Rownanie kwadratowe(w załączniku wzory)Wyznacz miejsca zerowe funkcji FPotrzebne na zaraz :P.... Question from @bladzioszek

November 2018 1 10 Report

Rownanie kwadratowe(w załączniku wzory)
$k) 2x^{2}+3=-7x$
$l) x+10=3 x^{2}$
$ł) 16 x^{2} +24 \sqrt{3x} =-18$
$m) 2x+18 x^{2} =0$
$n) 36 x^{2} =6x$
$o) 24x- x^{2} =144$
$p)1+9 x^{2} +6x=0$
$r)1- x^{2} =1-a$
$s) x(x+1)=9x-16$
Wyznacz miejsca zerowe funkcji F
$a) f(x)= x^{2} -2x-15$
$b) f(x)=3+5x-2 x^{2}$
Potrzebne na zaraz :P

Roma Równania kwadratowe:
k)
$2x^2+3=-7x \\\
2x^2+7x+3=0 \\\
\Delta = 7^2 - 4 \cdot 2 \cdot 3 = 49-24=25 \\\
\sqrt{\Delta} = \sqrt[25} = 5 \\\\
x_1 = \frac{-7-5}{2 \cdot 2} =\frac{-12}{4} =-3 \\\\
x_2 = \frac{-7+5}{2 \cdot 2} =\frac{-2}{4} =-\frac{1}{2} \\\\
Zatem: \\\
x = -3 \ \vee \ x =-\frac{1}{2}$

l)
$x+10=3 x^2 \\\
-3x^2+x+10=0 \\\
\Delta = 1^2 - 4 \cdot (-3) \cdot 10 = 1+120=121 \\\
\sqrt{\Delta} = \sqrt{121} = 11 \\\\
x_1 = \frac{-1-11}{2 \cdot (-3)} =\frac{-12}{-6} =2 \\\\
x_2 = \frac{-1+11}{2 \cdot (-3)} =\frac{10}{-6} =-\frac{5}{3}=-1\frac{2}{3} \\\\
Zatem: \\\
x =-1\frac{2}{3} \ \vee \ x =2$

ł)
$16 x^2 +24 \sqrt{3x} =-18$
Tak zapisane równanie nie ma rzeczywistych pierwiastków, chyba że chodzi o równanie:
$16x^2 +24 \sqrt{3}x =-18 \\\
16x^2 +24 \sqrt{3}x +18=0 \ /:2 \\\
8x^2 +12\sqrt{3}x +9=0 \\\
\Delta = (12\sqrt{3})^2 - 4 \cdot 8 \cdot 9= 432-288=144 \\\
\sqrt{\Delta} = \sqrt{144} = 12 \\\\
x_1 = \frac{-12\sqrt{3}-12}{2 \cdot 8} =\frac{4 \cdot (-3\sqrt{3}-3)}{16} =\frac{-3\sqrt{3}-3}{4} \\\\
x_2 = \frac{-12\sqrt{3}+12}{2 \cdot 8} =\frac{4 \cdot (-3\sqrt{3}+3)}{16} =\frac{-3\sqrt{3}+3}{4} \\\\
Zatem: \\\
x =\frac{-3\sqrt{3}-3}{4} \ \vee \ x =\frac{-3\sqrt{3}+3}{4}$

m)
$2x+18 x^2=0 \\\
2x \cdot (1 + 9x) = 0 \\\
2x = 0 \ \vee \ 1 +9x = 0 \\\\
2x = 0 \ / :2 \\\
x = 0 \\\\
1 +9x = 0 \\\
9x = - 1 \ /:9 \\\
x = -\frac{1}{9} \\\\
Zatem: \\\
x =-\frac{1}{9} \ \vee \ x = 0$

n)
$36x^2=6x \\\
36x^2-6x=0 \\\
6x \cdot (6x - 1) = 0 \\\
6x = 0 \ \vee 6x -1 = 0 \\\\
6x = 0 \ / :6 \\\
x = 0 \\\\
6x-1=0 \\\
6x = 1 \ /:6 \\\
x = \frac{1}{6} \\\\
Zatem: \\\
x = 0 \ \vee \ x = \frac{1}{6}$

o)
$24x- x^2 =144 \\\
-x^2 +24x-144 = 0 \ / \cdot (-1) \\\
x^2 -24x +144 = 0 \\\
(x - 12)^2 = 0 \ /^{\sqrt{}} \\\
x - 12 = 0 \\\
x = 12$

p)
$1+9x^2 +6x=0 \\\
9x^2 +6x+1=0 \\\
(3x +1)^2 = 0 \ /^{\sqrt{}}\\\
3x + 1 = 0 \\\
3x = -1 \ /:3 \\\
x = -\frac{1}{3}$

r)
$1- x^2 =1-a \\\
1- x^2 -1+a =0 \\\
-x^2 +a = 0 \ / \cdot(-1) \\\
x^2 - a = 0 \\\
x^2 = a \\\
x = \sqrt{a} \ \vee \ x = - \sqrt{a}$

s)
$x(x+1)=9x-16 \\\
x^2 +x - 9x +16 = 0 \\\
x^2 -8x +16 = 0 \\\
(x -4)^2 = 0 \ /^{\sqrt{}} \\\
x-4 = 0 \\\
x = 4$

Miejsca zerowe funkcji:
a)
$f(x)= x^2-2x-15 \\\
f(x) = 0 \\\
x^2-2x-15 =0 \\\
\Delta = (-2)^2 - 4 \cdot 1 \cdot (-15)= 4+60=64 \\\
\sqrt{\Delta} = \sqrt{64} = 8 \\\\
x_1 = \frac{2-8}{2 \cdot 1} =\frac{-6}{2} =-3 \\\\
x_2 = \frac{2+8}{2 \cdot 1} =\frac{10}{2} =5$

Miejsca zerowe funkcji f to: $- 3 \ i \ 5$ .

b)
$f(x)=3+5x-2 x^2 \\\
f(x) = 0 \\\
3+5x-2x^2 = 0 \\\
-2x^2+5x+3 = 0 \\\
\Delta = 5^2 - 4 \cdot (-2) \cdot 3= 25+24=49 \\\
\sqrt{\Delta} = \sqrt{49} = 7 \\\\
x_1 = \frac{-5-7}{2 \cdot (-2)} =\frac{-12}{-4} =3 \\\\
x_2 = \frac{-5+7}{2 \cdot (-2)} =\frac{2}{-4} =-\frac{1}{2}$

Miejsca zerowe funkcji f to: $-\frac{1}{2} \ i \ 3$ .

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