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prohojiy [21]
1 year ago
9

Solve. Check for extraneous solutions. √x²+3 =x+1

Mathematics
1 answer:
Andre45 [30]1 year ago
6 0

x = 1 is not an extraneous solution.

✓(x² + 3) = x + 1

Taking square on both the sides of the equation, we get

(✓(x² + 3))² = (x + 1)²

x² + 3 = (x + 1)²

Using the identity (a+b)²= a² + b² + 2ab on the right hand side of the equation, we get

x² + 3 = x² + (1)² + 2(x)(1)

x² + 3 = x² + 1 + 2x

2x = 2

x = 1

Confirming the value is extraneous solution or not

Putting the root x = 1 in the original equation ✓(x² + 3) = x + 1, we get

✓(1² + 3) = 1 + 1

✓4 = 2

2 = 2

⇒ LHS = RHS

∴ x = 1 is not an extraneous solution.

Know more about Extraneous Solutions: -brainly.com/question/20532561

#SPJ4

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gregori [183]

The maximum earning is v(r,b)

Let v(r,b) be the expected value of the game for the player, assuming optimal play, if the remaining deck has r red cards and b black cards.

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The stopping rule is simple: Stop when v(r,b)=0.

To explain the recursion . . .

If r,b>0, and the player elects to play a card, then:

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