Multiple choice

The condition for the roots of equation $ \displaystyle x^{2}-lx+m=0 $ to differ by one is

  1. $ \displaystyle l^{2}=4m+1 $
  2. $ \displaystyle l^{2}+m^{2}=1 $
  3. $ \displaystyle m^{2}=4l+1 $
  4. $l=m+1$
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A Correct answer
Explanation

If roots are a and a+1, then sum = 2a+1 = l and product = a(a+1) = m. From the sum, a = (l-1)/2. Substitute into product: ((l-1)/2)((l+1)/2) = m => (l^2-1)/4 = m => l^2-1 = 4m => l^2 = 4m+1.

AI explanation

Let the roots be t and t+1. By Vieta's formulas, the sum of the roots gives t + (t+1) = l, so 2t + 1 = l. The product of the roots gives t(t+1) = m, which expands to t^2 + t = m. From the sum equation, t = (l - 1) / 2. Substituting this into the product equation yields ((l - 1) / 2)^2 + (l - 1) / 2 = m. Multiplying by 4 and expanding gives l^2 - 2l + 1 + 2l - 2 = 4m. Simplifying this results in l^2 - 1 = 4m, or l^2 = 4m + 1.

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