How do you solve for a²+b²=289 when b+a=23

1 Use Product Rule: {x}^{a}{x}^{b}={x}^{a+b}x

​a

​​ x

​b

​​ =x

​a+b

​​  

{a}^{2}+bb=289a

​2

​​ +bb=289

2 Use Product Rule: {x}^{a}{x}^{b}={x}^{a+b}x

​a

​​ x

​b

​​ =x

​a+b

​​  

{a}^{2}+{b}^{2}=289a

​2

​​ +b

​2

​​ =289

3 Subtract {b}^{2}b

​2

​​  from both sides

{a}^{2}=289-{b}^{2}a

​2

​​ =289−b

​2

​​  

4 Take the square root of both sides

a=\pm \sqrt{289-{b}^{2}}a=±√

​289−b

​2

​​  

​

​​  

5 Rewrite 289-{b}^{2}289−b

​2

​​  in the form {a}^{2}-{b}^{2}a

​2

​​ −b

​2

​​ , where a=17a=17 and b=bb=b

a=\pm \sqrt{{17}^{2}-{b}^{2}}a=±√

​17

​2

​​ −b

​2

​​

6 Use Difference of Squares: {a}^{2}-{b}^{2}=(a+b)(a-b)a

​2

​​ −b

​2

​​ =(a+b)(a−b)

a=\pm \sqrt{(17+b)(17-b)}a=±√

​(17+b)(17−b)

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step-by-step explanation: