SBMPTN Zone : Persamaan Kuadrat

Berikut ini adalah kumpulan soal mengenai persamaan kuadrat tipe SBMPTN. Jika ingin bertanya soal, silahkan gabung ke grup Facebook atau Telegram.

Tipe:

Standar SBMPTN HOTS

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No. 1

Jika m dan n merupakan akar-akar dari persamaan kuadrat {x^2-6x+2=0}, maka persamaan kuadrat baru dengan akar-akar \left(\dfrac1m+\dfrac1n\right)^{mn} dan (mn)^{\left(\frac1m+\frac1n\right)} adalah....

1. {x^2-17x+72=0}
2. {x^2-13x+36=0}
3. {x^2-8x+16=0}

4. {x^2-5x+6=0}
5. {x^2-2x+6=0}

a=1, b=-6, c=2

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

$\begin{array}{rl}mn& ={\displaystyle \frac{c}{a}}\\ & ={\displaystyle \frac{2}{1}}\\ & =2\end{array}$

Misal p=\left(\dfrac1m+\dfrac1n\right)^{mn} dan q=(mn)^{\left(\frac1m+\frac1n\right)}

$\begin{array}{rl}p& ={({\displaystyle \frac{1}{m}}+{\displaystyle \frac{1}{n}})}^{mn}\\ & ={\left({\displaystyle \frac{m+n}{mn}}\right)}^{mn}\\ & ={\left({\displaystyle \frac{6}{2}}\right)}^2\\ & =3^2\\ & =9\end{array}$

$\begin{array}{rl}q& =(mn{)}^{(\frac{1}{m}+\frac{1}{n})}\\ & =2^3\\ & =8\end{array}$

Persamaan kuadrat barunya adalah
$\begin{array}{rl}{x}^2-(p+q)x+pq& =0\\ x^2-(9+8)x+9\cdot 8& =0\\ x^2-17x+72& =0\end{array}$

Jadi, persamaan kuadrat barunya adalah {x^2-17x+72=0}.
JAWAB: A

No. 2

Salah satu akar persamaan kuadrat {ax^2+(a+1)x+(a-1)=0}, a\gt0 adalah x_1. Jika akar lainnya {x_2=2x_1}, maka konstanta a= ....

1. 2
2. 1
3. -1

4. -2
5. -3

$$\begin{array}{rl}{x}_1+x_2& =-{\displaystyle \frac{a+1}{a}}\\ x_1+2x_1& =-{\displaystyle \frac{a+1}{a}}\\ 3x_1& =-{\displaystyle \frac{a+1}{a}}\\ x_1& =-{\displaystyle \frac{a+1}{3a}}\end{array}$$\)

$\begin{array}{rl}{x}_1{x}_2& ={\displaystyle \frac{a-1}{a}}\\ x_1\left(2x_1\right)& ={\displaystyle \frac{a-1}{a}}\\ 2{x_1}^2& ={\displaystyle \frac{a-1}{a}}\\ 2{(-{\displaystyle \frac{a+1}{3a}})}^2& ={\displaystyle \frac{a-1}{a}}\\ 2\left({\displaystyle \frac{a^2+2a+1}{9a^2}}\right)& ={\displaystyle \frac{a-1}{a}}\\ 2(a^2+2a+1)& =9a^2\left({\displaystyle \frac{a-1}{a}}\right)\\ 2a^2+4a+2& =9a(a-1)\\ 2a^2+4a+2& =9a^2-9a\\ 7a^2-13a-2& =0\\ (7a+1)(a-2)& =0\end{array}$
a=-\dfrac17\lt0(TM) atau a=2.

Jadi, a=2.
JAWAB: A

No. 3

Misalkan x_1 dan x_2 merupakkan akar-akar persamaan {px^2+qx-1=0}, p\neq0. Jika {\dfrac1{x_1}+\dfrac1{x_2}=-1} dan {x_1=-\dfrac32x_2}, maka {p+q=} ....

