Question

find the equation whose roots are 2
,-1/2​

Answer 1

Step-by-step explanation:

α+β=a,αβ=b

α+β=a,αβ=blet equation whose roots are 2α+1,2β+1

α+β=a,αβ=blet equation whose roots are 2α+1,2β+1be x

α+β=a,αβ=blet equation whose roots are 2α+1,2β+1be x 2

α+β=a,αβ=blet equation whose roots are 2α+1,2β+1be x 2 +Ax+B=0

α+β=a,αβ=blet equation whose roots are 2α+1,2β+1be x 2 +Ax+B=0 then −A=(2α+1)+(2β+1)=0

α+β=a,αβ=blet equation whose roots are 2α+1,2β+1be x 2 +Ax+B=0 then −A=(2α+1)+(2β+1)=0 −A=2(a)+2=2(a+1)

α+β=a,αβ=blet equation whose roots are 2α+1,2β+1be x 2 +Ax+B=0 then −A=(2α+1)+(2β+1)=0 −A=2(a)+2=2(a+1) B=(2α+1)(2β+1)

α+β=a,αβ=blet equation whose roots are 2α+1,2β+1be x 2 +Ax+B=0 then −A=(2α+1)+(2β+1)=0 −A=2(a)+2=2(a+1) B=(2α+1)(2β+1) =4αβ+2α+2β+1

α+β=a,αβ=blet equation whose roots are 2α+1,2β+1be x 2 +Ax+B=0 then −A=(2α+1)+(2β+1)=0 −A=2(a)+2=2(a+1) B=(2α+1)(2β+1) =4αβ+2α+2β+1 =4b+2a+1

α+β=a,αβ=blet equation whose roots are 2α+1,2β+1be x 2 +Ax+B=0 then −A=(2α+1)+(2β+1)=0 −A=2(a)+2=2(a+1) B=(2α+1)(2β+1) =4αβ+2α+2β+1 =4b+2a+1 ∴ required equation is

α+β=a,αβ=blet equation whose roots are 2α+1,2β+1be x 2 +Ax+B=0 then −A=(2α+1)+(2β+1)=0 −A=2(a)+2=2(a+1) B=(2α+1)(2β+1) =4αβ+2α+2β+1 =4b+2a+1 ∴ required equation is x

α+β=a,αβ=blet equation whose roots are 2α+1,2β+1be x 2 +Ax+B=0 then −A=(2α+1)+(2β+1)=0 −A=2(a)+2=2(a+1) B=(2α+1)(2β+1) =4αβ+2α+2β+1 =4b+2a+1 ∴ required equation is x 2

α+β=a,αβ=blet equation whose roots are 2α+1,2β+1be x 2 +Ax+B=0 then −A=(2α+1)+(2β+1)=0 −A=2(a)+2=2(a+1) B=(2α+1)(2β+1) =4αβ+2α+2β+1 =4b+2a+1 ∴ required equation is x 2 −2(a+1)x+4b+2a+1=0

Answer 2

Answer:

$2x {}^{2} - 3x - 2$

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