# If x_1 and X-2 are roots of equation 5x^2 - 7x - 11=0,find (1+x_1)(1+x_2)?

oldnick | (Level 1) Valedictorian

Posted on

`5x^2-7x-11=0`

We know that:

`x_1 + x_2= 7/5`                        (1)

and     `x_1 xx x_2=-11/5`                (2)

From (1):

`1+x_1+x_2= 12/5`

So:

`1+x_1=12/5-x_2`       `1+x_2=12/5-x_1`

Then:    `(1+x_1)(1+x_2)=(12/5-x_1)(12/5-x_2)=`

`=144/25 -12/5(x_1+x_2) + x_1x_2` `=144/25 -12/5 xx 7/5- 11/5=`

`=60/25+11/5= 12/5 -11/5=1/5`

Proof:

`Delta= 269 >0`   has two differents real solutions:

`x=(7+-sqrt(269))/10`

`1+x_1= (17+sqrt(269))/10`

`1+x_2= (17-sqrt(269))/10`

`(1+x_1)(1+x_2)=(17+sqrt(269))/10 xx (17-sqrt(269))/10=` `(289-269)/100=` `=20/100=1/5`

As we've  just found.

pramodpandey | College Teacher | (Level 3) Valedictorian

Posted on

We have given

`5x^2-7x-11=0`

We  know relation between roots and coefficients.

`ax^2+bx+c=0`  then

`x_1 +x_2= -b/a`

`x_1 xx x_2 =c/a`

Thus

`x_1+x_2=7/5`

`x_1 xx x_2=-11/5`

`So`

`(1+x_1)(1+x_2)=1+(x_1+x_2)+x_1 x_2`

`=1+(7/5)-11/5`

`=(5+7-11)/5`

`=1/5`

Ans.