1. -7
2. -5
3. 0

4. 5
5. 7

$\begin{array}{rl}{\displaystyle \frac{1}{x_1}}+{\displaystyle \frac{1}{x_2}}& =-1\\ {\displaystyle \frac{x_1+x_2}{x_1{x}_2}}& =-1\\ {\displaystyle \frac{-{\displaystyle \frac{3}{2}}{x}_2+x_2}{(-{\displaystyle \frac{3}{2}}{x}_2)x_2}}& =-1\\ {\displaystyle \frac{-{\displaystyle \frac{1}{2}}{x}_2}{-{\displaystyle \frac{3}{2}}{x_2}^2}}& =-1\\ {\displaystyle \frac{1}{3x_2}}& =-1\\ 1& =-3x_2\\ x_2& =-{\displaystyle \frac{1}{3}}\end{array}$

$\begin{array}{rl}{x}_1& =-{\displaystyle \frac{3}{2}}{x}_2\\ & =-{\displaystyle \frac{3}{2}}(-{\displaystyle \frac{1}{3}})\\ & ={\displaystyle \frac{1}{2}}\end{array}$

$\begin{array}{rl}(3x+1)(2x-1)& =0\\ 6x^2-x-1& =0\end{array}$

$\begin{array}{rl}p+q& =6+(-1)\\ & =5\end{array}$

Jadi, {p+q=5}.
JAWAB: B

No. 4

Hasil jumlah akar-akar persamaan yang dinyatakan dengan $\left|\begin{array}{cc}2x-2& x+1\\ x+2& x\end{array}\right|=1$ adalah ....

$\begin{array}{rl}\left|\begin{array}{cc}2x-2& x+1\\ x+2& x\end{array}\right|& =1\\ 2x^2-2x-(x^2+2x+2)& =1\\ 2x^2-2x-x^2-2x-2-1& =0\\ x^2-4x-3& =0\end{array}$

$\begin{array}{rl}{x}_1+x_2& ={\displaystyle \frac{-b}{a}}\\ & ={\displaystyle \frac{-(-4)}{1}}\\ & =4\end{array}$

Jadi, hasil jumlah akar-akarnya adalah 4.

No. 5

Akar-akar persamaan kuadrat {x^2-2x+n=0} adalah p dan q dengan {2p+q=6}. Persamaan kuadrat baru yang akar-akarnya pq dan {p+q} adalah ....

1. {x^2-6x-10=0}
2. {x^2+6x-16=0}
3. {x^2-6x+2=0}

4. {x^2+6x-4=0}
5. {x^2-4x-2=0}

Ganesha Operation

{a=1}, {b=-2}, {c=n}

$\begin{array}{rl}p+q& =-{\displaystyle \frac{b}{a}}\\ & =-{\displaystyle \frac{-2}{1}}\\ & =2\end{array}$

$\begin{array}{rl}2p+q& =6\\ p+q& =2-\\ p& =4\end{array}$

$\begin{array}{rl}p+q& =2\\ 4+q& =2\\ q& =-2\end{array}$

$\begin{array}{rl}pq& =(4)(-2)\\ & =-8\end{array}$

$\begin{array}{rl}p+q+pq& =2+(-8)\\ & =-6\end{array}$

$\begin{array}{rl}(p+q)(pq)& =(2)(-8)\\ & =-16\end{array}$

Persamaan kuadrat barunya,
$\begin{array}{rl}{x}^2-(-6)x+(-16)& =0\\ x^2+6x-16& =0\end{array}$

Jadi, persamaan kuadrat barunya adalah {x^2+6x-16=0}.
JAWAB: B

No. 6

Jika x memenuhi persamaan {\sqrt{\dfrac1{125}}}^{(5x)}=\dfrac{5^{(2x-3)}}{\sqrt{5\sqrt{5\sqrt5}}}{\sqrt{\dfrac1{625}}}^{(x-1)} dan y memenuhi persamaan ^9\negthinspace\log\left({^3\negthinspace\log y}\right)={^3\negthinspace\log\left({^9\negthinspace\log y}\right)}, maka persamaan kuadrat yang memiliki akar-akar 4x dan \sqrt{y} adalah

