\alpha \: + \: \beta )^2 \: = \: 81 \\ \\ \sf \: \: \therefore \: \: \alpha \: + \: \beta \: = \: \pm9$




$\textsf{The quadratic equation will be :} \\ \\ \sf \implies x^2 \: - \: ( Sum \: of \: zeroes )x \: + \: Product \: of \: zeroes \: = \: 0 \\ \\ \sf \implies {x}^{2} \: - \: ( \alpha \: + \: \beta)x \: + \: \alpha \beta \: = \: 0 \\ \\ \sf \implies {x}^{2} \: - \: ( \pm9)x \: + \: 18\: = \: 0 \\ \\ \sf \implies {x}^{2} \: \pm \: 9x \: + \: 18 \: = \: 0$



$\underline{\textsf{The required quadratic equations are : }} \\ \\ \sf \implies x^2 \: + \: 9x \: + \: 18 \: = \: 0 \\ \sf \qquad \qquad \: \: \: \: \: Or \\ \sf \implies x^2 \: - \: 9x \: + \: 18 \: = \: 0$

Answer 2

let two roots be a and b

As difference of roots are 3

So a - b= 3

As difference between cubes are 189

So a^3 - b^3 = 189

As a^3 -b^3 = ( a- b)( a^2 + ab+ b^2)

= ( a-b)( (a+ b)^2 -2ab + ab))

= ( a-b)( ( a+b)^2 - ab))

= ( a-b)(( a-b)^2 +4ab - ab))

= ( a-b)(( a-b)^2 + 3ab)

Put a - b= 3

So its 3(3^2 + 3ab)= 189

3( 9 +3ab)= 189

9( 3 + ab)= 189

3+ ab= 21
ab= 18

So a- b= 3

(a-b)^2 = 9

( a+b)^2 -4ab= 9

(a+b)^2 -4(18)= 9

(a+b)^2 - 72 = 9

(a+b)^2 = 81

(a+b)= +-9

So quadratic equation is

x^2 -( a+b) x + ab

x^2 -( +-9) x + 18

x^2 +- 9x + 18

So equations are

x^2 + 9 x +18

and
x^2 - 9 x +18


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