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

Remaining Attempts : 3

Answer 1

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

We need to find two numbers, \(a\) and \(b\), such that:

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

Let's consider the factors of \(42\):

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

Now we will check these pairs to see which pair adds up to \(-13\):

1. \(1 + 42 = 43\) (not suitable)
2. \(-1 + (-42) = -43\) (not suitable)
3. \(2 + 21 = 23\) (not suitable)
4. \(-2 + (-21) = -23\) (not suitable)
5. \(3 + 14 = 17\) (not suitable)
6. \(-3 + (-14) = -17\) (not suitable)
7. \(6 + 7 = 13\) (not suitable)
8. \(-6 + (-7) = -13\) (suitable)

Thus, the two numbers that can be used to split the middle term are \(-6\) and \(-7\).

So, the answer is:
-6, -7