Kopertina e librit Matematika 12

Zgjidhja e ushtrimit 5

Zgjidhja e ushtrimit 5 të mësimit 1.4A në librin Matematika 12 nga shtëpia botuese Botime Pegi me autorë Brian Jefferson, David Bowles, Eddie Mullan, Garry Wiseman, John Rayneau, Katie Wood, Mike Heylings, Paul Williams dhe Rob Wagner.

Pyetja

Veçoni katrorin e binomit për të gjetur zgjidhjet e sakta për këto ekuacione të fuqisë së dytë.

  1. x2−2x=0x^2 - 2x = 0
  2. 3−4x−x2=03 - 4x - x^2 = 0
  3. x2−14x+33=0x^2 - 14x + 33 = 0
  4. x2+8x+10=0x^2 + 8x + 10 = 0
  5. x2−6x+9=0x^2 - 6x + 9 = 0
  6. x2+10x+24=0x^2 + 10x + 24 = 0
  7. x2+22x+118=0x^2 + 22x + 118 = 0
  8. x2−16x+54=0x^2 - 16x + 54 = 0
  9. 4x2−12x+2=04x^2 - 12x + 2 = 0
  10. 9x2+12x−2=09x^2 + 12x - 2 = 0
  11. x2+11x+3=0x^2 + 11x + 3 = 0
  12. 9x2−30x−32=09x^2 - 30x - 32 = 0

