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)

Question

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 these two numbers as \(a\) and \(b\). We need:

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

To find the numbers, we can think of the pairs of factors of 42:

• (1, 42)
• (2, 21)
• (3, 14)
• (6, 7)

Since we need the sum to be -13, we consider pairs of factors with negative signs:

• (-1, -42) → sum is -43
• (-2, -21) → sum is -23
• (-3, -14) → sum is -17
• (-6, -7) → sum is -13

The pair that sums to -13 is (-6, -7).

So, to complete the grouping step, we can write:

\[x^2 - 13x + 42 = x^2 + (-6)x + (-7)x + 42 = 0\]

Thus, the numbers needed to complete this step are -7 and -6. The answer is:

-7, -6