√x+7 | = | x+1 | ⅅ={x∈ℝ|x≥ −7} | |
√x+7 | = | x+1 | |quadrieren | |
x+7 | = | x2+2x+1 | |−x−7 | |
0 | = | x2 + x − 6 | ||
x1,2 | = | −1/2±√1/4+6 | ||
x1=2 | x2 = −3 |
√2+7 | = | 2+1 | ⇒ | √9=3 | x1=2 | ✓ |
√−3+7 | = | √4 | ≠ | −3+1 = −2 | x2=−3 | f |
L={2} | ||||||
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Lösung ansehen .
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Beispiel 8 - 33:
3+√x+3 | = | √3x+6+2 | ⅅ = {x∈ℝ | x≥−2} |
3+√x+3 | = | √3x+6+2 | |−2 |
1+√x+3 | = | √3x+6 | |Quadrieren |
12+2√x+3+(x+3) | = | 3x+6 | |−4−x |
2√x+3 | = | 2x+2 | |:2 |
√x+3 | = | x+1 | |Quadrieren |
x+3 | = | x2+2x+1 | |−x−3 |
0 | = | x2+x−2 | |
x1=−2 | x2=1 | ||
Probe:
3+√−2+3 | = | 4 | ≠ | √3·(−2)+6+2 | = | 2 |
3+√1+3 | = | 5 | = | √3+6+2 | = | 5 |
L={1} | ||||||