Question 139466: Find the value of b for which the equation has exactly one real root. (Hint- think discriminate0
x^2+bx+36=0
Answer by scott8148(6628)   (Show Source):
You can put this solution on YOUR website! if b is twice the square root of the constant term, then the quadratic is a perfect square (one real root)
b=2sqrt(36) __ b=2(+- 6) __ b=+- 12
if the discriminant equals zero, then there is one real root
b^2-4ac=0 __ b^2=4ac __ b=+- sqrt(4ac) __ b=+- sqrt(4*1*36) __ b=+- sqrt(144) __ b=+- 12
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