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

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

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Answer 1 · 2018-04-14T08:45:02

a = $\pm 4$ $+-4$

Explanation:

1/ $a^{2}$ $a^2$ + 1 = 17
2/ $a^{2}$ $a^2$ = 16
3/ Take the square root both side
4/ a = $\pm 4$ $+-4$

Answer 2 · 2018-04-14T09:14:57

$a \pm 4$ $a+-4$

Explanation:

$rearrange and equate to zero$ $"rearrange and equate to zero"$

$subtract 17 from both sides$ $"subtract 17 from both sides"$

$\Rightarrow a^{2} - 16 = 0$ $rArra^2-16=0$

$a^{2} - 16 is a difference of squares$ $a^2-16" is a "color(blue)"difference of squares"$

$[a^{2} - b^{2} = (a - b) (a + b)]$ $[a^2-b^2=(a-b)(a+b)]$

$\Rightarrow (a - 4) (a + 4) = 0$ $rArr(a-4)(a+4)=0$

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

$a - 4 = 0 \Rightarrow a = 4$ $a-4=0rArra=4$

$a + 4 = 0 \Rightarrow a = - 4$ $a+4=0rArra=-4$

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

2 Answers

Explanation:

Explanation:

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