How do you solve $\sqrt{2 a^{2} - 121} = a$ $sqrt(2a^2-121)=a$ and check your solution?

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Answer 1 · 2017-01-26T21:44:20

$a = \pm 11$ $a=+-11$

$a = \sqrt{2 a^{2} - 121}$ $a=sqrt(2a^2-121)$

Square both sides

$a^{2} = 2 a^{2} - 121$ $a^2=2a^2-121$

Minus $a^{2}$ $a^2$ and add $121$ $121$ to both sides

$a^{2} = 121$ $a^2=121$

Square root both sides

$a = \pm \sqrt{121} = \pm 11$ $a=+-sqrt(121)=+-11$

Now to check:

$a = \sqrt{2 \times 11^{2} - 121} = \sqrt{242 - 121} = \sqrt{121} = \pm 11$ $a=sqrt(2xx11^2-121)=sqrt(242-121)=sqrt(121)=+-11$

How do you solve √2a2−121=asqrt(2a^2-121)=a and check your solution?