Esercizio n° 1

Risolvi le seguente equazioni purie e spurie.

1)(2x+1)(3x-5) =x² – 5

2)3+\sqrt{2} x(x+3)=3(\sqrt{2} x + 1) -4x

3)(2-x)²+3x(1+x)=5-x

4)\frac{1-2x}{2} + \frac{2+4x ^{2}}{3} = \frac{x ^{2}}{6} – x

5)(\sqrt{2}x-\sqrt{3})(\sqrt{2}x+\sqrt{3}) = (x + \sqrt{2})² – 2x\sqrt{2}

6)\frac{(2x-1)(2x+1)-4(x-1)}{3} = 1 – \frac{4}{3}x

7)x(x-2)+1=(1-x)(1+x)

8)2-x(1+3x)=  [ 7-(1 -5x) ]x + 2(1 -x)

9)x² + 3 – { 1 -[ 2 – x² – x] }= 4 +2x + x²

10)\frac{1-3x}{5} + 1 – \frac{(2-x)(2 +x)}{3} = x – \frac{1}{5} + \frac{1 + x^{2}}{15}

11)(2x+1)(x-3) = (1 -x)(4 -x)

12)x² – 8 = \frac{(x+1)(x-1)}{8}

13)(x+4)² +1= 8x

14)(2x – \frac{1}{3})(2x + \frac{1}{3})- (2x + \frac{1}{3})² +4x(x +\frac{1}{3})=0 

15)2x(x -3)+\frac{1}{2}(\frac{1}{3}x² -1) + \frac{2x - x^{2}}{6} = – \frac{1}{3}( 17x + \frac{21}{2})

Esercizio n° 2

Risolvi le seguenti equazioni di secondo grado complete.

1)x(3x-1) – 8 -2x(1 – x)

2)\sqrt{5}x² – 4x – \sqrt{5} =0

3)2x² – 3x + 20=0

4)x² – \frac{5}{3} x + \frac{25}{36}=0

5)x² – \sqrt{2}x – 4 = 0

6)x(x – 9) = \frac{19}{4}

Esercizio n° 3

Risolvi le seguenti equazioni di secondo grado applicando la formula ridotta.

1)x² +6x – 7=0

2)10y²+8y+5=0

3)16x²+24x+9

Esercizio n° 4

Risolvi le seguenti equazioni di secondo grado il cui discriminante è riconducibile al quadrato di un binomio.

1)x² + 2x – 2\sqrt{2} – 2

2)2x² – 4x – 1 + 2\sqrt{2}=0

3)x(x-2)= 4\sqrt{3}(2 – x)

 

SVOLGIMENTO

Esercizio n° 1

Risolvi le seguente equazioni purie e spurie.

1)(2x+1)(3x-5) =x² – 5

6x² -10x + 3x – 5= x² – 5

6x² – ×² – 10x + 3x =0

5x² – 7x =0

x(5x – 7 )=0                   

x=0      e       5x-7=0 ⇒ x= \frac{7}{5}             (0;\frac{7}{5})

2)3+\sqrt{2} x(x+3)=3(\sqrt{2} x + 1) -4x

3 +x ²\sqrt{2} +3x \sqrt{2}= 3x \sqrt{2} + 3 – 4x

x ²\sqrt{2} + 4x =0

x(x\sqrt{2} + 4)=0

x=0        e          x\sqrt{2} + 4=0⇒  x\sqrt{2} = -4 quindi x = –\frac{4}{\sqrt{2}}= -2\sqrt{2}         (0;-2\sqrt{2}  )

3)(2-x)²+3x(1+x)=5-x

4 + x² – 4x + 3x + 3x² = 5 – x

4x² – 1 =0

x²= \frac{1}{4} ⇒ x= ±\sqrt{\frac{1}{4}} = ±\frac{1}{2}       (-\frac{1}{2}; +\frac{1}{2})

