Question

If a and b are the roots of the quadratic equation 2r + 6x -7 = 0, form the equation with the following causes.


`(1)/(2 \alpha + 1) (1)/(2 \ \beta + 1 )`
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Answer 1

Explanation:

Assuming the given equation is actually
`2x^2+6x-7=0`.

Then
`\alpha +\beta=-(6)/(2) =-3,and,\alpha \beta=-(7)/(2)`

The sum of roots of the new equation is
`(1)/(2\alpha+1) +(1)/(2\beta +1) =(2(\alpha+\beta+1))/(4\alpha \beta+2(\alpha +\beta)+1)`

`\implies (1)/(2\alpha+1) +(1)/(2\beta +1) =(2(-3+1))/(4*-3.5+2(-3)+1) =(4)/(19)`

The product of the roots of the new equation is
`(1)/(2\alpha +1)*(1)/(2\beta+1)=(1)/(4\alpha \beta+2(\alpha+\beta)+1)`

`\implies (1)/(2\alpha +1)*(1)/(2\beta+1)=(1)/(4*-3.5+2(-3)+1) =-(1)/(19)`

The new equation is given by:

`x^2-(sum\:of\:roots)x+product\:of\:roots=0`

`x^2-(4)/(19) x-(1)/(19) =0`

`19x^2-x-1=0`