Exercise Zone : Persamaan Kuadrat

Berikut ini adalah kumpulan soal mengenai Persamaan Kuadrat. Jika ingin bertanya soal, silahkan gabung ke grup Telegram, Signal, Discord, atau WhatsApp.

Tipe:

Standar SBMPTN HOTS

No.

Himpunan penyelesaian dari x^2-7x + 12 = 0 adalah....

1. (3,4)
2. (2,3)
3. (2,4)

4. (3,6)
5. (4,6)

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$$\begin{array}{rl}{x}^2-7x+12& =0\\ (x-3)(x-4)& =0\end{array}$$
x=3 dan x=4

Jadi, himpunan penyelesaian dari x^2-7x + 12 = 0 adalah (3,4). JAWAB: A

No.

Jika \alpha dan \beta adalah akar-akar dari persamaan kuadrat {x^2-6x + 6 = 0}, maka nilai dari {\alpha^2 + \beta^2} adalah ....

1. 6
2. 9
3. 12

4. 18
5. 24

$$\begin{array}{rl}\alpha +\beta & =-{\displaystyle \frac{-b}{a}}\\ & =-{\displaystyle \frac{-6}{1}}\\ & =6\end{array}$$

$$\begin{array}{rl}\alpha \beta & ={\displaystyle \frac{c}{a}}\\ & ={\displaystyle \frac{6}{1}}\\ & =6\end{array}$$

$$\begin{array}{rl}{\alpha }^2+{\beta }^2& ={(\alpha +\beta )}^2-2\alpha \beta \\ & =(6{)}^2-2(6)\\ & =36-12\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 24}}}}\end{array}$$

Jadi, \alpha^2 + \beta^2=24. JAWAB: E

No.

Nilai-nilai m agar persamaan kuadrat {(m-5)x^2-4mx+(m-2)=0} mempunyai akar-akar positif adalah ....

1. {m\leq-\dfrac{10}3}
2. {m\leq-\dfrac{10}3} atau m\gt5
3. 1\leq m\lt2

4. m=0
5. 2\leq m\lt5

Ganesha Operation

Syarat-syarat akar positif:

• -\dfrac{b}a\gt0
• \dfrac{c}a\gt0
• D\gt0

• -\dfrac{b}a\gt0
  $$\begin{array}{rl}-{\displaystyle \frac{-4m}{m-5}}& >0\\ {\displaystyle \frac{m}{m-5}}& >0\end{array}$$
  m\lt0 atau m\gt5
  
  
• \dfrac{c}a\gt0
  \dfrac{m-2}{m-5}\gt0
  m\lt2 atau m\gt5
  
  
• D\geq0
  $$\begin{array}{rl}(-4m{)}^2-4(m-2)(m-5)& \ge 0\\ 16m^2-4m^2+28m-40& \ge 0\\ 12m^2+28m-40& \ge 0\\ 3m^2+7m-10& \ge 0\\ (3m+10)(m-1)& \ge 0\end{array}$$
  m\leq-\dfrac{10}3 atau m\geq1

Jadi, m\leq-\dfrac{10}3 atau m\gt5. JAWAB: B

No.

Persamaan {9x^2-48x+c=0} akar-akarnya sama. Nilai c adalah

1. 4
2. 8
3. 16

4. 48
5. 64

$$\begin{array}{rl}D& =0\\ b^2-4ac& =0\\ (-48{)}^2-4(9)c& =0\\ (2\cdot 3\cdot 8{)}^2-4(9)c& =0\\ 64-c& =0\\ c& =64\end{array}$$

Jadi, c=64. JAWAB: E

No.

