Question

4
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Q:9. If & and ß are zews of x² - 5x+6 . For on a new
equation whose zaos of (2&-1) , (2B-1)​

Answer 1

Step-by-step explanation:

I think question is like this

α and β are roots of x² - 5x + 6 = 0 , then we have

to find A new equation whose roots are

( 2α - 1 ) and ( 2β -1 ) .

Solution---> ATQ,

x² - 5x + 6 = 0

Roots of this equation is α and β

So , Sum of roots

= - coefficient of x / Coefficient of x²

=> α + β = - ( - 5 ) / 1

=> α + β = 5

Product of roots

= Constant term / Coefficient of x²

=> Product of roots = 6 / 1 = 6

Now roots of required equation are

( 2α - 1 ) and ( 2β - 1 )

Sum of roots = ( 2α - 1 ) + ( 2β - 1 )

= 2 ( α + β ) - 2

= 2 ( 5) - 2

= 10 - 2 = 8

Product of roots = ( 2α - 1 ) ( 2β - 1 )

= 4αβ - 2α - 2β + 1

= 4αβ - 2 ( α + β ) +1

= 4 ( 6 ) - 2 ( 5 ) + 1

= 24 - 10 + 1

= 15

Now required equation is

x² - ( Sum of roots ) x + ( product of roots ) = 0

=> x² - ( 8 ) x + ( 15 ) =0

=> x² - 8x + 15 = 0