Question

If α & β are the zeroes of the quadratic polynomial f(x) =x² -3x+2, then how do you find a quadratic polynomial whose zeroes are 1/ (2 α + β) and 1/ (α +2 β)?

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{New\:Quadratic\to 20x^{2}-9x+1=0}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies f(x) = {x}^{2} - 3x + 2 \\ \\ \tt: \implies Zeroes \: of \: polyonmial = \alpha \: and \: \beta \\ \\ \tt: \implies New \: zeroes = \frac{1}{2 \alpha + \beta } \: and \: \frac{1}{2 \beta + \alpha } \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies New \: quadratic =?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies {x}^{2} - 3x + 2 = 0 \\ \\ \tt: \implies {x}^{2} - 2x - x + 2 = 0 \\ \\ \tt: \implies x(x - 2) - 1(x - 2) = 0 \\ \\ \tt: \implies x = 2 \: and \: 1 \\ \\ \green{\tt: \circ \: \alpha = 1 \: \: \: \: \: \: \beta = 2} \\ \\ \bold{New \: zeroes : } \\ \green{\tt: \implies \frac{1}{2 \alpha + \beta } = \frac{1}{4} } \\ \\ \green{\tt: \implies \frac{1}{2 \beta + \alpha } = \frac{1}{5}} \\ \\ \bold{For \:New \: Quadratic : } \\ \tt: \implies {x}^{2} - (sum \: of \: zeroes)x + (product \: of \: zeroes) = 0 \\ \\ \tt: \implies {x}^{2} - ( \frac{1}{4} + \frac{1}{5} )x + ( \frac{1}{4} \times \frac{1}{5} ) = 0 \\ \\ \tt: \implies {x}^{2} - \frac{9}{20}x + \frac{1}{20} = 0 \\ \\ \green{\tt: \implies {20x}^{2} - 9x + 1 = 0}$

Answer 2

Answer :

Given quadratic polynomial : x² - 3x + 2.

Finding its zeroes -

Splliting the middle term :

⇒ x² - 3x + 2 = 0

⇒ x² - 1x - 2x + 2 = 0

⇒ x (x - 1) - 2 (x - 1) = 0

⇒ (x - 2) (x - 1) = 0

⇒ x - 2 = 0 and x - 1 = 0

Hence, x = 2 and 1.

So, the zeroes of the polynomial are alpha = 2 and beta = 1.

Now, Let us find the values of 1/(2 alpha + beta) and 1/(alpha + 2 beta).

⇒ 1/(2 alpha + beta)

⇒ 1/(2 × 2 + 1)

⇒ 1/(4 + 1)

⇒ 1/5

⇒ 1/(alpha + 2 beta)

⇒ 1/(2 + 2 × 1)

⇒ 1/(2 + 2)

⇒ 1/4

Sum of zeroes = 1/5 + 1/4

= (4 + 5)/20

= 9/20

Product of zeroes = 1/5 × 1/4

= 1/20

Now, Quadratic Equation = x² - (Sum)x + Product.

➸ x² - (9/20)x + 1/20

➸ 20x² - 9x + 1

Hence, Quadratic Equation = 20x² - 9x + 1.