Download App

MATHS 2 Marks Questions

Quadratic Equations · Rajasthan Board (RBSE) STD 11 Science (Rajasthan - English Medium). 24 questions.

Questions

2 Marks Questions

Take a timed test

24 questions · self-marked practice — reveal the answer and mark yourself.

Question 12 Marks
Show the following quadratic equation by factorization method:
$17x^2 - 8x + 1 = 0$
Answer
$17x^2 - 8x + 1 = 0$
We will apply discriminant rule,
$\text{x}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}\ ...(\text{A})$
Where $D = b^2 - 4ac$
$= (-8)^2 - 4.17.1$
$= 64 - 68$
$= -4$
From $(A)$
$\text{x}=\frac{-(-8)\pm\sqrt{-4}}{2.17}$
$=\frac{8\pm2\text{i}}{34}$
$=\frac{4\pm\text{i}}{17}$
$\therefore\text{x}=\frac{4}{17}+\frac{\text{i}}{17},\frac{4}{17}-=\frac{\text{i}}{17}$
View full question & answer→
Question 22 Marks
Show the following quadratic equation by factorization method:
$x^2 - x + 1 = 0$
Answer
$x^2 - x + 1 = 0$
Now, completing the squares, we get
$\Big(\text{x}+\frac{1}{2}\Big)^2+\frac{3}{4}=0$
$\Rightarrow\Big(\text{x}+\frac{1}{2}\Big)^2-\Big(\frac{\sqrt{3}}{2}\text{i}\Big)^2=0$
$\Rightarrow\Big(\text{x}+\frac{1}{2}+\frac{\sqrt{3}}{2}\text{i}\Big)\Big(\text{x}+\frac{1}{2}-\frac{\sqrt{3 }}{2}\text{i}\Big)=0$
$\Rightarrow\Big(\text{x}+\frac{1}{2}+\frac{\sqrt{3}}{2}\text{i}\Big)=0\ \text{or } \Big(\text{x}+\frac{1}{2}-\frac{\sqrt{3}}{2}\text{i}\Big)=0$
$\therefore\text{x}=\frac{-1}{2}+\frac{-1}{2}+\frac{\sqrt{3}}{2}\text{ i},\frac{-1}{2}-\frac{\sqrt{3}}{2}\text{ i}$
View full question & answer→
Question 32 Marks
Show the following quadratic equation by factorization method:
$x^2+ (1 − 2i) x − 2i = 0$
Answer
$x^2+ (1 − 2i) x − 2i = 0$
$\Rightarrow x^2 + x - 2i - 2i = 0$
$\Rightarrow x (x + 1) - 2i (x + 1) = 0$
$\Rightarrow (x + 2i) (x + 1) = 0$
$\Rightarrow x = 2i, -1$
 
View full question & answer→
Question 42 Marks
Show the following quadratic equation by factorization method:
$\text{x}^2-(2\sqrt{3}+3\text{i}))\text{x}+6\sqrt{3}\text{ i}=0$
Answer
$\text{x}^2-(2\sqrt{3}+3\text{i}))\text{x}+6\sqrt{3}\text{ i}=0$
$\Rightarrow\text{x}^2-2\sqrt{3\text{x}}-3\text{ix}+6\sqrt{3}\text{i}=0$
$\Rightarrow\text{x}(\text{x}-2\sqrt{3})-3\text{i}(\text{x}-2\sqrt{3})=0$
$\Rightarrow(\text{x}-3\text{i})(\text{x}-2\sqrt{3})=0$
$\Rightarrow\text{x}=3\text,2\sqrt{3}$
View full question & answer→
Question 52 Marks
Show the following quadratic equation:$(2 + i) x^2 - (5 - i) x + 2 (1 - i) = 0$
Answer
$(2 + i) x^2 - (5 - i) x + 2 (1 - i) = 0$
$\Rightarrow (2 + i) x^2 - 2x - (3 - i) x + 2 (1 - i) = 0$
$\Rightarrow x [2 + ix - 2] - (1 - i) [(2 + i) x - 2] = 0$
$\Rightarrow [x - (1 - i)] [(2 + i) x - 2] = 0$
Either $[x - (1 - i)] = 0$ or $[(2 + i) x - 2] = 0$
$\Rightarrow\text{x}=1- \text{i }\text{ or }\text{x}=\frac{2}{2+\text{i}}$
$\Rightarrow\text{x}=1-\text{i}\text{ or }\text{x}=\frac{2\times2-\text{i}}{(2+\text{i})(2-\text{i})}$
$\text{x}=\frac{4-2\text{i}}{4+1}=\frac{4}{5}-\frac{2}{5}\text{i}$
Thus
$\text{x}=1-\text{i},\frac{4}{5}-\frac{2}{5}\text{i}$
View full question & answer→
Question 62 Marks
Show the following quadratic equation:
$\text{x}^2-(3\sqrt{2}+2\text{i})\text{x}+6\sqrt{2}\text{ i}=0$
Answer
$\text{x}^2-(3\sqrt{2}+2\text{i})\text{x}+6\sqrt{2}\text{ i}=0$
$\Rightarrow\text{x}^2-3\sqrt{2}\text{x}-2\text{ix}+\sqrt{2}\text{i}=0$
$\Rightarrow\text{x}(\text{x}-3\sqrt{2})-2\text{i}(\text{x}-3\sqrt{2})=0$
$\Rightarrow(\text{x}-2\text{i})(\text{x}-3\sqrt{2})=0$
$\Rightarrow\text{x}=2\text{i}\ \text{or }3\sqrt{2}$
View full question & answer→
Question 72 Marks
Show the following quadratic equation by factorization method:
$x^2 + 10ix - 21 = 0$
Answer
$x^2 + 10ix - 21 = 0$
$\Rightarrow x^2 + 10ix + 21i^2 = 0$
$\Rightarrow x^2 + 7ix + 3ix + 21i^2 = 0$
$\Rightarrow (x + 3i) (x + 7i) = 0$
$\therefore\text{x}=-3\text{i},-7\text{i}$
View full question & answer→
Question 82 Marks
Show the following quadratic equation by factorization method:
$5x^2 + 6x + 2 = 0$
Answer
$5x^2 + 6x + 2 = 0$
We will apply discriminant rule,
$\text{x}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}\ ...(\text{A})$
Where $D = b^2 - 4ac$
$= 36 - 40$
$= 4 - 8$
$= -4$
From $(A)$
$\text{x}=-\frac{-(-6)\pm\sqrt{-4}}{2.5}$
$=\frac{6\pm2\text{i}}{10}$
$=\frac{3\pm\text{i}}{5}$
$\therefore\text{x}=\frac{3}{5}+\frac{\text{i}}{5},\frac{3}{5}-\frac{\text{i}}{5}$
View full question & answer→
Question 92 Marks
Show the following quadratic equation by factorization method:
$x^2 + 4x + 7 = 0$
Answer
$x^2 + 4x + 7 = 0$
We will apply discriminant rule,
$\text{x}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}\ ...(\text{A})$
Where $D = b^2 - 4ac = (-4)^2 - 4.1.7 = -12$
From $(A)$
$\text{x}=-\frac{(-4)\pm\sqrt{-12}}{2}$
$=\frac{4\pm2\sqrt{3}\text{i}}{2}$
$=2\pm\sqrt{3}\text{i}$
$\therefore\text{x}=2+\sqrt{3}\text{i},2-\sqrt{3}\text{i}$
View full question & answer→
Question 102 Marks
Show the following quadratic equation by factorization method:
$x^2 + 2x + 2 = 0$
Answer
$x^2 + 2x + 2 = 0$
We will apply discriminant rule,
$\text{x}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}\ ...(\text{A})$
Where $ D = b^2 - 4ac$
$= 2^2 - 4.1.2$
$= 4 - 8$
$= -4$
From (A)
$\text{x}=-\frac{(-2)\pm\sqrt{-4}}{2}$
$=\frac{-2\pm2\text{i}}{2}$
$=-1\pm\text{i}$
$\therefore\text{x}=-1+\text{i},-1-\text{i}$
View full question & answer→
Question 112 Marks
Show the following quadratic equation by factorization method:
$4x^2 + 1 = 0$
Answer
$4x^2 + 1 = 0$
$\Rightarrow (2x)^2 - i^2 = 0 [\because i^2 = -1]$
$\Rightarrow (2x + i) (2x - i) = 0$
$\Rightarrow$ Either $2x + i = 0$ or $2x - i = 0$
$\Rightarrow\text{x}=\frac{-\text{i}}{2}\ \text{or }\text{x}=\frac{\text{i}}{2}$
$\therefore\text{x}=\frac{-\text{i}}{2},\frac{\text{i}}{2}$
View full question & answer→
Question 122 Marks
Show the following quadratic equation by factorization method:
$x^2 + x + 1 = 0$
Answer
$x^2 + x + 1 = 0$
We will apply discriminant rule,
$\text{x}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}\ ...(\text{A})$
Where $D = b^2 - 4ac$
$= (-1)^2 - 4.1.1$
$= 1 - 4$
$= -3$
From $(A)$
$\text{x}=\frac{-1\pm\sqrt{-3}}{2}$
$=\frac{-1\pm\sqrt{3\text{i}}}{2}$
$\therefore\text{x}=\frac{-1}{2}+\frac{\sqrt{3}}{2}\ \text{i},\frac{-1}{2}-\frac{\sqrt{3}}{2}\ \text{i}$
View full question & answer→
Question 132 Marks
Show the following quadratic equation:
$ix^2 - 4x - 4i = 0$
Answer
$ix^2 - 4x - 4i = 0$
$\Rightarrow ix^2 + 4i^2x + 4i^3 = 0$
$\Rightarrow x^2 + 4ix + 4i^2 = 0$
$\Rightarrow x^2 + 2ix + 2ix + 4i^2 = 0$
$\Rightarrow x (x + 2i) + 2i (x + 2i) = 0$
$\Rightarrow (x + 2i) (x + 2i)$
$\therefore\text{x}=-2\text{i},-2\text{i}$
View full question & answer→
Question 142 Marks
Show the following quadratic equation by factorization method:
$13x^2 - 7x + 1 = 0$
Answer
$13x^2 - 7x + 1 = 0$
We will apply discriminant rule,
$\text{x}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}\ ...(\text{A})$
Where $D = b^2 - 4ac$
$= 7^2 - 4.13.1$
$= 49 - 52$
$= -3$
From (A)
$\text{x}=\frac{-7\pm\sqrt{-3}}{2.13}$
$=\frac{-7\pm2\sqrt{3}\text{ i}}{26}$
$\therefore\text{x}=\frac{-7}{26}\pm\frac{\sqrt{3}}{26}\ \text{i}$
View full question & answer→
Question 152 Marks
Show the following quadratic equation:
$x^2− (5 − i) x +(18 + i) = 0$
Answer
$x^2− (5 − i) x +(18 + i) = 0$
$\Rightarrow x^2 - 5x - ix + 18 + i = 0$
$\Rightarrow x^2- (3 - 4i) x - (2 + 3i) x + (18 + i) = 0$
$\Rightarrow x (x - (3 - 4i)) - (2 + 3i) (x - (3 - 4i)) = 0$
$\Rightarrow (x - (2 + 3i)) (x - (3 - 4i)) = 0$
$\Rightarrow x = 2 + 3i$ or $3 - 4i$
View full question & answer→
Question 162 Marks
Show the following quadratic equation by factorization method:
$x^2 - x + 1 = 0$
Answer
$x^2 - x + 1 = 0$
We will apply discriminant rule,
$\text{x}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}\ ...(\text{A})$
Where $D = b^2 - 4ac$
$= (-1)^2 - 4.1.1$
$= 1 - 4$
$= -3$
From (A)
$\therefore\text{x}=\frac{-(-1)\pm\sqrt{-3}}{2}$
$=\frac{1\pm\sqrt{3\text{i}}}{2}$
$\therefore\text{x}=\frac{1}{2}+\frac{\sqrt{3}}{2}\ \text{i},\frac{1}{2}-\frac{\sqrt{3}}{2}\ \text{i}$
View full question & answer→
Question 172 Marks
Show the following quadratic equation by factorization method:
$21x^2 + 9x + 1 = 0$
Answer
$21x^2 + 9x + 1 = 0$
Comparing the given equation with the general form
$ax^2 + bx + c = 0,$ we get $a = 21, b = 9, c = 1$
Substituting a and b in,
$\alpha=\frac{-\text{b}+\sqrt{\text{b}^2-4\text{ac}}}{2\text{a}}$ and $\beta=\frac{-\text{b}-\sqrt{\text{b}^2-4\text{ac}}}{2\text{a}}$
$\alpha=\frac{-9+\sqrt{81+84}}{42}$ and $\beta=\frac{-9-\sqrt{81-84}}{42}$
$\Rightarrow\alpha=\frac{-9+\sqrt{-3}}{42}$ and $\beta=\frac{-9-\sqrt{-3}}{42}$
$\Rightarrow\alpha=\frac{-9+\text{i}\sqrt{3}}{42}$ and $\beta=\frac{-9-\text{i}\sqrt{3}}{42}$
The roots are $\text{x}=\frac{-9}{42}\pm\frac{\text{i}\sqrt{3}}{42}$
View full question & answer→
Question 182 Marks
Show the following quadratic equation:
$x^2 - (2 + i) x - (1 - 7i) = 0$
Answer
$x^2 - (2 + i) x - (1 - 7i) = 0$
$\Rightarrow x^2 - (2 + i) x - (1 - 7i) = 0$
$\Rightarrow x^2 - (3 - i) x + (1 - 2i) x - (1 - 7i) = 0$
$\Rightarrow x (x - (3 - i)) + (1 - 2i) (x - (3 - i)) = 0$
$\Rightarrow [ x + (1 - 2i)] [x - (3 - i)] = 0$
$\Rightarrow x = -1 + 2i, 3 - i$
View full question & answer→
Question 192 Marks
Show the following quadratic equation by factorization method:
$6x^2 - 17ix - 12 = 0$
Answer
$6x^2 - 17ix - 12 = 0$
$\Rightarrow 6x^2 - 17ix + 12i^2 = 0$
$\Rightarrow 6x^2 - 9ix - 8ix + 12i^2 = 0$
$\Rightarrow 3x (2x - 3i) -4i (2x - 3i) = 0$
$\Rightarrow (3x - 4i) (2x - 3i) = 0$
 $\Rightarrow\text{x}=\frac{4}{3}\text{i}\ \text{or }\frac{3}{2}\text{i}$ 
View full question & answer→
Question 202 Marks
Show the following quadratic equation by factorization method:
$4x^2 - 12x + 25 = 0$
Answer
$4x^2 - 12x + 25 = 0$
Now, completing the squares, we get
$(2x - 3)^2 + 16 = 0$
$\Rightarrow (2x - 3)^2 - 4i^2 = 0$
$\Rightarrow (2x - 3 + 4i) = 0$ or $(2x - 3 - 4i) = 0$
$\therefore\text{x}=\frac{3}{2}+2\text{i},\frac{3}{2}-2\text{i}$
View full question & answer→
Question 212 Marks
Show the following quadratic equation by factorization method:
$2x^2 + x + 1 = 0$
Answer
$2x^2 + x + 1 = 0$
We will apply discriminant rule,
$\text{x}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}\ ...(\text{A})$
Where $D = b^2 - 4ac$
$= 1^2 - 4.2.1$
$= 1 - 8$
$= -7$
From (A)
$\text{x}=\frac{-1\pm\sqrt{-7}}{2.2}$
$=\frac{-1\pm\sqrt{7}\text{ i}}{4}$
$\therefore\text{x}=\frac{-1}{4}\pm\frac{\sqrt{7}}{4}\ \text{i}$
View full question & answer→
Question 222 Marks
Show the following quadratic equation:
$x^2 - x + (1 + i) = 0$
Answer
$x^2 - x + (1 + i) = 0$
$x^2 - ix - (1 - i) x + i (1 - i) = 0$
$(x - i) (x - (1 - i)) = 0$
$x = i, 1 - i$
View full question & answer→
Question 232 Marks
Show the following quadratic equation:
$x^2 + 4ix - 4 = 0$
Answer
$x^2 + 4ix - 4 = 0$
$\Rightarrow x^2 + 4ix + 4i^2 = 0$
$\Rightarrow x^2 + 2ix + 2ix + 4i^2 = 0$
$\Rightarrow x (x + 2i) + 2i (x + 2i) = 0$
$\Rightarrow (x + 2i) (x + 2i) = 0$
$\Rightarrow x = -2i, -2i$
View full question & answer→
Question 242 Marks
Show the following quadratic equation by factorization method:
$27x^2 - 10x + 1 = 0$
Answer
$27x^2 - 10x + 1 = 0$
We will apply discriminant rule,
$\text{x}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}\ ...(\text{A})$
Where $D = b^2 - 4ac$
$= (-10)^2 - 4.27.1$
$= 100 - 108$
$= -4$
From $(A)$
$\text{x}=\frac{-(-10)\pm\sqrt{-8}}{54}$
$=\frac{10\pm2\sqrt{2}\text{i}}{54}$
$\frac{5\pm\sqrt{2}\text{i}}{27}$
$\therefore\text{x}=\frac{5}{27}+\frac{\sqrt{2\text{i}}}{27},\frac{5}{27}-\frac{\sqrt{2}}{27}\text{i}$
View full question & answer→
Generate Paper FreeAll Quadratic Equations topics