Exercise Zone : Persamaan Kuadrat
Table of Contents
Tipe:
No.
Himpunan penyelesaian dari- (3,4)
- (2,3)
- (2,4)
- (3,6)
- (4,6)
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
x^2-7x+12&=0\\
(x-3)(x-4)&=0
\end{aligned}\)
x = 3 dan x = 4
Jadi, himpunan penyelesaian dari x2 − 7x + 12 = 0 adalah (3,4).
JAWAB: A
JAWAB: A
No.
Jika α dan β adalah akar-akar dari persamaan kuadrat- 6
- 9
- 12
- 18
- 24
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
\alpha+\beta&=-\dfrac{-b}a\\
&=-\dfrac{-6}1\\
&=6
\end{aligned}\)
\(\begin{aligned} \alpha\beta&=\dfrac{c}a\\ &=\dfrac{6}1\\ &=6 \end{aligned}\)
\(\begin{aligned} \alpha\beta&=\dfrac{c}a\\ &=\dfrac{6}1\\ &=6 \end{aligned}\)
\(\begin{aligned}
\alpha^2+\beta^2&=\left(\alpha+\beta\right)^2-2\alpha\beta\\
&=(6)^2-2(6)\\
&=36-12\\
&=\boxed{\color{blue}\boxed{24}}
\end{aligned}\)
Jadi, α2 + β2 = 24 .
JAWAB: E
JAWAB: E
No.
Nilai-nilai m agar persamaan kuadrat- \({m\leq-\dfrac{10}3}\)
- \({m\leq-\dfrac{10}3}\) atau
m > 5 1 ≤ m < 2
m = 0 2 ≤ m < 5
ALTERNATIF PENYELESAIAN
Syarat-syarat akar positif:
- \(-\dfrac{b}a\gt0\)
- \(\dfrac{c}a\gt0\)
D > 0
- \(-\dfrac{b}a\gt0\)
\(\begin{aligned} -\dfrac{-4m}{m-5}&\gt0\\[6pt] \dfrac{m}{m-5}&\gt0 \end{aligned}\)
m < 0 ataum > 5
- \(\dfrac{c}a\gt0\)
\(\dfrac{m-2}{m-5}\gt0\)
m < 2 ataum > 5
D ≥ 0
\(\begin{aligned} (-4m)^2-4(m-2)(m-5)&\geq0\\ 16m^2-4m^2+28m-40&\geq0\\ 12m^2+28m-40&\geq0\\ 3m^2+7m-10&\geq0\\ (3m+10)(m-1)&\geq0 \end{aligned}\)
\(m\leq-\dfrac{10}3\) ataum ≥ 1
Jadi, \({m\leq-\dfrac{10}3}\) atau m > 5 .
JAWAB: B
JAWAB: B
No.
Persamaan- 4
- 8
- 16
- 48
- 64
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
D&=0\\
b^2-4ac&=0\\
(-48)^2-4(9)c&=0\\
(2\cdot3\cdot8)^2-4(9)c&=0\\
64-c&=0\\
c&=64
\end{aligned}\)
Jadi, c = 64.
JAWAB: E
JAWAB: E
No.
Akar-akar persamaan kuadratx2 − 6x + 25 = 0 x2 − 3x + 5 = 0 x2 − 5x + 3 = 0
x2 − 29 = 0 x2 + 10 = 0
ALTERNATIF PENYELESAIAN
CARA BIASA
\(x_1+x_2=-\dfrac{b}a=-\dfrac{3}1=-3\)\(x_1x_2=\dfrac{c}a=\dfrac{-5}1=-5\)
Misal
\(\begin{aligned} p+q&=2x_1+3+2x_2+3\\ &=2\left(x_1+x_2\right)+6\\ &=2(-3)+6\\ &=-6+6\\ &=0 \end{aligned}\)
\(\begin{aligned} pq&=(2x_1+3)(2x_2+3)\\ &=4x_1x_2+6x_1+6x_2+9\\ &=4x_1x_2+6(x_1+x_2)+9\\ &=4(-5)+6(-3)+9\\ &=-20-18+9\\ &=-29 \end{aligned}\)
Persamaan kuadrat barunya,
\(\begin{aligned} x^2-(p+q)x+pq&=0\\ x^2-(0)x+(-29)&=0\\ x^2-29&=0 \end{aligned}\)
CARA CEPAT
\(\begin{aligned} x'&=2x+3\\ x&=\dfrac{x'-3}2 \end{aligned}\)Persamaan kuadrat barunya,
\(\begin{aligned} \left(\dfrac{x-3}2\right)^2+3\left(\dfrac{x-3}2\right)-5&=0\\ \dfrac{x^2-6x+9}4+\dfrac{3x-9}2-5&=0\qquad\color{red}{\times4}\\ x^2-6x+9+6x-18-20&=0\\ x^2-29&=0 \end{aligned}\)
Jadi, ersamaan kuadrat yang memiliki akar-akar 2x1 + 3 dan 2x2 + 3 adalah x2 − 29 = 0 .