1. {a^2-2a + 1}
2. {a^2 + 2a + 1}
3. {a^2 -4a + 3}

4. {a^2 + 10a + 9}
5. {a^2-10a + 9}

$\begin{array}{rl}{\sqrt{{\displaystyle \frac{1}{125}}}}^{(5x)}& ={\displaystyle \frac{5^{(2x-3)}}{\sqrt{5\sqrt{5\sqrt{5}}}}}{\sqrt{{\displaystyle \frac{1}{625}}}}^{(x-1)}\\ {\sqrt{{\displaystyle \frac{1}{5^3}}}}^{(5x)}& ={\displaystyle \frac{5^{(2x-3)}}{5^{\frac{1}{2}}\cdot 5^{\frac{1}{2\cdot 2}}\cdot 5^{\frac{1}{2\cdot 2\cdot 2}}}}{\sqrt{{\displaystyle \frac{1}{5^4}}}}^{(x-1)}\\ {\sqrt{5^{-3}}}^{(5x)}& ={\displaystyle \frac{5^{(2x-3)}}{5^{\frac{1}{2}}\cdot 5^{\frac{1}{4}}\cdot 5^{\frac{1}{8}}}}{\sqrt{5^{-4}}}^{(x-1)}\\ {\left(5^{-\frac{3}{2}}\right)}^{(5x)}& ={\displaystyle \frac{5^{(2x-3)}}{5^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}}}}{\left(5^{-2}\right)}^{(x-1)}\\ 5^{-\frac{15x}{2}}& ={\displaystyle \frac{5^{(2x-3)}}{5^{\frac{7}{8}}}}\left(5^{-2x+2}\right)\\ 5^{-\frac{15x}{2}}& =5^{(2x-3)-\frac{7}{8}+(-2x+2)}\\ 5^{-\frac{15x}{2}}& =5^{-\frac{15}{8}}\\ -{\displaystyle \frac{15x}{2}}& =-{\displaystyle \frac{15}{8}}\\ x& ={\displaystyle \frac{1}{4}}\\ 4x& =1\end{array}$

$\begin{array}{rl}{}^9\log \left({}^3\log y\right)& ={}^3\log \left({}^9\log y\right)\\ {}^9\log \left({}^3\log y\right)& ={}^{3^2}\log {\left({}^{3^2}\log y\right)}^2\\ {}^9\log \left({}^3\log y\right)& ={}^9\log {\left({\displaystyle \frac{1}{2}}{}^3\log y\right)}^2\\ {}^3\log y& ={\displaystyle \frac{1}{4}}{\left({}^3\log y\right)}^2\\ 1& ={\displaystyle \frac{1}{4}}{}^3\log y\\ {}^3\log y& =4\\ y& =3^4\\ & =81\\ \sqrt{y}& =9\end{array}$

Persamaan kuadrat yang memiliki akar-akar 1 dan 9 adalah
$\begin{array}{rl}(x-1)(x-9)& =0\\ x^2-10x+9& =0\end{array}$

Jadi, persamaan kuadrat yang memiliki akar-akar 4x dan \sqrt{y} adalah x^2-10x+9=0.
JAWAB: E

No. 7

Persamaan kuadrat {x^2+2px+p+2=0} mempunyai akar riil, tidak nol dan bertanda sama. Nilai p yang memenuhi adalah

1. p\leq-1
2. p\leq2
   
3. p\leq-2 atau p\geq-1
   

4. -2\lt p\leq-1 atau p\geq2
5. p\lt-2 atau -1\leq p\leq2
   

Akar riil : D\geq0
$\begin{array}{rl}D& \ge 0\\ b^2-4ac& \ge 0\\ (2p{)}^2-4(1)(p+2)& \ge 0\\ 4p^2-4p-8& \ge 0\\ p^2-p-2& \ge 0\\ (p+1)(p-2)& \ge 0\end{array}$
pembuat nol: p=-1 dan p=2


Tidak nol dan bertanda sama:
$\begin{array}{rl}{x}_1{x}_2& >0\\ {\displaystyle \frac{c}{a}}& >0\\ {\displaystyle \frac{p+2}{1}}& >0\\ p+2& >0\\ p& >-2\end{array}$


iriskan kedua garis bilangan:

-2\lt p\leq-1 atau p\geq2

Jadi, -2\lt p\leq-1 atau p\geq2.
JAWAB: D

No. 8

Jika jumlah kebalikan akar-akar persamaan {x^2-8x+(c-1)=0} sama dengan jumlah kuadrat akar-akar real persamaan {x^2-2x-c=0}, maka nilai c sama dengan

1. -3
2. -1
3. \dfrac12

4. 2
5. 3

SKMP Juli 2020

Syarat mempunyai akar real adalah D\geq0.
$\begin{array}{rl}D& \ge 0\\ (-2{)}^2-4(1)(-c)& \ge 0\\ 4+4c& \ge 0\\ 4c& \ge -4\\ c& \ge -1\end{array}$

Misal akar-akar x^2-8x+(c-1)=0 adalah x_1 dan x_2.
x_1+x_2=-\dfrac{b}a=-\dfrac{-8}1=8
x_1x_2=\dfrac{c}a=\dfrac{c-1}1=c-1

Misal akar-akar x^2-2x-c=0 adalah x_3 dan x_4.
x_3+x_4=-\dfrac{b}a=-\dfrac{-2}1=2
x_3x_4=\dfrac{c}a=\dfrac{-c}1=-c

