Question

for what value of r the quadratic equation rx ^2 +4x-4=0 has real root​

Answer 1

Answer:

Step-by-step explanation:

For real roots we know that the Discriminant of the equation must be greater than 0 i.e. ,

b^2+4ac>=0 , where a=coefficient of x^2 term, b = coefficient of x term

                                  and c = constant of the equation

this implies that,

by b^2+ 4ac >=0

=>  (4)^2 + 4(r)(-4)>=0

=>   8- 16r >=0

=>   -16r>=-8

=> r>=0.5

since r can have a value greater than equal to 0.5 so we assume its value to be 0.5

now put its value and solve the equation,

=> 0.5x^2+4x-4=0

=> by quadratic formula, the roots are

    x = 0.898 or x = -8.898

Hope this helps

Answer 2

Please mark as brainlist