How do you solve $n^{2} - 5 = - 4$ $n^2-5=-4$ ?

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How do you solve $n^{2} - 5 = - 4$ $n^2-5=-4$ ?

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Answer 1 · 2016-09-01T18:49:54

$n = \pm 1$ $n=+-1$

Explanation:

To solve a quadratic equation we require to equate it to zero.

The first step is therefore to add 4 to both sides of the equation.

$rArrn^2-5+4=cancel(-4)+cancel(4)=0$

$rArrn^2-1=0" is the equation to be solved"$

Now $n^2-1$ is a $color(blue)"difference of squares"$

$rArr(n-1)(n+1)=0$

solve: $n-1=0rArrn=1$

solve $n+1=0rArrn=-1$

Thus the solutions to the equation are $n=+-1$

Answer 2 · 2016-09-04T16:45:45

$n = +-1$

Explanation:

Although this is a quadratic equation, it is a special case because there is no 'n' term.

Isolate the $n^2$ term.

$n^2 = -4+5$

$n^2 = 1$

$n = +-1$

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How do you solve $n^{2} - 5 = - 4$ $n^2-5=-4$ ?

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How do you solve $n^{2} - 5 = - 4$ $n^2-5=-4$ ?