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Zgjidhja

  1. x2−2x=0⇒x2−2x+1−1=0⇒(x−1)2−1=0⇒(x−1)2=1⇒x−1=1⇒x−1=±1⇒x=1±1⇒x=2x^2 - 2x = 0 \rArr x^2-2x+1-1=0 \rArr (x - 1)^2 - 1 = 0 \rArr (x - 1)^2 = 1 \rArr x-1=\sqrt{1} \rArr x - 1 = \pm 1 \rArr x = 1 \pm 1 \rArr x= 2 ose x=0x=0
  2. 3−4x−x2=0⇒−(x2+4x−3)=0⇒−(x2+4x+4−4−3)⇒−[(x+2)2−4−3]=0⇒[(x+2)2−7]=0⇒x+2=±7⇒x=−2±73 - 4x - x^2 = 0 \rArr -(x^2+4x-3) = 0 \rArr -(x^2+4x+4-4-3) \rArr -[(x + 2)^2 - 4 - 3] = 0 \rArr [(x + 2)^2 - 7] = 0 \rArr x + 2 = \pm \sqrt{7} \rArr x = -2 \pm \sqrt{7} ⇒\rArr x=7−2x=\sqrt7-2 ose x=−2−7x=-2-\sqrt7
  3. x2−14x+33=0⇒x2−14x+49−49+33=0⇒(x−7)2−49+33=0⇒(x−7)2−16=0⇒(x−7)2=16⇒x−7=±4⇒x=7±4⇒x=7+4=11x^2 - 14x + 33 = 0 \rArr x^2-14x+49-49+33=0 \rArr (x - 7)^2 - 49 + 33 = 0 \rArr (x - 7)^2 - 16 = 0 \rArr (x-7)^2=16 \rArr x - 7 = \pm 4 \rArr x = 7 \pm 4 \rArr x=7+4 = 11 ose x=7−4=3x=7-4 =3
  4. x2+8x+10=0⇒x2+8x+16−16+10=0⇒(x+4)2−16+10=0⇒(x+4)2−6=0⇒(x+4)2=6⇒x+4=±6⇒x=−4±6x^2 + 8x + 10 = 0 \rArr x^2+8x+16-16+10=0 \rArr (x + 4)^2 - 16 + 10 = 0 \rArr (x + 4)^2 - 6 = 0 \rArr (x+4)^2=6 \rArr x + 4 = \pm \sqrt{6} \rArr x = -4 \pm \sqrt{6} ⇒\rArr x=6−4x=\sqrt6-4 ose x=−4−6x=-4-\sqrt6
  5. x2−6x+9=0⇒(x−3)2=0⇒x−3=0⇒x=3x^2 - 6x + 9 = 0 \rArr (x - 3)^2 = 0 \rArr x - 3 = 0 \rArr x = 3
  6. x2+10x+24=0⇒x2+10+25−25+24=0⇒(x+5)2−25+24=0⇒(x+5)2=1⇒x+5=±1⇒x=−5±1⇒x=−5+1=−4=−6x^2 + 10x + 24 = 0 \rArr x^2+10+25-25+24=0 \rArr (x + 5)^2 - 25 + 24 = 0 \rArr (x + 5)^2 = 1 \rArr x + 5 = \pm 1 \rArr x = -5 \pm 1 \rArr x=-5+1=-4 = -6 ose x=−5−1=−6x=-5-1=-6
  7. x2+22x+118=0⇒x2+22x+121−121+118=0⇒(x+11)2−121+118=0⇒(x+11)2=3⇒x+11=±3⇒x=−11±3⇒x=3−11x^2 + 22x + 118 = 0 \rArr x^2+22x+121-121+118=0 \rArr (x + 11)^2 - 121 + 118 = 0 \rArr (x + 11)^2 = 3 \rArr x + 11 = \pm \sqrt{3} \rArr x = -11 \pm \sqrt{3} \rArr x=\sqrt3-11 ose x=−11−3x=-11-\sqrt3
  8. x2−16x+54=0⇒x2−16x+64−64+54=0⇒(x−8)2−64+54=0⇒(x−8)2−10=0⇒x−8=±10⇒x=8±10⇒x=8+10x^2 - 16x + 54 = 0 \rArr x^2-16x+64-64+54=0 \rArr (x - 8)^2 - 64 + 54 = 0 \rArr (x - 8)^2 - 10 = 0 \rArr x - 8 = \pm \sqrt{10} \rArr x = 8 \pm \sqrt{10} \rArr x=8+\sqrt{10} ose x=8−10x=8-\sqrt{10}
  9. 4x2−12x+2=0⇒4x2−12x+9−9+2=0⇒(2x−3)2−9+2=0⇒(2x−3)2−7=0⇒2x−3=±7⇒x=3±724x^2 - 12x + 2 = 0 \rArr 4x^2-12x+9-9+2=0 \rArr (2x - 3)^2 - 9 + 2 = 0 \rArr (2x - 3)^2 - 7 = 0 \rArr 2x - 3 = \pm \sqrt{7} \rArr x = \dfrac{3 \pm \sqrt{7}}{2} ⇒\rArr x=3+72x=\dfrac{3+\sqrt7}{2} ose x=3−72x=\dfrac{3-\sqrt7}{2}
  10. 9x2+12x−2=0⇒9x2+12x+4−4−2=0⇒(3x+2)2−4−2=0⇒(3x+2)2−6=0⇒3x+2=±6⇒x=−2±639x^2 + 12x - 2 = 0 \rArr 9x^2+12x+4-4-2=0 \rArr (3x + 2)^2 - 4 - 2 = 0 \rArr (3x + 2)^2 - 6 = 0 \rArr 3x +2 = \pm \sqrt{6} \rArr x = \dfrac{-2 \pm \sqrt{6}}{3} ⇒\rArr x=6−23x=\dfrac{\sqrt6-2}{3} ose x=−2−63x=\dfrac{-2-\sqrt6}{3}
  11. x2+11x+3=0⇒x2+11x+11222−11222+3=0⇒(x+112)2−1214+3=0⇒(x+112)2−1094=0⇒x+112=±1094⇒x=−11±1092x^2 + 11x + 3 = 0 \rArr x^2+11x+\dfrac{11^2}{2^2}-\dfrac{11^2}{2^2}+3=0 \rArr (x + \dfrac{11}{2})^2 - \dfrac{121}{4} + 3 = 0 \rArr (x + \dfrac{11}{2})^2 - \dfrac{109}{4} = 0 \rArr x + \dfrac{11}{2} = \pm \sqrt{\dfrac{109}{4}} \rArr x = \dfrac{-11 \pm \sqrt{109}}{2} ⇒\rArr x=109−112x=\dfrac{\sqrt{109}-11}{2} ose x=−11−1092x=\dfrac{-11-\sqrt{109}}{2}
  12. 9x2−30x−32=0⇒9x2−30x+25−25−32=0⇒(3x−5)2−25−32=0⇒(3x−5)2−57=0⇒3x−5=±57⇒x=5±5739x^2 - 30x - 32 = 0 \rArr 9x^2-30x+25-25-32=0 \rArr (3x - 5)^2 - 25 - 32 = 0 \rArr (3x - 5)^2 - 57 = 0 \rArr 3x - 5 = \pm \sqrt{57} \rArr x = \dfrac{5 \pm \sqrt{57}}{3} ⇒\rArr x=5+573x=\dfrac{5+\sqrt{57}}{3} ose x=5−573x=\dfrac{5-\sqrt{57}}{3}