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Zorluk: ZorRational and Radical Expressions and Equations

When solving the radical equation 3x+10+x+2=2\sqrt{3x+10} + \sqrt{x+2} = 2 for all real values of xx, squaring both sides yields two potential solutions. What is the numerical value of the extraneous solution?

Cevap:The numerical value of the extraneous solution is 【2】.

Cevap

The extraneous solution is x=2x = 2.
The value x=2x = 2 is the extraneous solution because substituting it into the original equation yields 16+4=6≠2\sqrt{16} + \sqrt{4} = 6 \neq 2, showing it does not satisfy the original equation despite being generated by the algebraic squaring process.

Adım Adım Çözüm

1
Isolate one of the radical terms in the equation.
3x+10=2−x+2\sqrt{3x+10} = 2 - \sqrt{x+2}
Isolating one radical simplifies the squaring process.
2
Square both sides of the equation.
3x+10=4−4x+2+(x+2)3x+10 = 4 - 4\sqrt{x+2} + (x+2), which simplifies to 2x+4=−4x+22x+4 = -4\sqrt{x+2}.
Squaring eliminates the isolated radical.
3
Divide both sides by 2 and square both sides again.
x+2=−2x+2⇒(x+2)2=4(x+2)x+2 = -2\sqrt{x+2} \Rightarrow (x+2)^2 = 4(x+2), which simplifies to x2−4=0x^2 - 4 = 0.
This eliminates the remaining radical term.
4
Solve the quadratic equation for xx.
x=2x = 2 or x=−2x = -2
Factoring (x−2)(x+2)=0(x-2)(x+2) = 0 gives the potential solutions.
5
Check both potential solutions in the original equation.
For x=−2x = -2: 4+0=2\sqrt{4} + \sqrt{0} = 2, which is true. For x=2x = 2: 16+4=6≠2\sqrt{16} + \sqrt{4} = 6 \neq 2, which is false.
This identifies which solution is extraneous.

Anahtar Kavram

Solving radical equations and verifying solutions to identify extraneous roots
Tahmini Süre:2m 0s
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Konular

Rational and Radical Expressions and Equations

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