Question

For what value of m,mx²+8x-2=0 has real root?

Answer 1

HEY FRIEND YOUR ANSWER IS HERE.

WHEN M = 10 THEN THE EQUATION WILL

HAS REAL ROOT

CONSIDER M= 10

10X^2+8X-2 = 0

10×2 =20

SO THE FACOTRS OF 20 ARE (10,2) (5,4) (20,1)

HERE WE NEED -20 AND ON SUBTRACTING

WE HAVE TO GET 8

SO LET US CONSIDER (10,-2)

THE ROOTS OF EQUATION ARE (X+1) AND

(5X-1)

I HOPE THIS WILL HELP YOU

PLEASE MAKE AS BRAINLIEST

Answer 2

here,
a=m, b=8, c=-2

for the real root, the value of root over b^2-4ac should be more than or equal to 0.
ie. 64-4×m×(-2)≥0
64+8m≥0
8m≥-64
m≥-8
ie. for all value of m where m is less than or equal to 0. the value of the roots will be real