SNBT Zone : Persamaan Kuadrat [2]
Table of Contents
Tipe:
No.
Jika x1 dan x2 adalah akar-akar persamaan kuadrat- 20
- 21
- 22
- 23
- 24
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
x_1+x_2&=-\dfrac{b}a\\[8pt]
&=-\dfrac11\\
&=-1
\end{aligned}\)
\(\begin{aligned} x_1x_2&=\dfrac{c}a\\[8pt] &=\dfrac{-3}1\\ &=-3 \end{aligned}\)
\(\begin{aligned} x^2+x-3&=0\\ x^2+x&=3 \end{aligned}\)
\(\begin{aligned} 4{x_1}^2+3{x_2}^2+2x_1+x_2&=2{x_1}^2+2{x_1}^2+2{x_2}^2+{x_2}^2+2x_1+x_2\\ &=2{x_1}^2+2{x_2}^2+2{x_1}^2+2x_1+{x_2}^2+x_2\\ &=2\left({x_1}^2+{x_2}^2\right)+2\left({x_1}^2+x_1\right)+{x_2}^2+x_2\\ &=2\left(\left({x_1}+{x_2}\right)^2-2x_1x_2\right)+2\left({x_1}^2+x_1\right)+{x_2}^2+x_2\\ &=2\left(\left(-1\right)^2-2(-3)\right)+2\left(3\right)+3\\ &=2\left(1+6\right)+6+3\\ &=2\left(7\right)+9\\ &=14+9\\ &=\boxed{\boxed{23}} \end{aligned}\)
\(\begin{aligned} x_1x_2&=\dfrac{c}a\\[8pt] &=\dfrac{-3}1\\ &=-3 \end{aligned}\)
\(\begin{aligned} x^2+x-3&=0\\ x^2+x&=3 \end{aligned}\)
\(\begin{aligned} 4{x_1}^2+3{x_2}^2+2x_1+x_2&=2{x_1}^2+2{x_1}^2+2{x_2}^2+{x_2}^2+2x_1+x_2\\ &=2{x_1}^2+2{x_2}^2+2{x_1}^2+2x_1+{x_2}^2+x_2\\ &=2\left({x_1}^2+{x_2}^2\right)+2\left({x_1}^2+x_1\right)+{x_2}^2+x_2\\ &=2\left(\left({x_1}+{x_2}\right)^2-2x_1x_2\right)+2\left({x_1}^2+x_1\right)+{x_2}^2+x_2\\ &=2\left(\left(-1\right)^2-2(-3)\right)+2\left(3\right)+3\\ &=2\left(1+6\right)+6+3\\ &=2\left(7\right)+9\\ &=14+9\\ &=\boxed{\boxed{23}} \end{aligned}\)
Jadi, hasil dari 4x12 + 3x22 + 2x1 + x2 adalah 23.
JAWAB: D
JAWAB: D
No.
Misalkan dua persamaan kuadrat mempunyai satu akar yang sama, yaitu 1 dan akar-akar lainnya berkebalikan. Jika salah satu persamaan itu adalah6x2 + 5x + 1 = 0 6x2 − 5x − 1 = 0 6x2 − 5x + 1 = 0
5x2 − 6x + 2 = 0 5x2 − 6x + 1 = 0
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
x^2-ax+5&=0\\
1^2-a(1)+5&=0\\
1-a+5&=0\\
6-a&=0\\
a&=6
\end{aligned}\)
\(\begin{aligned} x^2-6x+5&=0\\ (x-1)(x-5)&=0 \end{aligned}\)
x = 1 dan x = 5
Persamaan kedua mempunyai akar 1 dan \(\dfrac15\).
\(\begin{aligned} (x-1)\left(x-\dfrac15\right)&=0&\color{red}{\times5}\\ (x-1)(5x-1)&=0\\ 5x^2-6x+1&=0 \end{aligned}\)
\(\begin{aligned} x^2-6x+5&=0\\ (x-1)(x-5)&=0 \end{aligned}\)
x = 1 dan x = 5
Persamaan kedua mempunyai akar 1 dan \(\dfrac15\).
\(\begin{aligned} (x-1)\left(x-\dfrac15\right)&=0&\color{red}{\times5}\\ (x-1)(5x-1)&=0\\ 5x^2-6x+1&=0 \end{aligned}\)
Jadi, persamaan kuadrat lainnya adalah 5x2 − 6x + 1 = 0 .
JAWAB: E
JAWAB: E
No.
