Question

find the zero t^2 -8x
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Answer 1

Step-by-step explanation:

${t}^{2} - 8x = 0 \\ = > {t}^{2} = 8x \\ = > t = \sqrt{8x} \\ = > t = 2 \sqrt{2} \sqrt{x}$

Answer 2

Answer:

this question have missed something

I think it is in t²-8t

t×t=8×t

t=8×t/t

t=8×1

t=8