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ivanzaharov [21]
3 years ago
8

Explain why the quadratic relation y=2x^2+3x+5 would have no zeros

Mathematics
1 answer:
alex41 [277]3 years ago
5 0

Complete the square to rewrite the quadratic:

2 <em>x</em>² + 3 <em>x</em> + 5 = 2 (<em>x</em>² + 3/2 <em>x</em>) + 5

... = 2 (<em>x</em>² + 3/2 <em>x</em> + 9/16 - 9/16) + 5

... = 2 (<em>x</em>² + 3/2 <em>x</em> + (3/4)²) + 5 - 9/8

... = 2 (<em>x</em> + 3/4)² + 31/8

Any real number squared becomes non-negative, so the quadratic expression has a minimum value of 31/8, which is greater than 0, and so there are no (real) <em>x</em> for which <em>y</em> = 0.

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