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Example of using the completing square

Question

Prove that no matter what the value of x is, the value of 3x2 - 7x + 5 is always larger than zero.

Solution

3x2 - 7x + 5
= 3[x2 - (7/3)x] + 5
= 3[x2 - (7/3)x + (7/6)2 - (7/6)2] + 5
= 3[x2 - (7/3)x + (7/6)2 - 49/36] + 5
= 3[x2 - (7/3)x + (7/6)2] - 3 × 49/36 + 5
= 3[x2 - (7/3)x + (7/6)2] - 49/12 + 5
= 3[x2 - (7/3)x + (7/6)2] - 49/12 + 60/12
= 3[x2 - (7/3)x + (7/6)2] + 11/12
= 3(x - 7/6)2 + 11/12

If x < 0 and |x| is a very large number, then the value of (x - 7/6)2 is still a positive number and the value of 11/12 is also a positive number. So the value of 3(x - 7/6)2 + 11/12 is always a positive number. Therefore, the value of 3x2 - 7x + 5 is a positive number no matter what the value of x is.