JAWAB: D
JAWAB: D
No.
Dengan menggunakan rumus ABC, tentukan Himpunan Penyelesaian (HP) dari persamaan berikut!\(x^2=\dfrac12x+5\)
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
x^2&=\dfrac12x+5\qquad{\color{red}\times2}\\
2x^2&=x+10\\
2x^2-x-10&=0
\end{aligned}\)
a = 2 , b = −1 , c = −10
\(\begin{aligned} x_{1,2}&=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\[8pt] &=\dfrac{-(-1)\pm\sqrt{(-1)^2-4(2)(-10)}}{2(2)}\\[8pt] &=\dfrac{1\pm\sqrt{1+80}}4\\[8pt] &=\dfrac{1\pm\sqrt{81}}4\\[8pt] &=\dfrac{1\pm9}4 \end{aligned}\)
\(\begin{aligned} x_{1,2}&=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\[8pt] &=\dfrac{-(-1)\pm\sqrt{(-1)^2-4(2)(-10)}}{2(2)}\\[8pt] &=\dfrac{1\pm\sqrt{1+80}}4\\[8pt] &=\dfrac{1\pm\sqrt{81}}4\\[8pt] &=\dfrac{1\pm9}4 \end{aligned}\)
\(\begin{aligned}
x_1&=\dfrac{1+9}4\\[8pt]
&=\dfrac{10}4\\[8pt]
&=\dfrac52
\end{aligned}\)
\(\begin{aligned}
x_2&=\dfrac{1-9}4\\[8pt]
&=\dfrac{-8}4\\[8pt]
&=-2
\end{aligned}\)
Jadi, Hp = \(\left\{-2,\dfrac52\right\}\)
No.
Jika x1 dan x2 adalah akar-akar persamaan kuadrat- −40
- −42
- −44
- −46
- −48
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
x_1+x_2&=-\dfrac{b}a\\[8pt]
&=-\dfrac21\\[8pt]
&=-2
\end{aligned}\)
\(\begin{aligned} x_1x_2&=\dfrac{c}a\\[8pt] &=\dfrac{-4}1\\[8pt] &=-4 \end{aligned}\)
\(\begin{aligned} {x_1}^3x_2+x_1{x_2}^3&=x_1x_2\left({x_1}^2+{x_2}^2\right)\\ &=x_1x_2\left((x_1+x_2)^2-2x_1x_2\right)\\ &=-4\left((-2)^2-2(-4)\right)\\ &=-4(4+8)\\ &=\boxed{\boxed{-48}} \end{aligned}\)
\(\begin{aligned} x_1x_2&=\dfrac{c}a\\[8pt] &=\dfrac{-4}1\\[8pt] &=-4 \end{aligned}\)
\(\begin{aligned} {x_1}^3x_2+x_1{x_2}^3&=x_1x_2\left({x_1}^2+{x_2}^2\right)\\ &=x_1x_2\left((x_1+x_2)^2-2x_1x_2\right)\\ &=-4\left((-2)^2-2(-4)\right)\\ &=-4(4+8)\\ &=\boxed{\boxed{-48}} \end{aligned}\)
Jadi, nilai x13x2 + x1x23 adalah −48.
JAWAB: E
JAWAB: E
No.