$\begin{array}{rl}{\displaystyle \frac{1}{x_1}}+{\displaystyle \frac{1}{x_2}}& ={x_3}^2+{x_4}^2\\ {\displaystyle \frac{x_1+x_2}{x_1{x}_2}}& ={(x_3+x_4)}^2-2x_3{x}_4\\ {\displaystyle \frac{8}{c-1}}& ={\left(2\right)}^2-2(-c)\\ {\displaystyle \frac{8}{c-1}}& =4+2c\\ 8& =4c+2c^2-4-2c\\ 8& =2c^2+2c-4\\ 2c^2+2c-12& =0\\ c^2+c-6& =0\\ (c+3)(c-2)& =0\end{array}$
c=-3\lt-1 (TM) atau c=2

Jadi, c=2.
JAWAB: D

No. 9

Agar grafik fungsi {y=x^2+2x-5} seluruhnya di bawah grafik fungsi {y=2x^2-ax+2a+4} maka nilai a

1. {-4\lt a\lt8}
2. {-8\lt a\lt4}
   
3. {a\lt-8} atau {a\gt-4}
   

4. {a\lt8}
5. {a\gt-4}
   

$\begin{array}{rl}{x}^2+2x-5& =2x^2-ax+2a+4\\ x^2-(a+2)x+9& =0\end{array}$

$\begin{array}{rl}D& <0\\ b^2-4ac& <0\\ (-(a+2){)}^2-4(1)(9)& <0\\ a^2+4a+4-36& <0\\ a^2+4a-32& <0\\ (a+8)(a-4)& <0\\ \boxed{{\displaystyle \boxed{{\displaystyle -8<a<4}}}}\end{array}$

Jadi, -8\lt a\lt4.
JAWAB: B

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No. 10

Diketahui x_1 dan x_2 merupakan akar-akar persamaan {x^2 + 5x + a = 0} dengan x_1 dan x_2 kedua-duanya tidak sama dengan nol. x_1, 2x_2, dan -3x_1x_2 masing-masing merupakan suku pertama, suku kedua, dan suku ketiga dari deret geometri dengan rasio positif, maka nilai a sama dengan....

1. -6
2. 2
3. 6

4. -6 atau 6
5. 2 atau 3

$\begin{array}{rl}{\left(2x_2\right)}^2& =x_1\cdot (-3x_1{x}_2)\\ 4{x_2}^2& =-3{x_1}^2{x}_2\\ 4x_2& =-3{x_1}^2\\ x_2& =-{\displaystyle \frac{3}{4}}{x_1}^2\end{array}$

$\begin{array}{rl}{x}_1+x_2& =-{\displaystyle \frac{b}{a}}\\ x_1-{\displaystyle \frac{3}{4}}{x_1}^2& =-{\displaystyle \frac{5}{1}}\\ -{\displaystyle \frac{3}{4}}{x_1}^2+x_1& =-5\\ -{\displaystyle \frac{3}{4}}{x_1}^2+x_1+5& =0\\ 3{x_1}^2-4x_1-20& =0\\ (3x_1-10)(x_1+2)& =0\end{array}$
x_1=\dfrac{10}3 atau x_1=-2

• Untuk x_1=\dfrac{10}3,
  $\begin{array}{rl}{x}_2& =-{\displaystyle \frac{3}{4}}{x_1}^2\\ & =-{\displaystyle \frac{3}{4}}{\left({\displaystyle \frac{10}{3}}\right)}^2\\ & =-{\displaystyle \frac{3}{4}}\left({\displaystyle \frac{100}{9}}\right)\\ & =-{\displaystyle \frac{25}{3}}\end{array}$
  
  $\begin{array}{rl}r& ={\displaystyle \frac{2x_2}{x_1}}\\ & ={\displaystyle \frac{2(-{\displaystyle \frac{25}{3}})}{{\displaystyle \frac{10}{3}}}}<0\end{array}$
  kontradiktif
• Untuk x_1=-2,
  $\begin{array}{rl}{x}_2& =-{\displaystyle \frac{3}{4}}{x_1}^2\\ & =-{\displaystyle \frac{3}{4}}(-2{)}^2\\ & =-3\end{array}$

$\begin{array}{rl}{x}_1{x}_2& ={\displaystyle \frac{c}{a}}\\ (-2)(-3)& ={\displaystyle \frac{a}{1}}\\ 6& =a\\ a& =\boxed{{\displaystyle \boxed{{\displaystyle 6}}}}\end{array}$

Jadi, a=6.
JAWAB: C

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