How do you solve #a^ { 2} + 1= 17#?

2 Answers
WADU HEK
Apr 14, 2018

a = #+-4#

Explanation:

1/ #a^2# + 1 = 17
2/ #a^2# = 16
3/ Take the square root both side
4/ a = #+-4#

Jim G.
Apr 14, 2018

#a+-4#

Explanation:

#"rearrange and equate to zero"#

#"subtract 17 from both sides"#

#rArra^2-16=0#

#a^2-16" is a "color(blue)"difference of squares"#

#[a^2-b^2=(a-b)(a+b)]#

#rArr(a-4)(a+4)=0#

#"equate each factor to zero and solve for x"#

#a-4=0rArra=4#

#a+4=0rArra=-4#

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