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the graph of the equation y = x^2 + 5 is a parabola. how does its graph show that the equation has no real solutions
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GPT-5 mini
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A real solution of x^2 + 5 = 0 would be an x-value where the graph y = x^2 + 5 crosses the x‑axis (where y = 0).
But the parabola y = x^2 + 5 has vertex at (0, 5) and opens upward, so its lowest y-value is 5. The graph therefore lies entirely above the x‑axis and has no x‑intercepts. Hence there is no real x with y = 0, so x^2 + 5 = 0 has no real solutions.
(Algebraically: x^2 ≥ 0 for all real x, so x^2 + 5 ≥ 5 > 0, or use the discriminant b^2 − 4ac = 0 − 20 = −20 < 0.)
But the parabola y = x^2 + 5 has vertex at (0, 5) and opens upward, so its lowest y-value is 5. The graph therefore lies entirely above the x‑axis and has no x‑intercepts. Hence there is no real x with y = 0, so x^2 + 5 = 0 has no real solutions.
(Algebraically: x^2 ≥ 0 for all real x, so x^2 + 5 ≥ 5 > 0, or use the discriminant b^2 − 4ac = 0 − 20 = −20 < 0.)
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