b(n)=sqrt(2n-8)

Lösung

b (n) = 2 n - 8

n = \frac{1 + - 8 b ^{2} + 1}{b ^{2}} {0 < b \leq \frac{1}{2 2}}, n = \frac{1 - - 8 b ^{2} + 1}{b ^{2}} {0 < b \leq \frac{1}{2 2}}

Schritte zur Lösung

b (n) = 2 n - 8

Quadriere beide Seiten:

b^{2} n^{2} = 2 n - 8

b^{2} n^{2} = 2 n - 8

Löse

b^{2} n^{2} = 2 n - 8 : n = \frac{1 + - 8 b ^{2} + 1}{b ^{2}}, n = \frac{1 - - 8 b ^{2} + 1}{b ^{2}}

n = \frac{1 + - 8 b ^{2} + 1}{b ^{2}}, n = \frac{1 - - 8 b ^{2} + 1}{b ^{2}}

Überprüfe die Lösungen:

n = \frac{1 + - 8 b ^{2} + 1}{b ^{2}} {0 < b \leq \frac{1}{2 2}}, n = \frac{1 - - 8 b ^{2} + 1}{b ^{2}} {0 < b \leq \frac{1}{2 2}}

Die Lösungen sind

n = \frac{1 + - 8 b ^{2} + 1}{b ^{2}} {0 < b \leq \frac{1}{2 2}}, n = \frac{1 - - 8 b ^{2} + 1}{b ^{2}} {0 < b \leq \frac{1}{2 2}}

5 x + 46 = x + 8

3 x - 4 = 0

0 = - 2 + a - 7

4 12 x^{2} - 35 = x

x + 12 - x - 8 = 2