If alpha and beta are the roots of ac²+bx+c then find1 I alpha + 1/ B

1.	If alpha and beta are the roots of ac²+bx+c then find1 I alpha + 1/ Beta. I will Mark As branilist those who answer correct pls don't take free points.
Answer» Answer : 1/α + 1/ß = -b/c NOTE: ★ The possible VALUES of the variable which satisfy the EQUATION are called its roots or solutions . ★ A quadratic equation can have atmost two roots . ★ The general form of a quadratic equation is given as ; ax² + BX + c = 0 ★ If α and ß are the roots of the quadratic equation ax² + bx + c = 0 , then ; • Sum of roots , (α + ß) = -b/a • Product of roots , (αß) = c/a Solution : Here , The given quadratic equation is : ax² + bx + c = 0 . Also , It is given that , α and ß are the roots of the given quadratic equation . Thus , The sum of roots will be given as ; α + ß = -b/a Also , The product of roots will be given as ; α•ß = c/a Now , => 1/α + 1/ß = (ß + α)/α•ß => 1/α + 1/ß = (α + ß)/α•ß => 1/α + 1/ß = (-b/a) / (c/a) => 1/α + 1/ß = (-b/a) × (a/c) => 1/α + 1/ß = -b/c Hence , 1/α + 1/ß = -b/c

If alpha and beta are the roots of ac²+bx+c then find1 I alpha + 1/ Beta. I will Mark As branilist those who answer correct pls don't take free points.

Answer»

Answer :

1/α + 1/ß = -b/c

NOTE:

★ The possible VALUES of the variable which satisfy the EQUATION are called its roots or solutions .

★ A quadratic equation can have atmost two roots .

★ The general form of a quadratic equation is given as ; ax² + BX + c = 0

★ If α and ß are the roots of the quadratic equation ax² + bx + c = 0 , then ;

• Sum of roots , (α + ß) = -b/a

• Product of roots , (αß) = c/a

Solution :

Here ,

The given quadratic equation is :

ax² + bx + c = 0 .

Also ,

It is given that , α and ß are the roots of the given quadratic equation .

Thus ,

The sum of roots will be given as ;

α + ß = -b/a

Also ,

The product of roots will be given as ;

α•ß = c/a

Now ,

=> 1/α + 1/ß = (ß + α)/α•ß

=> 1/α + 1/ß = (α + ß)/α•ß

=> 1/α + 1/ß = (-b/a) / (c/a)

=> 1/α + 1/ß = (-b/a) × (a/c)

=> 1/α + 1/ß = -b/c