4)\frac{1-2x}{2} + \frac{2+4x ^{2}}{3} = \frac{x ^{2}}{6} – x

\frac{3(1 - 2x)+2(2 + 4x ^{2})}{6} = \frac{x ^{2}-6x}{6}

3 – 6x + 4 + 8x² = x² – 6x

8x² – x² + 7 =0

7x²= – 7 

x²= – 1     IMPOSSIBILE

5)(\sqrt{2}x-\sqrt{3})(\sqrt{2}x+\sqrt{3}) = (x + \sqrt{2})² – 2x\sqrt{2}

2x² – 3= x² + 2 + 2\sqrt{2}x 2\sqrt{2}x

2x² –  x²  – 3 – 5 =0

x²-5 =0         x²= 5         x= ± \sqrt{5}

6)\frac{(2x-1)(2x+1)-4(x-1)}{3} = 1 – \frac{4}{3}x

\frac{4x ^{2}-1-4x+4}{3} = \frac{3-4x}{3}

4x² + 34x = 3  4x

4x²=0    x=0      due radici entrambe nulle

7)x(x-2)+1=(1-x)(1+x)

x² – 2x + 1 = 1 – x²

x² + x² – 2x =0

2x(x – 1)=0

x=0         x-1=0⇒ x=1        (0;1)

8)2- x(1+3x)=  [ 7-(1 -5x) ]x + 2(1 -x)

2 – x – 3x² = [ 7 – 1 + 5x] x + 2 – 2x

2 – x – 3x² = 6x + 5x² + 2 – 2x

5x² + 6x – 2x + 3x² + x=0

8x² + 5x =0

x(8x + 5)=0           x=0 e  8x + 5 =0⇒ x= -\frac{5}{8}        (0;-\frac{5}{8})

9)x² + 3 – { 1 -[ 2 – x² – x] }= 4 +2x + x²

x² + 3 – { 1 – 2 + x² + x } = 4 + 2x + x²

+ 3 +1 – x² – x = 4 + 2x +

-x² – x + 4 – 4 – 2x =0

x² + 3x =0      

x(x + 3)=0          x=0    e   x+3=0 ⇒x=-3        (0; -3)

10)\frac{1-3x}{5} + 1 – \frac{(2-x)(2 +x)}{3} = x – \frac{1}{5} + \frac{1 + x^{2}}{15}

\frac{3(1 - 3x) + 15 - 5(4 - x ^{2})}{15} = \frac{15x - 3 + 1 + x ^{2}}{15}

3 – 9x + 15 – 20 + 5x² = 15x – 3 + 1 + x²

5x² – x² – 9x – 15x – 2 + 3 – 1 =0

4x² – 24x =0           

 4x(x – 6)=0        4x=0 ⇒x=0       e  x – 6 =0⇒ x=6

11)(2x+1)(x-3) = (1 -x)(4 -x)

2x² – 6x + x – 3 = 4 – x4x + x²

2x² – x² = 7

x²= 7 ⇒  x=±\sqrt{7}

12)x² – 8 = \frac{(x+1)(x-1)}{8}

\frac{8x ^{2} - 64}{8} = \frac{x ^{2} - 1}{8}

7x² – 63 =0

x² = \frac{63}{7} =9   ⇒ x=± \sqrt{9} ⇒ x=±3

13)(x+4)² +1= 8x

x² + 16 + 8x + 1 = 8x

x² + 15=0

x² = – 15   IMPOSSIBILE

14)(2x – \frac{1}{3})(2x + \frac{1}{3})- (2x + \frac{1}{3})² +4x(x +\frac{1}{3})=0 

