Question

One of the steps needed to solve a quadratic equation by grouping is to split the middle term as a sum of two terms. Given the equation x2−13x+42=0
, which two numbers would be needed to complete this step? Enter the lesser value first.(1 point)
x2−13x+42=0

x2+
x+
x+42=0

Answer 1

To solve the quadratic equation \( x^2 - 13x + 42 = 0 \) by grouping, we need to split the middle term, \(-13x\), into two terms that add up to \(-13\) and multiply to \(42\) (the constant term).

We are looking for two numbers \(a\) and \(b\) such that:
1. \( a + b = -13 \)
2. \( a \cdot b = 42 \)

After considering factor pairs of \(42\), we find:
- \( (1, 42) \)
- \( (2, 21) \)
- \( (3, 14) \)
- \( (6, 7) \)

Out of these pairs, we need a pair that adds up to \(-13\). The appropriate pair is \(-6\) and \(-7\) since:

- \(-6 + (-7) = -13\)
- \(-6 \cdot -7 = 42\)

Thus, the two numbers are \(-6\) and \(-7\). To complete the step of splitting the middle term, we write:

\[
x^2 - 6x - 7x + 42 = 0
\]

So the answer is:
\(-7, -6\)