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)