To solve the quadratic equation \( x^2 - 13x + 42 = 0 \) by grouping, we need to split the middle term \(-13x\) into two terms such that their sum equals \(-13\) and their product equals the constant term \(42\).
Let's denote the two numbers we need as \(a\) and \(b\). The conditions we need are:
- \(a + b = -13\)
- \(a \cdot b = 42\)
To find \(a\) and \(b\), we can look for two numbers that multiply to \(42\) and add up to \(-13\).
The pairs of factors of \(42\) are:
- \(1 \cdot 42\)
- \(2 \cdot 21\)
- \(3 \cdot 14\)
- \(6 \cdot 7\)
We can take the negatives of these pairs because we are seeking a sum of \(-13\):
- \(-1\) and \(-42\) (sum is \(-43\))
- \(-2\) and \(-21\) (sum is \(-23\))
- \(-3\) and \(-14\) (sum is \(-17\))
- \(-6\) and \(-7\) (sum is \(-13\))
The correct pair \(a\) and \(b\) is \(-6\) and \(-7\).
Thus, the two numbers that need to be used to split the middle term are \(-6\) and \(-7\). Since you asked for the lesser value first:
Answer: -7, -6
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