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%0D%0A𝑥%0D%0A2%0D%0A−%0D%0A13%0D%0A𝑥%0D%0A+%0D%0A42%0D%0A=%0D%0A0%0D%0A , which two numbers would be needed to complete this step? Enter the lesser value first.(1 point)%0D%0Ax2−13x+42=0%0D%0A𝑥%0D%0A2%0D%0A−%0D%0A13%0D%0A𝑥%0D%0A+%0D%0A42%0D%0A=%0D%0A0%0D%0A%0D%0Ax2+%0D%0A𝑥%0D%0A2%0D%0A+%0D%0Ax+%0D%0A𝑥%0D%0A+%0D%0Ax+42=0

Answer 1

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

We are looking for two numbers \(a\) and \(b\) such that:

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

Let's find pairs of factors of \(42\):

- \(1\) and \(42\)
- \(2\) and \(21\)
- \(3\) and \(14\)
- \(6\) and \(7\)

Now let's consider the sums of these factor pairs:

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

Since we need them to sum to \(-13\), we can take the negatives of each:

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

The pair \(-6\) and \(-7\) successfully sums to \(-13\) and multiplies to \(42\).

Therefore, the two numbers that complete the grouping step are \(-6\) and \(-7\).

**Answer**: -7, -6