Akar-akar persamaan kuadrat {x^2+3x-5=0} adalah x_1 dan x_2. Persamaan kuadrat yang memiliki akar-akar {2x_1+3} dan {2x_2+3} adalah

1. {x^2-6x+25=0}
2. {x^2-3x+5=0}
3. {x^2-5x+3=0}

4. {x^2-29=0}
5. {x^2+10=0}

CARA BIASA

x_1+x_2=-\dfrac{b}a=-\dfrac{3}1=-3
x_1x_2=\dfrac{c}a=\dfrac{-5}1=-5

Misal p=2x_1+3 dan q=2x_1+3

$$\begin{array}{rl}p+q& =2x_1+3+2x_2+3\\ & =2(x_1+x_2)+6\\ & =2(-3)+6\\ & =-6+6\\ & =0\end{array}$$

$$\begin{array}{rl}pq& =(2x_1+3)(2x_2+3)\\ & =4x_1{x}_2+6x_1+6x_2+9\\ & =4x_1{x}_2+6(x_1+x_2)+9\\ & =4(-5)+6(-3)+9\\ & =-20-18+9\\ & =-29\end{array}$$

Persamaan kuadrat barunya,
$$\begin{array}{rl}{x}^2-(p+q)x+pq& =0\\ x^2-(0)x+(-29)& =0\\ x^2-29& =0\end{array}$$

CARA CEPAT

$$\begin{array}{rl}{x}^{\prime }& =2x+3\\ x& ={\displaystyle \frac{x^{\prime }-3}{2}}\end{array}$$

Persamaan kuadrat barunya,
$$\begin{array}{rl}{\left({\displaystyle \frac{x-3}{2}}\right)}^2+3\left({\displaystyle \frac{x-3}{2}}\right)-5& =0\\ {\displaystyle \frac{x^2-6x+9}{4}}+{\displaystyle \frac{3x-9}{2}}-5& =0\times 4\\ x^2-6x+9+6x-18-20& =0\\ x^2-29& =0\end{array}$$

Jadi, persamaan kuadrat yang memiliki akar-akar {2x_1+3} dan {2x_2+3} adalah {x^2-29=0}. JAWAB: D

No.

Dengan menggunakan rumus ABC, tentukan Himpunan Penyelesaian (HP) dari persamaan berikut!
x^2=\dfrac12x+5

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$$\begin{array}{rl}{x}^2& ={\displaystyle \frac{1}{2}}x+5\times 2\\ 2x^2& =x+10\\ 2x^2-x-10& =0\end{array}$$
a=2, b=-1, c=-10

$$\begin{array}{rl}{x}_{1,2}& ={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}\\ & ={\displaystyle \frac{-(-1)\pm \sqrt{(-1{)}^2-4(2)(-10)}}{2(2)}}\\ & ={\displaystyle \frac{1\pm \sqrt{1+80}}{4}}\\ & ={\displaystyle \frac{1\pm \sqrt{81}}{4}}\\ & ={\displaystyle \frac{1\pm 9}{4}}\end{array}$$

$$\begin{array}{rl}{x}_1& ={\displaystyle \frac{1+9}{4}}\\ & ={\displaystyle \frac{10}{4}}\\ & ={\displaystyle \frac{5}{2}}\end{array}$$

$$\begin{array}{rl}{x}_2& ={\displaystyle \frac{1-9}{4}}\\ & ={\displaystyle \frac{-8}{4}}\\ & =-2\end{array}$$

Jadi, Hp = \left\{-2,\dfrac52\right\}

No.

Jika x_1 dan x_2 adalah akar-akar persamaan kuadrat {x^2 + 2x-4 = 0}, maka nilai {{x_1}^3x_2+x_1{x_2}^3} adalah

1. -40
2. -42
3. -44

4. -46
5. -48

$$\begin{array}{rl}{x}_1+x_2& =-{\displaystyle \frac{b}{a}}\\ & =-{\displaystyle \frac{2}{1}}\\ & =-2\end{array}$$

$$\begin{array}{rl}{x}_1{x}_2& ={\displaystyle \frac{c}{a}}\\ & ={\displaystyle \frac{-4}{1}}\\ & =-4\end{array}$$

$$\begin{array}{rl}{x_1}^3{x}_2+x_1{x_2}^3& =x_1{x}_2({x_1}^2+{x_2}^2)\\ & =x_1{x}_2((x_1+x_2{)}^2-2x_1{x}_2)\\ & =-4((-2{)}^2-2(-4))\\ & =-4(4+8)\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -48}}}}\end{array}$$

Jadi,

Jadi, {x_1}^3x_2+x_1{x_2}^3=-48.
JAWAB: E

No.

