Kopertina e librit Matematika 12

Zgjidhja e ushtrimit 3

Zgjidhja e ushtrimit 3 të mësimit 2B në librin Matematika 12 nga shtëpia botuese Mediaprint me autorë Greg Attwood, Jack Barraclough, Ian Bettison, Alistair Macpherson, Bronwen Moran, Su Nicholson, Diane Oliver, Joe Petran, Keith Bledger, Harry Smith, Geoff Staley, Robert Ward-Penny dhe Dave Wilkins.

Pyetja

Për secilin nga ekuacionet që vijojnë, zgjidh një metodë të përshtatshme dhe gjej të gjitha zgjidhjet. Kur është e nevojshme, përgjigja të jepet me afërsi dy shifra dhjetore.

  1. x2+8x+12=0x^2 + 8x +12 = 0
  2. x2+9x−11=0x^2 + 9x - 11 = 0
  3. x2−9x−1=0x^2 - 9x - 1 = 0
  4. 2x2+5x+2=02x^2 + 5x + 2 = 0
  5. (2x+8)2=100(2x+8)^2 = 100
  6. 6x2+6=12x6x^2 + 6 = 12x
  7. 2x2−11=7x2x^2 - 11 = 7x
  8. x=8x−15x = \sqrt{8x - 15}

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Zgjidhja

  1. x2+2x+6x+12=0⇒x(x+2)+6(x+2)=0⇒(x+6)(x+2)=0→x+6=0⇒x=−6x^2 + 2x + 6x + 12 = 0 \rArr x(x+2)+6(x+2)=0 \rArr (x+6)(x+2)=0 \to x+6= 0 \rArr \boxed{x=-6} dhe x+2=0⇒x=−2x+2=0 \rArr \boxed{x=-2}
  2. 1xx2 = −9±92−4×1×(−11)2×1=−9±1252⇒\dfrac{-9 \pm \sqrt{9^2 - 4 \times 1 \times (-11)}}{2 \times 1} = \dfrac{-9 \pm \sqrt{125}}{2} \rArr x1=−9+1252≈−9+11.182=2.182=1.09\boxed{x_1 = \dfrac{-9+\sqrt{125}}{2} \approx \dfrac{-9 + 11.18}{2} = \dfrac{2.18}{2} = 1.09} dhe x2=−9−1252≈−9−11.182=−20.182=−10.09\boxed{x_2 = \dfrac{-9-\sqrt{125}}{2} \approx \dfrac{-9 - 11.18}{2} = \dfrac{-20.18}{2} = -10.09}
  3. 1xx2 = 9±(−9)2−4×1×(−1)2×1=9±852⇒\dfrac{9 \pm \sqrt{(-9)^2 - 4 \times 1 \times (-1)}}{2 \times 1} = \dfrac{9 \pm \sqrt{85}}{2} \rArr x1=9+852≈9+9.232=18.232≈9.12\boxed{x_1 = \dfrac{9 + \sqrt{85}}{2} \approx \dfrac{9 + 9.23}{2} = \dfrac{18.23}{2} \approx 9.12} dhe x2=9−852≈9−9.232=−0.232≈−0.12\boxed{x_2 = \dfrac{9 - \sqrt{85}}{2} \approx \dfrac{9 - 9.23}{2} = \dfrac{-0.23}{2} \approx -0.12}
  4. 1xx2 = −5±52−4×2×22×2=−5±94=−5±34\dfrac{-5 \pm \sqrt{5^2 - 4 \times 2 \times 2}}{2 \times 2} = \dfrac{-5 \pm \sqrt{9}}{4} = \dfrac{-5 \pm 3}{4} ⇒\rArr x1=−5+34=−24=−12\boxed{x_1 = \dfrac{-5+3}{4} = \dfrac{-2}{4} = -\dfrac{1}{2}} dhe x2=−5−34=−84=−2\boxed{x_2 = \dfrac{-5-3}{4} = \dfrac{-8}{4} = -2}
  5. (2x+8)2=100⇒2x+8=±10⇒2x=±10−8⇒x=±10−82\sqrt{(2x+8)^2} = \sqrt{100} \rArr 2x+8= \pm 10 \rArr 2x = \pm 10 -8 \rArr x = \dfrac{\pm 10 - 8}{2} →\to x=10−82=22=1\boxed{x = \dfrac{10-8}{2} = \dfrac{2}{2} = 1} dhe x=−10−82=−182=−9\boxed{x = \dfrac{-10-8}{2} = \dfrac{-18}{2} = -9}
  6. 6x2−12x+6=0⇒6(x2−2x+1)=0⇒x2−2x+1=0⇒x2−x−x−1=0⇒x(x−1)−1(x−1)=0⇒(x−1)(x−1)=06x^2 - 12x + 6 = 0 \rArr 6(x^2 - 2x + 1) = 0 \rArr x^2 - 2x + 1 = 0 \rArr x^2 - x - x - 1 = 0 \rArr x(x-1)-1(x-1)= 0 \rArr (x-1)(x-1)=0 →\to x−1=0⇒x=1x-1=0 \rArr \boxed{x=1}
  7. 2x2−7x−11=0⇒2x^2 - 7x - 11 = 0 \rArr 1xx2 = 7±(−7)2−4×2×(−11)2×2=7±1374\dfrac{7 \pm \sqrt{(-7)^2 - 4 \times 2 \times (-11)}}{2 \times 2} = \dfrac{7 \pm \sqrt{137}}{4} ⇒\rArr x1=7+1374≈7+11.74=18.74≈4.68\boxed{x_1 = \dfrac{7 + \sqrt{137}}{4} \approx \dfrac{7 + 11.7}{4} = \dfrac{18.7}{4} \approx 4.68} dhe x2=7−1374≈7−11.74=−4.74≈−1.18\boxed{x_2 = \dfrac{7 - \sqrt{137}}{4} \approx \dfrac{7 - 11.7}{4} = \dfrac{-4.7}{4} \approx -1.18}
  8. x2=(8x−15)2⇒x2=8x−15⇒x2−8x−15=0⇒x2−3x−5x−15=0⇒x(x−3)−5(x−3)=0⇒(x−5)(x−3)x^2 = (\sqrt{8x-15})^2 \rArr x^2 = 8x - 15 \rArr x^2 - 8x - 15 = 0 \rArr x^2 - 3x - 5x - 15 = 0 \rArr x(x-3)-5(x-3)=0 \rArr (x-5)(x-3) →\to x−5=0⇒x=5x-5 = 0 \rArr \boxed{x=5} dhe x−3=0⇒x=3x-3 = 0 \rArr \boxed{x=3}