Question

pleasr solve it with picture.... the circuled nos...f2 & f3

Answer 1

F-2) 3x^
${3x}^{2} - 17x + 10 = 0 \\ {3x}^{2} - 15x - 2x - 10 = 0 \\ 3x(x - 5) - 2(x - 5) = 0 \\ (3x - 2)(x - 5) = 0 \\ x = \frac{2}{3} \: \: \: \: x = 5 \\ substituting \: the \: values \: of \: x \: in \: {2}^{nd} \: equation \: we \: get \: the \: values \: of \: \gamma \: as \: \gamma = \frac{26}{3} \: and \: 0 \\ sum \: of \: \gamma = \frac{26}{3} + 0 = \frac{26}{3}$

Answer 2

Heya User,

--> 3x² - 17x + 10 = 0

--> 3x² - 15x - 2x + 10 = 0
--> ( 3x - 2 )( x - 5 ) = 0
--> x = 5 ; 2/3 || 

--> Now, we take the polynomial function --> p(x) = x² - 5x + λ 

--> p(5) = 0 => λ = -25 + 25 = 0 || λ = 0 ;

--> p( 2/3 ) = 0 => λ = 5 ( 2/3 ) - ( 4/9 ) = ( 30 - 4 ) / 9 = 26 / 9

---> λ is either 26/9 or 0 => The sum of possible values = 26/9 ...

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Second Qn. --> 

--> 2a²x² - 2abx + b² = 0
--> Discriminant = 4a²b² - 4(2a²)(b²) = 4a²b² - 8a²b² = -4a²b²

However, since a, b are real --> 4a²b² ≥ 0 --> -4a²b² ≤ 0

=> Discriminant ≤ 0 --> The roots are imaginary ....

--> p²x² + 2pqx + q² = 0
=> ( px + q )² = 0
=> x = -q / p  <----- The only root is -q/p --> real 

=> Both the eqn.s have no common root ^_^