How do I solve the following equation: Find the values of k for which the roots of the equation (k+4)x^2+(k+1)x+1=0 are real and equal?
Md Ariful Islam
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,
এমবিএ(মার্কেটিং)
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Dr. Md Rushad Rahman
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পিজিটি(মেডিসিন)
Having a double root means that your espression is a perfect square. It also means that the discriminate is zero. Using either of these will give an equation in k that you can solve for the answer. A square takes the form (ax+b)^2= a^2 x^2 + 2abx + b^2. Obviously, then b = +1 or -1, leaving a=+/- (k+1)/2 = sqrt(k+4). Setting the discriminant of the original quadratic to zero gives (k+1)^2 - 4 (k+4) = 0. Whichever way you prefer you should get k=-3 (x^2 -2x +1 = (x-1)^2) and k=5 (9x^2 + 6x +1 = (3x+1)^2).
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