Question

if α,β are the zeroes of 2xsquare - 5x + 3 then find.
(1). α square + β square
(2). 1 upon 2α + 1 upon 2β
(3). α upon β + β upon α​

Answer 1

Given :

• α,β are root of 2x² - 5x + 3

To find :

(1). α² + β²

(2). (1/2α) + (1/2β)

(3). (α/β) + (β/α)

Solution :

First of all we need to find out the root of given quadratic equations.

➝ 2x² - 5x + 3 = 0

By splitting middle term

➩ 2x² - (2+3)x + 3 = 0

➩ 2x² - 2x - 3x + 3 = 0

➩ 2x(x-1) - 3(x-1) = 0

➩ (2x - 3)(x-1) = 0

This gives

Either (2x-3) = 0 or (x-1) = 0

Either 2x = 3 or x = 1

Either x = 3/2 or x = 1

Let , α = 3/2 than, β = 1

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(1) α² + β²

➩ α² + β² = (3/2)² + (1)²

➩ α² + β² = (9/4) + 1

➩ α² + β² = (9/4) + (4/4)

➩ α² + β² = (9+4)/4

➩ α² + β² = 13/4

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(2) (1/2α) + (1/2β)

$\sf \implies \: \dfrac{1}{2 \alpha } + \dfrac{1}{2 \beta} \:=\: \dfrac{1}{2 \times \frac{3}{2} } + \dfrac{1}{2 \times 1}$

$\sf \implies \: \dfrac{1}{2 \alpha } + \dfrac{1}{2 \beta} \:=\: \: \dfrac{1}{ \: 3 } + \dfrac{1}{2 }$

$\sf \implies \: \dfrac{1}{2 \alpha } + \dfrac{1}{2 \beta} \:=\:\dfrac{(1 \times 2) + (1 \times 3)}{6 }$

$\sf \implies \: \dfrac{1}{2 \alpha } + \dfrac{1}{2 \beta} \:=\:\dfrac{(2 + 3)}{6 }$

$\sf \implies\dfrac{1}{2 \alpha } + \dfrac{1}{2 \beta} \:=\: \: \bold{ \dfrac{5}{6 } }$

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(3) (α/β) + (β/α)

$\sf \implies\dfrac{\alpha}{ \beta } + \dfrac{ \beta}{\alpha} \:=\: \dfrac{ \: \: \dfrac{3}{2} \: \: }{1} \: + \: \dfrac{ \: \: 1 \: \: }{ \dfrac{3}{2} }$

$\sf \implies\dfrac{\alpha}{ \beta } + \dfrac{ \beta}{\alpha} \:=\: \: \: \dfrac{3}{2} \: + \: \dfrac{2}{3 }$

$\sf \implies\dfrac{\alpha}{ \beta } + \dfrac{ \beta}{\alpha} \:=\: \: \: \dfrac{(3 \times 3) + (2 \times 2)}{6} \:$

$\sf \implies\dfrac{\alpha}{ \beta } + \dfrac{ \beta}{\alpha} \:=\: \: \: \dfrac{ 9 + 4}{6} \:$

$\sf \implies\dfrac{\alpha}{ \beta } + \dfrac{ \beta}{\alpha} \:=\: \: \: \bold{ \dfrac{ 13}{6} \: }$

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ANSWER :

(1). α² + β² = 13/4

(2). (1/2α) + (1/2β) = 5/6

(3). (α/β) + (β/α) = 13/6