Question

QUKA
(QUIZ SUKA SUKA)

Sebuah fungsi kuadrat [tex]\rm x^2+px+1=0\:\: memiliki \:akar^2\:yaitu\:(a,b)[/tex] dan [tex]\rm x^2+qx+1=0\:\: memiliki \:akar^2\:yaitu\:(c,d)[/tex].
Tentukan f(p,q) !!
jika diketahui (a-c)(b-c)(a+d)(b+d)=f(p,q)=......

#happyWeekend
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Answer 1

Verified answer

f(p, q) = q² – p².
 

Penjelasan dengan langkah-langkah:

Dari kedua persamaan kuadrat yang diberikan beserta akar-akarnya, dapat kita ambil bahwa:

• a + b = –p dan ab = 1
• c + d = –q dan cd = 1

sehingga:

[tex]\begin{aligned}f(p,q)&=\bf(a-c)(b-c)(a+d)(b+d)\\\vphantom{\Big|}&=\bf\left(ab-(a+b)c+c^2\right)\left(ab+(a+b)d+d^2\right)\\\vphantom{\Big|}&\ \ \rightarrow\left\{ab=1\,\land\,a+b=-p\right\}\\\vphantom{\Big|}&\bf=\left(1+pc+c^2\right)\left(1-pd+d^2\right)\\\vphantom{\Bigg|}&\ \ \rightarrow\left\{cd=1\ \Rightarrow\ c=\frac{1}{d}\,\land\,d=\frac{1}{c}\right\}\\\vphantom{\Bigg|}&\bf=\left(1+\frac{p}{d}+\frac{c}{d}\right)\left(1-\frac{p}{c}+\frac{d}{c}\right)\end{aligned}[/tex]

[tex]\begin{aligned}\vphantom{\Bigg|}&\bf=\left(\frac{c+d+p}{d}\right)\left(\frac{c+d-p}{c}\right)\\\vphantom{\Bigg|}&\bf=\frac{(c+d)^2-p^2}{cd}\\\vphantom{\Big|}&\ \ \rightarrow\left\{cd=1\,\land\,c+d=-q\right\}\\\vphantom{\Bigg|}&\bf=\frac{(-q)^2-p^2}{1}\\f(p,q)&=\boxed{\vphantom{\Big|}\bf\,q^2-p^2\,}\end{aligned}[/tex]

 
 
[tex]\overline{\begin{array}{l}\small\textsf{Duc In Altum}\\\small\text{bertolaklah\;ke\;tempat}\\\small\text{yang\;lebih\;dalam}\end{array}}[/tex]