Question

Determine the nature of the solutions of the equation. x^2-14x+49=0 What does the equation have?
Is my answer correct "one real solution"?

Answer 1

tey only one real solution...

Answer 2

x^2-14x+49=0
(x-7)^2=0
x=7

Answer 3

just calculate b^2 - 4ac i.e 14^2 - 4(1)(49)

if it is positive i.e greater than zero, you will have two distinct real roots
if it is zero, you will have two real equal roots
if it is negative i.e. less than zero, you will have imaginary roots

here b^2 - 4ac = 196 - 196 = 0 , so it has tow real equal roots

Answer 4

Hope this clears it for you.....

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