If b^2-4ac=0 and ';a'; cannot =0, then the graph of y=ax^2+bx+c has only one intercept. True or False?

Please explain how you got either answer.If b^2-4ac=0 and ';a'; cannot =0, then the graph of y=ax^2+bx+c has only one intercept. True or False?
Solution :





Nota bene : 3 cases





(1) b^2 -4ac %26gt; 0 (positive discriminant) =%26gt; 2 real solutions.


(2) b^2 -4ac = 0 (discriminant is zero) =%26gt; 1 real solution


(3) b^2 -4ac %26lt; 0 (negative discriminant) =%26gt; only 2 imaginary solutions


(no real solution)





answer :





False for cases (1) and (3).


True for case (2)





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