4x² – \frac{1}{9} – (4x² + \frac{1}{9} + \frac{4}{3}x) + 4x² + \frac{4}{3}x =0

4x² – \frac{1}{9} – (4x² + \frac{1}{9} + \frac{4}{3}x) + 4x² + \frac{4}{3}x =0

4x² – \frac{1}{9} – 4x² – \frac{1}{9} – \frac{4}{3}x+ 4x² + \frac{4}{3}x =0

4x² – \frac{2}{9}= 0

4x² = \frac{2}{9} ⇒ x²= \frac{1}{18}

x=± \sqrt{\frac{1}{18}}

15)2x(x -3)+\frac{1}{2}(\frac{1}{3}x² -1) + \frac{2x - x^{2}}{6} = – \frac{1}{3}( 17x + \frac{21}{2})

2x² – 6x + \frac{1}{6}x² – \frac{1}{2} + \frac{2x - x ^{2}}{6} = -\frac{17}{3}x – \frac{7}{2}

\frac{12x ^{2} - 36x + x ^{2} - 3 + 2x - x ^{2}}{6} = \frac{-34x - 21}{6}

12x² + x ² 36x + 2x + 34x – 3 + 21 =0

12x² + 18 =0

x² =-  \frac{18}{12} IMPOSSIBILE

 

Esercizio n° 2

Risolvi le seguenti equazioni di secondo grado complete.

1)x(3x-1) – 8 -2x(1 – x)

3x² – x – 8 – 2x + 2x²=0

5x² – 3x – 8 =0

Δ= b² – 4ac           Δ= 9 + 160= 169

 x= \frac{-b \pm \sqrt{b^{2}-4ac} }{2a}  

x_{{1}}=  \frac{+3 + \sqrt{169}}{10}\frac{+3 + 13}{10} = \frac{16}{10} = \frac{8}{5}

x_{{2}}\frac{+3 - \sqrt{169}}{10}\frac{+3 - 13}{10} = – \frac{10}{10} = – 1

2)\sqrt{5}x² – 4x – \sqrt{5} =0

Δ= b² – 4ac           Δ= 16 + 20 = 36

x= \frac{-b \pm \sqrt{b^{2}-4ac} }{2a}  

x_{{1}}=  \frac{+4+\sqrt{36}}{2\sqrt{5}}  = \frac{+4+6}{2\sqrt{5}}  = \frac{10}{2\sqrt{5}}  = \frac{5}{\sqrt{5}}  =  razionalizzando \sqrt{5}

x_{{2}}=\frac{+4-\sqrt{36}}{2\sqrt{5}} \frac{+4-6}{2\sqrt{5}}  =- \frac{2}{2\sqrt{5}}  = – \frac{1}{\sqrt{5}}

3)2x² – 3x + 20=0

Δ= b² – 4ac           Δ= 9 – 80 < 0 IMPOSSIBILE

4)x² – \frac{5}{3} x + \frac{25}{36}=0

Δ= b² – 4ac           Δ= \frac{25}{9} – \frac{100}{36} = 0

X =\frac{\frac{5}{3}}{2} = \frac{5}{3} • \frac{1}{2} = \frac{5}{6}     x_{{1}}=x_{{2}}=\frac{5}{6}   RADICE DOPPIA

5)x² – \sqrt{2}x – 4 = 0

Δ= b² – 4ac           Δ=(-\sqrt{2})² + 16 = 18

x= \frac{-b \pm \sqrt{b^{2}-4ac} }{2a}  

x_{{1}}=\frac{\sqrt{2}+ \sqrt{18}}{2} = \frac{\sqrt{2}+ 3\sqrt{2}}{2} = \frac{4\sqrt{2}}{2} = 2\sqrt{2}

x_{{2}}\frac{\sqrt{2 }- \sqrt{18}}{2} = \frac{\sqrt{2 }- 3\sqrt{2}}{2} = \frac{-2\sqrt{2}}{2} = – \sqrt{2}

6)x(x – 9) = \frac{19}{4}

x² – 9x – \frac{19}{4}

Δ= b² – 4ac           Δ= 81 + 19 = 100

x= \frac{-b \pm \sqrt{b^{2}-4ac} }{2a}  

x_{{1}}\frac{+9 + \sqrt{100}}{2} = \frac{+9+ 10}{2} = \frac{19}{2}

x_{{2}}=\frac{+9 - \sqrt{100}}{2} = \frac{+9- 10}{2} = – \frac{1}{2}

Esercizio n° 3

Risolvi le seguenti equazioni di secondo grado applicando la formula ridotta.

