Gawanya mlinganyo mzima kwa mgawo wa x2, ambao ni 4.

Mathematics

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Haya, nitakusaidia kutatua maswali haya.

1. Tatua 4x2−16x−9=04x^2 - 16x - 9 = 0 kwa kutumia njia ya kukamilisha mraba.

  1. Gawanya mlinganyo mzima kwa mgawo wa x2x^2, ambao ni 4.

    x2−4x−94=0x^2 - 4x - \frac{9}{4} = 0
  2. Hamisha nambari isiyobadilika (constant term) upande wa kulia.

    x2−4x=94x^2 - 4x = \frac{9}{4}
  3. Kamilisha mraba upande wa kushoto.

    Ongeza (−42)2=(−2)2=4(\frac{-4}{2})^2 = (-2)^2 = 4 pande zote mbili.

    x2−4x+4=94+4x^2 - 4x + 4 = \frac{9}{4} + 4 x2−4x+4=94+164x^2 - 4x + 4 = \frac{9}{4} + \frac{16}{4} (x−2)2=254(x - 2)^2 = \frac{25}{4}
  4. Chukua mzizi wa mraba pande zote mbili.

    (x−2)2=±254\sqrt{(x - 2)^2} = \pm\sqrt{\frac{25}{4}} x−2=±52x - 2 = \pm\frac{5}{2}
  5. Tatua kwa xx.

    x=2±52x = 2 \pm \frac{5}{2}

    Kuna suluhisho mbili:

    x1=2+52=42+52=92x_1 = 2 + \frac{5}{2} = \frac{4}{2} + \frac{5}{2} = \frac{9}{2} x2=2−52=42−52=−12x_2 = 2 - \frac{5}{2} = \frac{4}{2} - \frac{5}{2} = -\frac{1}{2}

    Jibu la mwisho ni:

    x=92 au x=−12\boxed{x = \frac{9}{2} \text{ au } x = -\frac{1}{2}}

    2. Vipimo vya kitabu cha mazoezi ni 36.8cm36.8\text{cm} na 30.5cm30.5\text{cm}. Kokotoa tofauti inayowezekana kati ya vipimo hivyo viwili.

  6. Tambua vipimo vilivyotolewa.

    Kipimo cha kwanza ni 36.8cm36.8\text{cm}.
    Kipimo cha pili ni 30.5cm30.5\text{cm}.

  7. Tafuta tofauti kwa kutoa kipimo kidogo kutoka kwa kikubwa.

    Tofauti=36.8cm−30.5cm\text{Tofauti} = 36.8\text{cm} - 30.5\text{cm} Tofauti=6.3cm\text{Tofauti} = 6.3\text{cm}

    Jibu la mwisho ni:

    6.3cm\boxed{6.3\text{cm}}

    3. Tatua kwa xx katika mlinganyo (log⁡5x)2+log⁡5x=2(\log_5 x)^2 + \log_5 x = 2.

  8. Acha y=log⁡5xy = \log_5 x.

    Badilisha yy kwenye mlinganyo.

    y2+y=2y^2 + y = 2
  9. Panga upya mlinganyo kuwa mlinganyo wa quadratic.

    y2+y−2=0y^2 + y - 2 = 0
  10. Tatua mlinganyo wa quadratic kwa yy.

    Unaweza kutumia factorization.

    (y+2)(y−1)=0(y + 2)(y - 1) = 0

    Hivyo, y+2=0y + 2 = 0 au y−1=0y - 1 = 0.
    y=−2y = -2 au y=1y = 1.

  11. Badilisha log⁡5x\log_5 x kurudi kwa yy na utatue kwa xx.

    Kesi ya 1: y=−2y = -2

    log⁡5x=−2\log_5 x = -2

    Kwa kutumia ufafanuzi wa logarithm (by=xb^y = x), tunapata:

    x=5−2x = 5^{-2} x=152x = \frac{1}{5^2} x=125x = \frac{1}{25}

    Kesi ya 2: y=1y = 1

    log⁡5x=1\log_5 x = 1 x=51x = 5^1 x=5x = 5
  12. Hakikisha suluhisho ni halali (argument ya logarithm lazima iwe chanya).

    Kwa x=125x = \frac{1}{25}, log⁡5(125)\log_5 (\frac{1}{25}) ni halali.
    Kwa x=5x = 5, log⁡5(5)\log_5 (5) ni halali.

    Jibu la mwisho ni:

    x=125 au x=5\boxed{x = \frac{1}{25} \text{ au } x = 5}

    4. Fanya xx kuwa mada ya fomula a2x=ax2−mm\frac{a^2}{x} = a \sqrt{\frac{x^2 - m}{m}}.

  13. Gawanya pande zote mbili kwa aa (kwa kudhani a≠0a \neq 0).

    ax=x2−mm\frac{a}{x} = \sqrt{\frac{x^2 - m}{m}}
  14. Piga mraba pande zote mbili ili kuondoa mzizi wa mraba.

    (ax)2=(x2−mm)2\left(\frac{a}{x}\right)^2 = \left(\sqrt{\frac{x^2 - m}{m}}\right)^2 a2x2=x2−mm\frac{a^2}{x^2} = \frac{x^2 - m}{m}
  15. Zidisha msalaba.

    a2m=x2(x2−m)a^2 m = x^2 (x^2 - m) a2m=x4−mx2a^2 m = x^4 - mx^2
  16. Panga upya mlinganyo ili kuleta x2x^2 pamoja.

    x4−mx2−a2m=0x^4 - mx^2 - a^2 m = 0

    Huu ni mlinganyo wa quadratic kwa x2x^2. Acha u=x2u = x^2.

    u2−mu−a2m=0u^2 - mu - a^2 m = 0

    Tumia fomula ya quadratic u=−b±b2−4ac2au = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ambapo a=1,b=−m,c=−a2ma=1, b=-m, c=-a^2m.

    u=−(−m)±(−m)2−4(1)(−a2m)2(1)u = \frac{-(-m) \pm \sqrt{(-m)^2 - 4(1)(-a^2 m)}}{2(1)} u=m±m2+4a2m2u = \frac{m \pm \sqrt{m^2 + 4a^2 m}}{2} u=m±m(m+4a2)2u = \frac{m \pm \sqrt{m(m + 4a^2)}}{2}
  17. Badilisha x2x^2 kurudi kwa uu.

    x2=m±m(m+4a2)2x^2 = \frac{m \pm \sqrt{m(m + 4a^2)}}{2}

    Kwa kuwa x2x^2 lazima iwe chanya, tunachukua ishara chanya ya mzizi wa mraba.

    x2=m+m(m+4a2)2x^2 = \frac{m + \sqrt{m(m + 4a^2)}}{2}

    (Tunadhania m>0m > 0 na m+4a2>0m + 4a^2 > 0 ili mzizi wa mraba uwe halisi na x2x^2 iwe chanya).

  18. Chukua mzizi wa mraba pande zote mbili ili kupata xx.

    x=±m+m(m+4a2)2x = \pm\sqrt{\frac{m + \sqrt{m(m + 4a^2)}}{2}}

    Jibu la mwisho ni:

    x=±m+m(m+4a2)2\boxed{x = \pm\sqrt{\frac{m + \sqrt{m(m + 4a^2)}}{2}}}
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