6. the sum of the squares of two consecutive integers is 421. what could the integers be? list all…

Explanation:

Step1: Let the first integer be $n$.

The second consecutive integer is $n + 1$.

Step2: Set up the equation.

The sum of their squares is $n^{2}+(n + 1)^{2}=421$.

Step3: Expand the equation.

$n^{2}+n^{2}+2n + 1=421$. Combining like - terms gives $2n^{2}+2n+1 = 421$, then $2n^{2}+2n-420 = 0$. Divide through by 2: $n^{2}+n - 210=0$.

Step4: Solve the quadratic equation.

For a quadratic equation $ax^{2}+bx + c = 0$ (here $a = 1$, $b = 1$, $c=-210$), use the quadratic formula $n=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. First, calculate the discriminant $\Delta=b^{2}-4ac=(1)^{2}-4\times1\times(-210)=1 + 840=841$. Then $n=\frac{-1\pm\sqrt{841}}{2}=\frac{-1\pm29}{2}$.

Step5: Find the values of $n$.

When we take the plus sign: $n=\frac{-1 + 29}{2}=\frac{28}{2}=14$. The consecutive integer is $n + 1=15$. When we take the minus sign: $n=\frac{-1-29}{2}=\frac{-30}{2}=-15$. The consecutive integer is $n + 1=-14$.

Answer:

The pairs of consecutive integers are 14 and 15, - 15 and - 14.