Persamaan kuadrat yang akar-akarnya pangkat 3 dari akar-akar9x2 − 27x + 1 = 0 8x2 + 17x − 20 = 0 8x2 − 17x + 27 = 0
4x2 − 9x − 27 = 0 4x2 + 9x + 3 = 0
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
x_1+x_2&=-\dfrac{b}a\\
&=-\dfrac12
\end{aligned}\)
\(\begin{aligned} x_1x_2&=\dfrac{c}a\\ &=\dfrac32 \end{aligned}\)
Misalp = x13 dan q = x23
\(\begin{aligned} p+q&={x_1}^3+{x_2}^3\\ &=\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)\\ &=\left(-\dfrac12\right)^3-3\left(\dfrac32\right)\left(-\dfrac12\right)\\ &=-\dfrac18+\dfrac94\\ &=\dfrac{17}8 \end{aligned}\)
\(\begin{aligned} pq&={x_1}^3{x_2}^3\\ &=\left(x_1x_2\right)^3\\ &=\left(\dfrac32\right)^3\\ &=\dfrac{27}8 \end{aligned}\)
Persamaan kuadrat barunya adalah
\(\begin{aligned} x^2-(p+q)x+pq&=0\\ x^2-\left(\dfrac{17}8\right)x+\dfrac{27}8&=0\\ 8x^2-17x+27&=0 \end{aligned}\)
\(\begin{aligned} x_1x_2&=\dfrac{c}a\\ &=\dfrac32 \end{aligned}\)
Misal
\(\begin{aligned} p+q&={x_1}^3+{x_2}^3\\ &=\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)\\ &=\left(-\dfrac12\right)^3-3\left(\dfrac32\right)\left(-\dfrac12\right)\\ &=-\dfrac18+\dfrac94\\ &=\dfrac{17}8 \end{aligned}\)
\(\begin{aligned} pq&={x_1}^3{x_2}^3\\ &=\left(x_1x_2\right)^3\\ &=\left(\dfrac32\right)^3\\ &=\dfrac{27}8 \end{aligned}\)
Persamaan kuadrat barunya adalah
\(\begin{aligned} x^2-(p+q)x+pq&=0\\ x^2-\left(\dfrac{17}8\right)x+\dfrac{27}8&=0\\ 8x^2-17x+27&=0 \end{aligned}\)
Jadi, persamaan kuadrat yang akar-akarnya pangkat 3 dari akar-akar f(x) = 2x2 + x + 3 adalah 8x2 − 17x + 27 = 0 .
JAWAB: C
JAWAB: C
No.
Faktorkan persamaan kuadrat berikut.ALTERNATIF PENYELESAIAN
16×(−15) = −240
\(\begin{aligned}
16x^2-22x-15&=\dfrac1{16}(16x-30)(16x+8)\\
&=\dfrac1{16}\cdot2(8x-15)\cdot8(2x+1)\\
&=(8x-15)(2x+1)
\end{aligned}
\)
| −30 | + | 8 | −22 |
| −30 | \(\times\) | 8 | −240 |
Jadi, 16x2 − 22x − 15 = (8x − 15)(2x + 1).
No.
Tentukan akar-akar persamaan kuadrat berikut dengan cara memfaktorkan, melengkapkan kuadrat, dan rumus abc:ALTERNATIF PENYELESAIAN
Memfaktorkan
\(\begin{aligned} x^2+2x-3&=0\\ (x+3)(x-1)&=0 \end{aligned}\)Melengkapkan Kuadrat
\(\begin{aligned} x^2+2x-3&=0\\ x^2+2x&=3\\ x^2+2x+1&=3+1\\ (x+1)^2&=4\\ x+1&=\pm\sqrt4\\ x+1&=\pm2\\ x&=-1\pm2 \end{aligned}\)\(\begin{aligned}
x&=-1+2\\
&=1
\end{aligned}\)
\(\begin{aligned}
x&=-1-2\\
&=-3
\end{aligned}\)
Rumus ABC
\(\begin{aligned} x&=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\ &=\dfrac{-2\pm\sqrt{2^2-4(1)(-3)}}{2(1)}\\ &=\dfrac{-2\pm\sqrt{4+12}}2\\ &=\dfrac{-2\pm\sqrt{16}}2\\ &=\dfrac{-2\pm4}2\\ &=-1\pm2 \end{aligned}\)
\(\begin{aligned}
x&=-1+2\\
&=1
\end{aligned}\)
\(\begin{aligned}
x&=-1-2\\
&=-3
\end{aligned}\)
Jadi, akar-akar persamaan kuadrat tersebut adalah −3 dan 1.
Post a Comment