Question

if α, β are zeros of 3x² - 4x +1, find value of a) α/β+ β/α b) α²/β + β²/α​

Answer 1

Solution :-

Given : f(x) = 3x² - 4x + 1

Sum of zeros = α + β = 4/3

Product of zeros = α β = 1/3

a) α/β + β/α

= [α² + β²]/βα

= [(α + β)² - 2αβ]/ αβ

= [(4/3)² - 2 × 1/3]/[1/3]

= [16/9 - 2/3]/[1/6]

= [(16 - 6)/9]/[1/3]

= (10 × 3)/(1 × 9)

= 10/3 ans.

b) α²/β + β²/α

= [α³ + β³]/α β

= [(α + β)³ - 3αβ(α + β)]/αβ

= [(4/3)³ - 3 × 1/3(4/3)]/1/3

= [64/27 - 4/3]/[1/3]

= [(64 - 36)/27 ]/[1/3]

= (28 × 3)/27

= 28/9 ans.

Answer 2

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

$\sf Given : f(x) = 3x {}^{2} - 4x + 1 \\ \\ \sf </p><p></p><p>Sum \: of \: zeros = α + β \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{4}{3} \\ \\ \sf</p><p></p><p>Product \: of \: zeros = α β \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf= \dfrac{1}{3} \\ \\ \sf</p><p></p><p>a) \: \frac{α}{β} + \frac βα \\ \sf</p><p></p><p>= \dfrac{[α {}^{2} + β {}^{2} ]}{β \alpha} \\ \sf</p><p></p><p>= \dfrac{[{(α + β)}^{2} - 2αβ]}{αβ} \\ \sf</p><p></p><p>= \dfrac{ \bigg[{ \bigg( \dfrac43 \bigg)}^{2} - 2 × \dfrac 13 \bigg]}{\dfrac13} \\ \sf</p><p></p><p>= \dfrac{\bigg[ \dfrac{16}9 - \dfrac23\bigg]}{ \dfrac16} \\ \sf</p><p></p><p>= \dfrac{\bigg[ \dfrac{(16 - 6)}9\bigg]}{ \dfrac13} \\ \sf</p><p></p><p>= \dfrac{(10 × 3)}{(1 × 9)} \\ \sf</p><p></p><p>= \boxed{ \sf \dfrac{10}3} \\ \\ \sf </p><p></p><p>b) \dfrac{α {}^{2}}{β} + \dfrac{β {}^{2}}{α} \\ \sf</p><p></p><p>= \sf \dfrac{[α {}^{3} + β {}^{3} ]}{α β}\\ \sf</p><p></p><p>= \dfrac{[(α + β) {}^{3} - 3αβ(α + β)]}{αβ}</p><p>\\ \sf</p><p>= \dfrac{ \bigg[{\bigg( \dfrac{4}3\bigg)}^{3} - 3 × \dfrac13\bigg( \dfrac43\bigg)\bigg]}{ \dfrac13}</p><p>\\ \sf </p><p>= \dfrac{ \bigg[ \dfrac{64}{27} - \dfrac43\bigg]}{ \dfrac13}</p><p>\\ \sf</p><p>= \dfrac{\bigg[ \dfrac{(64 - 36)}{27} \bigg]}{ \dfrac{1}3}</p><p>\\ \sf</p><p>= \dfrac{(28 × 3)}{27}</p><p>\\ </p><p>= \boxed{ \sf \dfrac{28}{9}}$