If r and s are the solutions of quadratic equation mx2 + nx + c = 0, which of the following equations can have its roots equal to r2 and s2?
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m2x2 – n2x + 2mcx + c2 = 0
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n2x2 – m2x + 2mcx + c2 = 0
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m2x2 + n2x – 2mcx + c2 = 0
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mx2 – n2x + 2mcx + c2 = 0
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mx2 – nx + 2mcx + c = 0
If r, s are roots of mx^2 + nx + c = 0, then r+s = -n/m and rs = c/m. We want an equation with roots r^2, s^2. Sum = r^2+s^2 = (r+s)^2 - 2rs = n^2/m^2 - 2c/m = (n^2 - 2mc)/m^2. Product = r^2s^2 = c^2/m^2. The equation is x^2 - (Sum)x + Product = 0, which is x^2 - ((n^2-2mc)/m^2)x + c^2/m^2 = 0. Multiplying by m^2 gives m^2x^2 - (n^2-2mc)x + c^2 = 0, or m^2x^2 - n^2x + 2mcx + c^2 = 0.
For the original equation mx squared + nx + c = 0, the sum of the roots is r + s = -n/m and the product of the roots is rs = c/m. For the new equation with roots r squared and s squared, the sum of the roots is r squared + s squared, which equals (r + s) squared - 2rs = n squared/m squared - 2c/m. The product of the new roots is r squared s squared = c squared/m squared. An equation having these values as the sum and product of its roots is x squared - (sum)x + (product) = 0, which is x squared - (n squared - 2mc)/m squared times x + c squared/m squared = 0. Multiplying the entire equation by m squared yields m squared x squared - (n squared - 2mc)x + c squared = 0, which simplifies to m squared x squared - n squared x + 2mcx + c squared = 0.