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2 years ago

What are the solutions of 12 – x2 = 0?

Mathematics

1 answer:

Kruka [31]2 years ago

5 0

Answer:

x = ± 2 $\sqrt{3}$

Step-by-step explanation:

Given

12 - x² = 0 ( add x² to both sides )

12 = x² or

x² = 12 ( take the square root of both sides )

x = ± $\sqrt{12}$

= ± $\sqrt{4(3)}$ = ± 2 $\sqrt{3}$

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In this problem, we have the vertical leg as 16, the horizontal leg as x, and the hypotenuse as 20. Therefore, we can say that
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Therefore, we can plug into the formula to find x: $a = \sqrt{ {b}^{2} + {x}^{2} } \\ {a}^{2} = { \sqrt{ {b}^{2} + {x}^{2} } }^{2} \\ {a}^{2} = {b}^{2} + {x}^{2} \\ {a}^{2} - {b}^{2} = {x}^{2}$
We first change the variable c to x and square both sides of the equation. Then we subtract b^2 from both sides.
${x}^{2} = {a}^{2} - {b}^{2} \\ \sqrt{ {x}^{2} } = \sqrt{ {a}^{2} - {b}^{2} } \\ x = \sqrt{ {a}^{2} - {b}^{2} }$
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