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

Question

, 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:

$-7,-6$