Алгебра: F(x)=-4*x^2+13 Найти при F=5 F=3 F=0 F=-0.1 F=1/4 F=1 целая 1/2

ivanovartem450p08ajg

11.12.2020 02:41

Алгебра

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F(x)=-4*x^2+13
Найти при F=5
F=3
F=0
F=-0.1
F=1/4
F=1 целая 1/2

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LIquid0Horse

Алгебра

4,6(80 оценок)

$F(x) = - 4 {x}^{2} + 13$

$F(x) = 5 \\ - 4 {x}^{2} + 13 = 5 \\ - 4 {x}^{2} = - 8 \\ {x}^{2} = 2 \\ x = ± \sqrt{2}$

$F(x) = 3 \\ - 4 {x}^{2} + 13 = 3 \\ - 4 {x}^{2} = - 10 \\ {x}^{2} = \frac{10}{4} \\ x = ± \frac{ \sqrt{10} }{2}$

$F(x) = 0 \\ - 4 {x}^{2} + 13 = 0 \\ - 4 {x}^{2} = - 13 \\ {x}^{2} = \frac{13}{4} \\ x = ± \frac{ \sqrt{13} }{2}$

$F(x) = - 0.1 \\ - 4 {x}^{2} + 13 = - 0.1 \\ - 40 {x}^{2} + 130 = - 1 \\ - 40 {x}^{2} = - 131 \\ {x}^{2} = \frac{131}{40} \\ x = ± \sqrt{ \frac{131}{40} } \\ x = ± \frac{ \sqrt{1310} }{20}$

$F(x) = \frac{1}{4} \\ - 4 {x}^{2} + 13 = \frac{1}{4} \\ - 16 {x}^{2} + 52 = 1 \\ - 16 {x}^{2} = - 51 \\ {x}^{2} = \frac{51}{16} \\ x = ± \frac{ \sqrt{51} }{4}$

$F(x) = 2 \frac{1}{2} = \frac{5}{2} \\ - 4 {x}^{2} + 13 = \frac{5}{2} \\ - 8 {x}^{2} + 26 = 5 \\ - 8 {x}^{2} = - 21 \\ {x}^{2} = \frac{21}{8} \\ x = ± \frac{ \sqrt{21} \sqrt{2} }{2 \sqrt{2} \sqrt{2} } \\ x = ± \frac{ \sqrt{42} }{4}$