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 \(-13x\) in the quadratic equation \(x^2 - 13x + 42 = 0\) into two terms, we need to find two numbers that add up to \(-13\) and multiply to \(42\).

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

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

The factors of \(42\) that also satisfy the sum of \(-13\) can be determined by looking at the pairs of factors of \(42\):

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

Now, we can determine which pair of factors, when both are negative, will add up to \(-13\):

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

Thus, the two numbers that we need to complete the step are \(-6\) and \(-7\).

So, we can rewrite the equation as:

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

The lesser value first is:

\(-7, -6\)

So the answer is:

\(-7, -6\)