Question

exit ticket
solve each equation using square roots. no decimals
a.) $x^2 - 81 = 0$
b.) $-4a^2 - 13 = 3$
c.) $3b^2 - 5 = -29$

Explanation:

Step1: Isolate $x^2$ term

$x^2 - 81 = 0 \implies x^2 = 81$

Step2: Take square roots of both sides

$x = \pm\sqrt{81} = \pm9$

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Step1: Isolate $a^2$ term

$-4a^2 - 13 = 3 \implies -4a^2 = 16 \implies a^2 = -4$

Step2: Evaluate square root of negative

Since square of real number can't be negative, no real solutions.

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Step1: Isolate $b^2$ term

$3b^2 - 5 = -29 \implies 3b^2 = -24 \implies b^2 = -8$

Step2: Evaluate square root of negative

Since square of real number can't be negative, no real solutions.

Answer:

a.) $x = 9$ and $x = -9$
b.) No real solutions
c.) No real solutions