Question

Show that the roots of the equation 4x^2-48x+135=0 differ by 3

Answer 1

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Answer 2

Answer:

Step-by-step explanation:

we must now that in an quadratic equation, ax^2+bx+c=0, the sum of roots is equal to -b/a and product=c/a

now

we know that

(a-b)^2 = (a+b)^2 - 4ab

using the fact mentioned above

(a-b)^2 = (48/4)^2 - 4(135/4)

(a-b)^2 = 144 - 135

(a-b)^2 = 9

(a-b) = 3

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