Persamaan kuadrat yang akar-akarnya pangkat 3 dari akar-akar {f(x)=2x^2+x+3} adalah

1. {9x^2-27x+1=0}
2. {8x^2+17x-20=0}
3. {8x^2-17x+27=0}

4. {4x^2-9x-27=0}
5. {4x^2+9x+3=0}

$$\begin{array}{rl}{x}_1+x_2& =-{\displaystyle \frac{b}{a}}\\ & =-{\displaystyle \frac{1}{2}}\end{array}$$

$$\begin{array}{rl}{x}_1{x}_2& ={\displaystyle \frac{c}{a}}\\ & ={\displaystyle \frac{3}{2}}\end{array}$$

Misal p={x_1}^3 dan q={x_2}^3

$$\begin{array}{rl}p+q& ={x_1}^3+{x_2}^3\\ & ={(x_1+x_2)}^3-3x_1{x}_2(x_1+x_2)\\ & ={(-{\displaystyle \frac{1}{2}})}^3-3\left({\displaystyle \frac{3}{2}}\right)(-{\displaystyle \frac{1}{2}})\\ & =-{\displaystyle \frac{1}{8}}+{\displaystyle \frac{9}{4}}\\ & ={\displaystyle \frac{17}{8}}\end{array}$$

$$\begin{array}{rl}pq& ={x_1}^3{x_2}^3\\ & ={\left(x_1{x}_2\right)}^3\\ & ={\left({\displaystyle \frac{3}{2}}\right)}^3\\ & ={\displaystyle \frac{27}{8}}\end{array}$$

Persamaan kuadrat barunya adalah
$$\begin{array}{rl}{x}^2-(p+q)x+pq& =0\\ x^2-\left({\displaystyle \frac{17}{8}}\right)x+{\displaystyle \frac{27}{8}}& =0\\ 8x^2-17x+27& =0\end{array}$$

Jadi, persamaan kuadrat yang akar-akarnya pangkat 3 dari akar-akar {f(x)=2x^2+x+3} adalah {8x^2-17x+27=0}. JAWAB: B

No.

Faktorkan persamaan kuadrat berikut.
16x^2-22x-15

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16\times(-15)=-240

-30	+	8	-22
-30	\times	8	-240

$$\begin{array}{rl}16x^2-22x-15& ={\displaystyle \frac{1}{16}}(16x-30)(16x+8)\\ & ={\displaystyle \frac{1}{16}}\cdot 2(8x-15)\cdot 8(2x+1)\\ & =(8x-15)(2x+1)\end{array}$$

Jadi, 16x^2-22x-15=(8x-15)(2x+1).

No.

Tentukan akar-akar persamaan kuadrat berikut dengan cara memfaktorkan, melengkapkan kuadrat, dan rumus abc:
x^2+2x-3=0

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Memfaktorkan

$$\begin{array}{rl}{x}^2+2x-3& =0\\ (x+3)(x-1)& =0\end{array}$$
x=-3 dan x=1

Melengkapkan Kuadrat

$$\begin{array}{rl}{x}^2+2x-3& =0\\ x^2+2x& =3\\ x^2+2x+1& =3+1\\ (x+1{)}^2& =4\\ x+1& =\pm \sqrt{4}\\ x+1& =\pm 2\\ x& =-1\pm 2\end{array}$$

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$$\begin{array}{rl}x& =-1+2\\ & =1\end{array}$$

$$\begin{array}{rl}x& =-1-2\\ & =-3\end{array}$$

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Rumus ABC

a=1, b=2, c=-3
$$\begin{array}{rl}x& ={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}\\ & ={\displaystyle \frac{-2\pm \sqrt{2^2-4(1)(-3)}}{2(1)}}\\ & ={\displaystyle \frac{-2\pm \sqrt{4+12}}{2}}\\ & ={\displaystyle \frac{-2\pm \sqrt{16}}{2}}\\ & ={\displaystyle \frac{-2\pm 4}{2}}\\ & =-1\pm 2\end{array}$$

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$$\begin{array}{rl}x& =-1+2\\ & =1\end{array}$$

$$\begin{array}{rl}x& =-1-2\\ & =-3\end{array}$$

Jadi, akar-akar persamaan kuadrat x^2+2x-3=0 adalah -3 dan 1.