Question

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.​

Answer 1

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:-

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

Answer 2

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

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