Question

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 x^2−13x+42=0
, which two numbers would be needed to complete this step? Enter the lesser value first.(1 point)
x^2−13x+42=0

x^2+
x+
x+42=0

Answer 1

To split the middle term of the quadratic equation \( x^2 - 13x + 42 = 0 \), we need to find two numbers that multiply to the constant term \( 42 \) and add up to the coefficient of the middle term \( -13 \).

Let's denote the two numbers as \( a \) and \( b \). We need:

1. \( a + b = -13 \)
2. \( a \times b = 42 \)

To find these numbers, we can look for factor pairs of \( 42 \) and see which pair sums to \( -13 \):

- \( 1 \) and \( 42 \) → \( 1 + 42 = 43 \)
- \( 2 \) and \( 21 \) → \( 2 + 21 = 23 \)
- \( 3 \) and \( 14 \) → \( 3 + 14 = 17 \)
- \( 6 \) and \( 7 \) → \( 6 + 7 = 13 \)

The key is that we need negative pairs because they will add up to a negative number. Therefore, consider negative pairs:

- \( -6 \) and \( -7 \) → \( -6 + (-7) = -13 \)

Thus, the two numbers that will split the middle term are \( -6 \) and \( -7 \).

So, in the form you provided, it would be:

\( x^2 + (-6)x + (-7)x + 42 = 0 \)

The lesser value is \( -7 \).

**Final answer:** -7, -6