Kopertina e librit Matematika 12

Zgjidhja e ushtrimit 1

Zgjidhja e ushtrimit 1 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

Zgjidhni këto ekuacione të fuqisë së dytë me anë të faktorizimit.

  1. x2−18=0x^2 - 18 = 0
  2. 2x2−6=02x^2 - 6 = 0
  3. 4x2+5x=04x^2 + 5x = 0
  4. x2+23x+3=0x^2 + 2\sqrt{3}x + 3 = 0
  5. 2x2+5x−3=02x^2 + 5x - 3 = 0
  6. 3x2−23x+14=03x^2 - 23x + 14 = 0
  7. 16x2−24x+9=016x^2 - 24x + 9 = 0
  8. 18+x−4x2=018 + x - 4x^2 = 0

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Zgjidhja

  1. x2−18=0⇒(x−18)(x+18)=0⇒x=±32x^2 - 18 = 0 \rArr (x - \sqrt{18})(x + \sqrt{18}) = 0 \rArr x = \pm 3\sqrt{2}
  2. 2x2−6=0⇒2(x−3)(x+3)=0⇒x=±32x^2 - 6 = 0 \rArr 2(x - \sqrt{3})(x + \sqrt{3}) = 0 \rArr x = \pm \sqrt{3}
  3. 4x2+5x=0⇒4x(x+54)=0⇒4x=0⇒x=04x^2 + 5x = 0 \rArr 4x(x + \dfrac{5}{4}) = 0 \rArr 4x = 0 \rArr x=0 ose x+54=0⇒x=−54x + \dfrac{5}{4} = 0 \rArr x= -\dfrac{5}{4}
  4. x2+23x+3=0⇒x2+3x+3x+3=0⇒x(x+3)+3(x+3)=0⇒(x+3)(x+3)=0⇒x=−3x^2 + 2\sqrt{3}x + 3 = 0 \rArr x^2+\sqrt3x+\sqrt3x+3=0 \rArr x(x+\sqrt3)+\sqrt3(x+\sqrt3) = 0 \rArr (x + \sqrt{3})(x + \sqrt{3}) = 0 \rArr x = -\sqrt{3}
  5. 2x2+5x−3=0⇒2x2−x+6x−3=0⇒x(2x−1)+3(2x−1)=0⇒(2x−1)(x+3)=0⇒2x−1=0⇒2x=1⇒x=122x^2 + 5x - 3 = 0 \rArr 2x^2-x+6x-3=0 \rArr x(2x-1)+3(2x-1) = 0 \rArr (2x - 1)(x + 3) = 0 \rArr 2x-1=0 \rArr 2x=1 \rArr x = \dfrac{1}{2} ose x+3=0⇒x=−3x+3=0 \rArr x=-3
  6. 3x2−23x+14=0⇒3x2−21x−2x+14=0⇒3x(x−7)−2(x−7)=0⇒(3x−2)(x−7)=03x^2 - 23x + 14 = 0 \rArr 3x^2-21x-2x+14=0 \rArr 3x(x-7)-2(x-7)=0 \rArr (3x - 2)(x - 7) = 0  ⇒\rArr  3x−2=0⇒3x=2⇒x=233x-2=0 \rArr 3x=2 \rArr x=\dfrac{2}{3} ose x−7=0⇒x=7x-7=0 \rArr x=7
  7. 16x2−24x+9=0⇒16x2−12x−12x+9=0⇒4x(4x−3)−3(4x−3)=0⇒(4x−3)(4x−3)=016x^2 - 24x + 9 = 0 \rArr 16x^2-12x-12x+9=0 \rArr 4x(4x-3)-3(4x-3)=0 \rArr (4x - 3)(4x - 3) = 0 ⇒\rArr 4x−3=0⇒4x=3⇒x=344x-3=0 \rArr 4x=3 \rArr x=\dfrac{3}{4}
  8. 18+x−4x2=0⇒−4x2+x+18=0⇒−4x2−8x+9x+18=0⇒−4x(x+2)+9(x+2)=0⇒(9−4x)(x+2)=018 + x - 4x^2 = 0 \rArr -4x^2+x+18=0 \rArr -4x^2-8x+9x+18=0 \rArr -4x(x+2)+9(x+2)=0 \rArr (9 - 4x)(x+2) = 0 ⇒\rArr 9−4x=0⇒4x=9⇒x=949-4x=0 \rArr 4x=9 \rArr x=\dfrac{9}{4} ose x+2=0⇒x=−2x+2=0 \rArr x=-2