Anusiaa2viz
Rozwiązane

1. Oblicz sumę pierwiastków
(x-1)(x-2) = 2x2 - 4

2. Oblicz iloczyn pierwiastków
-2(2-x)2= (x-3)(x+3)

3. Oblicz sumę i iloczyn pierwiastków
x2-5x+4 = (6-x)(4x+2)

Daje naj!!!

Odpowiedź :

Catta1eyaviz

[tex]1. \\(x-1)(x-2)=2x^2-4\\x^2-2x-x+2=2x^2-4\\x^2-3x+2-2x^2+4=0\\-x^2-3x+6=0\\\Delta=(-3)^2-4*(-1)*6\\\Delta=9+24\\\Delta=33\\\sqrt{\Delta}=\sqrt{33}\\x_1=\frac{3-\sqrt{33}}{-2}=-\frac{3-\sqrt{33}}2\\x_2=\frac{3+\sqrt{33}}{-2}=-\frac{3+\sqrt{33}}2[/tex]

[tex]x_1+x_2=-\frac{3-\sqrt{33}}2+(-\frac{3+\sqrt{33}}2)=-\frac{3-\sqrt{33}}2-\frac{3+\sqrt{33}}{2}=\frac{-(3-\sqrt{33})-(3+\sqrt{33})}2=\frac{-3+\sqrt{33}-3-\sqrt{33}}2=\frac{-6}2=-3[/tex]

[tex]2. \\-2(2-x)^2=(x-3)(x+3)\\-2(4-4x+x^2)=x^2-9\\-8+8x-2x^2=x^2-9\\0=x^2-9+8-8x+2x^2\\0=3x^2-8x-1\\\Delta=(-8)^2-4*3*(-1)\\\Delta=64+12=76\\\sqrt{\Delta}=\sqrt{76}=2\sqrt{19}\\\\x_1=\frac{8-2\sqrt{19}}{6}=\frac{2(4-\sqrt{19})}6=\frac{4-\sqrt{19}}3\\x_2=\frac{8+2\sqrt{19}}6=\frac{2(4+\sqrt{19})}6=\frac{4+\sqrt{19}}3\\[/tex]

[tex]x_1*x_2=\frac{4-\sqrt{19}}3*\frac{4+\sqrt{19}}3=\frac{(4-\sqrt{19})(4+\sqrt{19})}9=\frac{4^2-(\sqrt{19})^2}9=\frac{16-19}9=\frac{-3}9=-\frac13[/tex]

3.

[tex]x^2-5x+4=(6-x)(4x+2)\\x^2-5x+4=24x+12-4x^2-2x\\x^2-5x+4-22x-12+4x^2=0\\5x^2-27x-8=0\\\Delta=(-27)^2-4*5*(-8)\\\Delta=729+160\\\Delta=889\\\sqrt{\Delta}=\sqrt{889}\\x_1=\frac{27+\sqrt{889}}{10}\\x_2=\frac{27-\sqrt{889}}{10}[/tex]

[tex]x_1+x_2=\frac{27+\sqrt{889}}{10}+\frac{27-\sqrt{889}}{10}=\frac{27+\sqrt{889}+27-\sqrt{889}}{10}=\frac{2*27}{10}=\frac{27}5=5\frac25=5,4[/tex]

[tex]x_1*x_2=\frac{27+\sqrt{889}}{10}*\frac{27-\sqrt{889}}{10}=\frac{(27+\sqrt{889})(27-\sqrt{889})}{100}=\frac{27^2-(\sqrt{889})^2}{100}=\frac{729-889}{100}=\frac{-160}{100}=-\frac{16}{10}=-1\frac6{10}=-1\frac3{5}=-1.6[/tex]

Basetlaviz

Jeżeli równanie kwadratowe  [tex]ax^{2}+bx + c = 0[/tex]  [tex](gdzie \ a \neq 0)[/tex] ma dwa rozwiązania  [tex]x_1, \ x_2,[/tex] to ze wzrów Viete'a:

[tex]x_1 + x_2 = -\frac{b}{a}, \ \ \ \ x_1\cdot x_2 = \frac{c}{a}[/tex]

Równanie kwadratowe ma rozwiązanie, gdy Δ ≥ 0.

1.

[tex](x-1)(x-2) = 2^{2}-4\\\\x^{2}-2x-x+2 = 2x^{2}-4\\\\x^{2}-2x^{2}-3x+2+4 = 0\\\\-x^{2}-3x+6 = 0\\\\a = -1, \ b = -3, \ c = 6\\\\\Delta = b^{2}-4ac = (-3)^{2}-4\cdot(-1)\cdot6 = 9+24 = 33 > 0\\\\x_1 + x_2 = -\frac{b}{a}=-\frac{-3}{-1}\\\\\boxed{x_1+x_2 = -3}[/tex]

2.

[tex]-2(2-x)^{2} = (x-3)(x+3)\\\\-2(4 - 4x + x^{2}) = x^{2}-9\\\\-8+8x-2x^{2}-x^{2}+9 = 0\\\\-3x^{2}+8x+1 = 0\\\\a = -3, \ b = 8, \ c = 1\\\\\Delta = b^{2}-4ac= 8^{2}-4\cdot(-3)\cdot1 = 64+12 = 76 > 0\\\\x_1\cdot x_2=\frac{c}{a} \\\\\boxed{x_1\cdot x_2 = -\frac{1}{3}}[/tex]

3.

[tex]x^{2}-5x+4 = (6-x)(4x+2)\\\\x^{2}-5x+4 = 24x+12 - 4x^{2}-2x\\\\x^{2}-5x+4 = -4x^{2}+22x+12\\\\x^{2}+4x^{2}-5x-22x+4-12 = 0\\\\5x^{2}-27x-8 = 0\\\\a = 5, \ b = -27, \ c = -8\\\\\Delta = b^{2}-4ac = (-27)^{2}-4\cdot5\cdot(-8) = 729+160 = 889 > 0\\\\x_1+x_2 = -\frac{b}{a} = -\frac{-27}{5}\\\\\boxed{x_1+x_2 = 5,4}\\\\x_1\cdot x_2 = \frac{c}{a}=\frac{-8}{5}\\\\\boxed{x_1\cdot x_2 = -1,6}[/tex]

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