1)x² +6x – 7=0

\frac{\Delta }{4} =( \frac{b}{2})^{2} -ac. = (\frac{6}{2}) ² +7 =  9 + 7 = 16

x=\frac{-\frac{b}{2} \pm \sqrt{(\frac{b}{2})^{2}-ac} }{a}

x_{{1}}=-3 +\sqrt{16} = -3 + 4 =1

x_{{2}}=-3 -\sqrt{16} = -3 – 4 = -7

2)10y²+8y+5=0

\frac{\Delta }{4} =( \frac{b}{2})^{2} -ac.= 16 – 50 < 0 IMPOSSIBILE

3)16x²+24x+9

\frac{\Delta }{4} =( \frac{b}{2})^{2} -ac.= (\frac{24}{12}) ^{2}– 144 = 144 – 144 = 0  

x=\frac{-\frac{b}{2} \pm \sqrt{(\frac{b}{2})^{2}-ac} }{a}

x_{{1}}x_{{2}}-\frac{12}{16} = -\frac{3}{4}

Esercizio n° 4

Risolvi le seguenti equazioni di secondo grado il cui discriminante è riconducibile al quadrato di un binomio.

1)x² + 2x – 2\sqrt{2} – 2

\frac{\Delta }{4}= (\frac{2 }{2})² – (– 2\sqrt{2} – 2)= 1 + 2\sqrt{2} + 2 questo discriminante può essere ricondotto a un quadrato=(1 + \sqrt{2}

x= -1 ± \sqrt{(1 + \sqrt{2}) ^{2}}

x_{{1}}= – 1 + 1 + \sqrt{2} = \sqrt{2}

x_{{2}}= -1 – 1 – \sqrt{2} = -2 –\sqrt{2}

2)2x² – 4x – 1 + 2\sqrt{2}=0

\frac{\Delta }{4}(-\frac{4}{2}) ^{2} – (2)(-1 + 2\sqrt{2}) = 4 + 2 – 4\sqrt{2} = (2 – \sqrt{2}

x_{{1}}\frac{+2 +\sqrt{(2-\sqrt{2}) ^{2}}}{2} = \frac{2 + 2 - \sqrt{2}}{2} = 4 – \sqrt{2}

x_{{2}}\frac{+2 -\sqrt{(2-\sqrt{2}) ^{2}}}{2}= \frac{2- 2 + \sqrt{2}}{2}\frac{\sqrt{2}}{2}

3)x(x-2)= 4\sqrt{3}(2 – x)

x² – 2x +4x \sqrt{3} – 8\sqrt{3}=0

x² + 2(-x + 2x\sqrt{3}) – 8\sqrt{3}=0

\frac{\Delta }{4}= (-1 + 2\sqrt{3})²+8\sqrt{3}  = 1 + 12 – 4\sqrt{3}+8\sqrt{3} = 13 + 4\sqrt{3} riconduciamo tale discriminante a un quadrato e lo scriviamo come 12 + 1 + 4\sqrt{3} = (2\sqrt{3} + 1)²

x_{{1}}= + 1 – 2\sqrt{3} + \sqrt{(2\sqrt{3}+1) ^{2}} = 1 – 2\sqrt{3} +2\sqrt{3} + 1= 2

x_{{2}}=+ 1 – 2\sqrt{3} – \sqrt{(2\sqrt{3}+1) ^{2}} = 1 – 2\sqrt{3} -2\sqrt{3} -1 = – 4\sqrt{3}

 

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