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=0x2−13x+42=0, which two numbers would be needed to complete this step? Enter the lesser value first.(1 point) x2−13x+42=0x2−13x+42=0 x2+x2+−13x+x+x+42=0

Answer 1

To solve the quadratic equation \( x^2 - 13x + 42 = 0 \) by grouping, we need to split the middle term \(-13x\) into two terms such that their sum equals \(-13x\) and their product equals the constant term, \(42\).

We are looking for two numbers that multiply to \(42\) and add up to \(-13\).

We can find these two numbers by considering the factors of \(42\):

1. \(1\) and \(42\) (sum: \(43\))
2. \(2\) and \(21\) (sum: \(23\))
3. \(3\) and \(14\) (sum: \(17\))
4. \(6\) and \(7\) (sum: \(13\))

To get \(-13\), we need to take negative values:

1. \(-1\) and \(-42\) (sum: \(-43\))
2. \(-2\) and \(-21\) (sum: \(-23\))
3. \(-3\) and \(-14\) (sum: \(-17\))
4. \(-6\) and \(-7\) (sum: \(-13\))

The numbers we need are \(-6\) and \(-7\), since:

• They multiply to \(42\): \[ -6 \times -7 = 42 \]

• They add up to \(-13\): \[ -6 + -7 = -13 \]

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

Given that we enter the lesser value first, the answer is:

\(-7, -6\)