Question

Alpa+Beta are zeroes of 3xsquare-5x+6=0 then alphasquare+betasquare=

Answer 1

Answer:

$\huge{\pink{\green{\sf{\therefore \alpha^{2}+\beta^{2}=4}}}}$

Step-by-step explanation:

$\huge{\pink{\green{\underline{\red{\sf{SOLUTION-}}}}}}$

▪First find the zeroes of this eqn.

• According to given question:

$\: \: \: \: \: \: \: {\orange{ \underline{given}}} \\ { \green{3 {x}^{2} - 5x + 6 = 0}} \\ { \green{ \alpha \: and \: \beta \: are \: the \: zeroes \: of \: this \: eqn }} \\ \\ { \red{ \underline{to \: find: }}} \\ { \purple{ { \alpha }^{2} + { \beta }^{2} = ?}}$

▪Using method Quadratic formula.

$\to 3 {x}^{2} - 5x + 6 \\ \to d = {b}^{2} - 4ac \\ \to d = ({ - 5})^{2} - 4 \times 3 \times 6 \\ \to d = 25 - 72 \\ \to d = - 47 \\ \\ \to x = \frac{ - ( - 5) ± \sqrt{47} }{2 \times 3} \\ \to x = \frac{5 ±\sqrt{47} }{6} \\ \\ { \green{ \to\alpha = \frac{5 + \sqrt{47} }{6} }} \\ { \green{ \to \beta = \frac{5 - \sqrt{47} }{6} }}$

▪We get the value of α and β.

$\to { \alpha }^{2} + { \beta }^{2} = ({ \alpha + \beta })^{2} - 2 \alpha \beta \\ \to { \alpha }^{2} + { \beta }^{2} = (\frac{5 + \sqrt{47} }{6} + \frac{5 - \sqrt{47} }{6} )^{2} - 2 \times \frac{5 + \sqrt{47} }{6} \times \frac{5 - \sqrt{47} }{6} \\ \to { \alpha }^{2} + { \beta }^{2} = ({ \frac{5 + \sqrt{47} + 5 - \sqrt{47} }{6} })^{2} - \frac{( {5})^{2} - ({ \sqrt{47} })^{2} }{18} \\ \to{ \alpha }^{2} + { \beta }^{2} = ({ \frac{10}{6} })^{2} - (\frac{25 - 47}{18} ) \\ \to{ \alpha }^{2} + { \beta }^{2} = \frac{25}{9} - (\frac{ - 22}{18} ) \\ \to { \alpha }^{2} + { \beta }^{2} = \frac{50 + 22}{18} \\ \to { \alpha }^{2} + { \beta }^{2} = \frac{72}{18} \\ { \green{\therefore { \alpha }^{2} + { \beta }^{2} = 4}}$

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