Math, asked by sabrina2365, 5 months ago

If α and β are the roots of a quadratic equation 6x² - 7x + 2 = 0, then find a quadratic equation whose roots are 5α and 5β.​

Answers

Answered by MaIeficent
56

Step-by-step explanation:

Given:-

  • A quadratic equation 6x² - 7x + 2 = 0

  • The zeroes of the equation are α and β.

To Find:-

  • The quadratic equation whose zeroes are 5α and 5β.

Concept used:-

For a quadratic equation ax² + bx + c = 0

\sf Sum\: of \: zeroes = \sf \dfrac{-b}{a}

\sf Product\:  of \: zeroes = \sf \dfrac{c}{a}

Solution:-

Comparing 6x² - 7x + 2 = 0 with ax² + bx + c = 0

Here:-

• a = 6  \:\:\:\: • b = - 7 \:\:\:\:  • c = 2

\sf Sum \: of \: zeroes = \dfrac{-(-7)}{6} = \dfrac{7}{6}

\implies \bf \alpha + \beta = \dfrac{7}{6}

\sf Product \: of \: zeroes = \dfrac{2}{6} = \dfrac{1}{3}

\implies \bf \alpha \beta = \dfrac{1}{3}

Now, we need to find the quadratic equation whose zeroes are 5α and 5β.

\sf Sum \: of \: zeroes = 5\alpha + 5\beta

\sf  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  = 5(\alpha + \beta)

\sf  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  = 5 \times \dfrac{7}{6}

\sf  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  = \dfrac{35}{6}

\sf Product \: of \: zeroes = 5\alpha \times 5\beta

\sf  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  = 25(\alpha \beta)

\sf  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:  \:  \:  \:  \:  \:  \:  = 25 \times \dfrac{1}{3}

\sf  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:\:  \:  \:  \:  \:  \:   \:  = \dfrac{25}{3}

General form of a quadratic equation is:-

\sf x^2 - (sum \: of \: zeroes)x + product \: of \: zeroes = 0

\sf \implies x^2 - \dfrac{35}{6}x + \dfrac{25}{3} = 0

\sf \implies 6x^2 - 35x + 50 = 0

Hence, Required quadratic equations is:-

\large\dashrightarrow \underline{\boxed{\bf 6x^2 - 35x + 50 = 0}}

Answered by ILLUSTRIOUS27
26

Equation 1-

 \bf{6 {x}^{2} - 7x + 2 = 0 }

Here,

  • a=6
  • b=-7
  • c=2

We know,

 \bf  \alpha  +  \beta  =  \frac{ - b}{a}  \\  \implies  \boxed{\bf  \alpha  +  \beta  =  \frac{7}{6}}

And also

 \bf \:  \alpha  \beta  =  \frac{c}{a}  \\  \bf \implies  \alpha  \beta  =  \frac{2}{6}  \\  \implies \boxed{ \bf \alpha  \beta  =  \frac{1}{3} }

Required equation is

 \bf \: k( {x}^{2}  - (s)x + (p)) = 0

  \bf \: sum \: of \: zeroes = 5 \alpha  + 5 \beta  \\  \implies \bf \: s = 5( \alpha  +  \beta ) \\  \implies \bf \: s = 5 \times  \frac{7}{6}  \\  \implies \ \boxed{ \bf s =  \frac{35}{6}}

 \bf \: product \: of \: zeroes = 5 \alpha  \times 5 \beta  \\ \bf \implies p = 25 \times  \frac{  1}{3}  \\  \implies \boxed{ \bf \: p =  \frac{ 25}{3} }

Put p and s in equation

\bf \: k( {x}^{2}  -  \frac{35}{6}x +  \frac{25}{3} )  \\  \bf  \implies \: k( \frac{6 {x}^{2} - 35 x + 50}{6} ) \\  \implies \boxed{\bf 6{x}^{2}  - 35x + 50 = 0} \rm where \: k = 6

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