Jika x1 dan x2 adalah akar-akar persamaan kuadrat- −12
- −14
- −16
- −18
- −20
ALTERNATIF PENYELESAIAN
a = 1, b = 2, c = −2
\(\eqalign{ x_1+x_2&=\dfrac{-b}a\\ &=\dfrac{-2}1\\ &=-2 }\)
\(\eqalign{ x_1x_2&=\dfrac{c}a\\ &=\dfrac{-2}1\\ &=-2 }\)
\(\eqalign{ x_1+x_2&=\dfrac{-b}a\\ &=\dfrac{-2}1\\ &=-2 }\)
\(\eqalign{ x_1x_2&=\dfrac{c}a\\ &=\dfrac{-2}1\\ &=-2 }\)
\(\eqalign{
{x_1}^3x_2 + x_1{x_2}^3&=x_1x_2\left({x_1}^2+{x_2}^2\right)\\
&=x_1x_2\left(\left(x_1+x_2\right)^2-2x_1x_2\right)\\
&=-2\left(\left(-2\right)^2-2(-2)\right)\\
&=-2\left(4+4\right)\\
&=-2(8)\\
&=\boxed{\boxed{-16}}
}\)
Jadi, x13x2 + x1x23 = −16 .
JAWAB: C
JAWAB: C
No.
Diketahui akar-akar persamaan- 104
- 102
- 112
- 98
- 96
ALTERNATIF PENYELESAIAN
\(x_1+x_2=\dfrac{-a}1=-a\)
\(x_1x_2=\dfrac{1-a}1=1-a\)
\(\eqalign{ \dfrac1{x_1}+\dfrac1{x_2}&=\dfrac54\\ \dfrac{x_1+x_2}{x_1x_2}&=\dfrac54\\ \dfrac{-a}{1-a}&=\dfrac54\\ -4a&=5-5a\\ a&=5 }\)
\(x_1x_2=\dfrac{1-a}1=1-a\)
\(\eqalign{ \dfrac1{x_1}+\dfrac1{x_2}&=\dfrac54\\ \dfrac{x_1+x_2}{x_1x_2}&=\dfrac54\\ \dfrac{-a}{1-a}&=\dfrac54\\ -4a&=5-5a\\ a&=5 }\)
\(\eqalign{
(a+3)(a+7)&=(5+3)(5+7)\\
&=(8)(12)\\
&=\boxed{\boxed{96}}
}\)
Jadi, (a + 3)(a + 7) = 96 .
JAWAB: E
JAWAB: E
No.
Akar-akar persamaan- −8
- −4
- 0
- 4
- 8
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
x_1+x_2&=-\dfrac{-a}2\\[4pt]
&=\dfrac{a}2
\end{aligned}\)
\(\begin{aligned} x_1x_2&=\dfrac{-2}2\\[4pt] &=-1 \end{aligned}\)
\(\begin{aligned} x_1^2-2x_1x_2+x_2^2&=-2a\\ x_1^2+x_2^2-2x_1x_2&=-2a\\ \left(x_1+x_2\right)^2-2x_1x_2-2x_1x_2&=-2a\\ \left(x_1+x_2\right)^2-4x_1x_2&=-2a\\ \left(\dfrac{a}2\right)^2-4(-1)&=-2a\\ \dfrac{a^2}4+4&=-2a&{\color{red}\times 4}\\[4pt] a^2+16&=-8a\\ a^2+8a+16&=0\\ (a+4)^2&=0\\ a&=\boxed{\boxed{-4}} \end{aligned}\)
\(\begin{aligned} x_1x_2&=\dfrac{-2}2\\[4pt] &=-1 \end{aligned}\)
\(\begin{aligned} x_1^2-2x_1x_2+x_2^2&=-2a\\ x_1^2+x_2^2-2x_1x_2&=-2a\\ \left(x_1+x_2\right)^2-2x_1x_2-2x_1x_2&=-2a\\ \left(x_1+x_2\right)^2-4x_1x_2&=-2a\\ \left(\dfrac{a}2\right)^2-4(-1)&=-2a\\ \dfrac{a^2}4+4&=-2a&{\color{red}\times 4}\\[4pt] a^2+16&=-8a\\ a^2+8a+16&=0\\ (a+4)^2&=0\\ a&=\boxed{\boxed{-4}} \end{aligned}\)
Jadi, a = −4.
JAWAB: B
JAWAB: B
No.
Misalkan p dan q adalah bilangan-bilangan real tidak nol dan persamaan kuadrat- 2
- 3
- 4
- 5
- 8
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
pq&=q\\
p&=1
\end{aligned}\)
\(\begin{aligned} p+q&=-p\\ 1+q&=-1\\ q&=-2 \end{aligned}\)
\(\begin{aligned} p+q&=-p\\ 1+q&=-1\\ q&=-2 \end{aligned}\)
\(\begin{aligned}
p^2-2q&=1^2-2(-2)\\
&=1+4\\
&=\color{blue}\boxed{\boxed{\color{black}5}}
\end{aligned}\)
Jadi, p2 − 2q = 5.
JAWAB: D
JAWAB: D
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