Verification by substitution
- Input
- a=2, b=-7, c=3
- Expected output
- x₁ = 3, x₂ = 0.5
Δ=25 is a perfect square; substituting x=3 and x=0.5 back into 2x²−7x+3, both give exactly zero.
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Bhaskara quadratic formula
The quadratic formula solves any ax²+bx+c=0 with a≠0: x = (−b ± √(b²−4ac)) / (2a). Once the roots come out, each one can be checked without redoing the whole calculation: substitute the value back into the original equation and see whether the result lands on zero.
Δ=25 is a perfect square; substituting x=3 and x=0.5 back into 2x²−7x+3, both give exactly zero.
Only the positive root makes physical sense: the object hits the ground about 4.13 s after launch; the negative root is a mathematical artifact of the parabola, not a real instant.
Δ=8=4×2, so √Δ=2√2≈2.828427125; the roots equal 1+√2 and 1−√2, since √2≈1.414213562.
The discriminant Δ = b²−4ac determines the nature of roots: Δ>0 → two distinct real roots; Δ=0 → double root; Δ<0 → no real roots.
It is a local teaching convention honoring the Indian mathematician Bhaskara II (1114-1185); the method of completing the square to solve second-degree equations, however, had already circulated for centuries before him. Outside Brazil, including English-speaking countries, the name used is simply "quadratic formula," with no personal attribution.
Mathematically yes, but in problems with a physical context (time, length, quantity), one root may not make practical sense and should be discarded. In the falling-object example, one of the two roots is a negative time, which does not correspond to any real instant, so only the positive root is used as the answer.
Substitute the value back into the original equation ax²+bx+c and check whether the result is exactly zero. For 2x²−7x+3=0, both x=3 and x=0.5 satisfy that condition: 2×9−21+3=0 and 2×0.25−3.5+3=0, confirming both roots without needing to redo the formula.
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