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JulsSmile [24]
3 years ago
6

How do extraneous solutions arise from radical equations?

Mathematics
2 answers:
Blababa [14]3 years ago
5 0
In general, extraneous solutions arise<span> when we perform non-invertible operations on both sides of an </span>equation<span>. (That is, they sometimes </span>arise, but not always.) ... Solvingequations<span> involving square roots involves squaring both sides of an</span>equation<span>. I hope that this helps you out, Have a wonderful day!!</span>
eduard3 years ago
3 0
In general, extraneous solutions arise<span> when we perform non-invertible operations on both sides of an </span>equation
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Factor -1/4 out of -1/2-5/4y
In-s [12.5K]
When we factor out something, we divide the factor by the equation. For example if we were to factor 3 out of 6, It will be 3(2). We got this answer by dividing 3 and 6. Using this concept:

\frac{-1}{4}[( \frac{-1}{2} / \frac{-1}{4}) - ( \frac{5}{4} y /  \frac{-1}{4})]

Simplifying:

\frac{-1}{4}(2 + 5y)

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I NEED HELP ASAP!!!! SOLVE THIS IT'S AN IMAGE!!!!
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Answer:

y = -7

Step-by-step explanation:

Put x = 2 in any one of the equation

5x +2y = -4

5*2 + 2y = -4

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6 0
3 years ago
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RoseWind [281]
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3 years ago
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Answer:

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