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1 vote
What must be added to

3x−7 to make x^2+4x−1?

By how much does
3−2a exceed a^2+a−6?

1 Answer

4 votes

Final answer:

The expression that we must add to 3x - 7 to result in x^2 + 4x - 1 is x^2 + x + 6. To answer the second question, we need to solve the quadratic inequality a^2 + 3a - 9 < 3 - 2a, which will provide the range for 'a' where 3 - 2a exceeds a^2 + a - 6.

Step-by-step explanation:

To answer the first student's question, we need to determine what must be added to 3x-7 to result in x^2+4x-1. We begin by setting up an equation:

3x - 7 + something = x^2 + 4x - 1.

To find out what that something is, we need to isolate it:

3x - 7 + something - 3x + 7 = x^2 + 4x - 1 - 3x + 7

Something = x^2 + x + 6

So the something we need to add is x^2 + x + 6.

To solve the second question, 'By how much does 3 - 2a exceed a^2 + a - 6?', we set up an inequality:

3 - 2a > a^2 + a - 6

Subtracting 3 - 2a from both sides gives us:

0 > a^2 + 3a - 9

This is a quadratic inequality, and its solution will give the range of values for which a^2 + 3a - 9 is less than 3 - 2a.

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