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Quadratic Equations · Gujarat Board (GSEB) STD 11 Science (GSEB - English Medium). 24 questions.

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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) $\text{x}=\frac{-(-8)\pm\sqrt{-4}}{2.17}$
$=\frac{8\pm2\text{i}}{34}$ $=\frac{4\pm\text{i}}{17}$
$\therefore\text{x}=\frac{4}{17}+\frac{\text{i}}{17},\frac{4}{17}-=\frac{\text{i}}{17}$
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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 $\Big(\text{x}+\frac{1}{2}\Big)^2+\frac{3}{4}=0$ $\Rightarrow\Big(\text{x}+\frac{1}{2}\Big)^2-\Big(\frac{\sqrt{3}}{2}\text{i}\Big)^2=0$ $\Rightarrow\Big(\text{x}+\frac{1}{2}+\frac{\sqrt{3}}{2}\text{i}\Big)\Big(\text{x}+\frac{1}{2}-\frac{\sqrt{3 }}{2}\text{i}\Big)=0$ $\Rightarrow\Big(\text{x}+\frac{1}{2}+\frac{\sqrt{3}}{2}\text{i}\Big)=0\ \text{or } \Big(\text{x}+\frac{1}{2}-\frac{\sqrt{3}}{2}\text{i}\Big)=0$ $\therefore\text{x}=\frac{-1}{2}+\frac{-1}{2}+\frac{\sqrt{3}}{2}\text{ i},\frac{-1}{2}-\frac{\sqrt{3}}{2}\text{ i}$
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Question 32 Marks
Show the following quadratic equation by factorization method: $x^2+(1-2 i) x-2 i=0$
Answer
$x^2+(1-2 i) x-2 i=0 \Rightarrow x^2+x-2 i-2 i=0$
$\Rightarrow x(x+1)-2 i(x+1)=0 \Rightarrow(x+2 i)(x+1)=0 \Rightarrow x=2 i,-1$
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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
$\text{x}^2-(2\sqrt{3}+3\text{i}))\text{x}+6\sqrt{3}\text{ i}=0$ $\Rightarrow\text{x}^2-2\sqrt{3\text{x}}-3\text{ix}+6\sqrt{3}\text{i}=0$ $\Rightarrow\text{x}(\text{x}-2\sqrt{3})-3\text{i}(\text{x}-2\sqrt{3})=0$ $\Rightarrow(\text{x}-3\text{i})(\text{x}-2\sqrt{3})=0$ $\Rightarrow\text{x}=3\text,2\sqrt{3}$
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Question 52 Marks
Show the following quadratic equation:$(2 + i) x^2 - (5 - i) x + 2 (1 - i) = 0$
Answer
$\begin{aligned} & (2+i) x^2-(5-i) x+2(1-i)=0 \Rightarrow(2+i) x^2-2 x-(3-i) x+2(1-i)=0 \Rightarrow x[2+i x-2]-(1-i)[(2+i) x-2]=0 \Rightarrow[x \\ & -(1-i)][(2+i) x-2]=0 \text { Either }[x-(1-i)]=0 \text { or }[(2+i) x-2]=0 \Rightarrow x=1-i \text { or } x=\frac{2}{2+i}\end{aligned}$ $\Rightarrow\text{x}=1-\text{i}\text{ or }\text{x}=\frac{2\times2-\text{i}}{(2+\text{i})(2-\text{i})}$
$\text{x}=\frac{4-2\text{i}}{4+1}=\frac{4}{5}-\frac{2}{5}\text{i}$ Thus $\text{x}=1-\text{i},\frac{4}{5}-\frac{2}{5}\text{i}$
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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
$\text{x}^2-(3\sqrt{2}+2\text{i})\text{x}+6\sqrt{2}\text{ i}=0$ $\Rightarrow\text{x}^2-3\sqrt{2}\text{x}-2\text{ix}+\sqrt{2}\text{i}=0$ $\Rightarrow\text{x}(\text{x}-3\sqrt{2})-2\text{i}(\text{x}-3\sqrt{2})=0$ $\Rightarrow(\text{x}-2\text{i})(\text{x}-3\sqrt{2})=0$ $\Rightarrow\text{x}=2\text{i}\ \text{or }3\sqrt{2}$
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Question 72 Marks
Show the following quadratic equation by factorization method: $x^2+10 ix -21=0$
Answer
$x^2+10 i x-21=0 \Rightarrow x^2+10 i x+21 i^2=0 \Rightarrow x^2+7 i x+3 i x+21 i^2=0 \Rightarrow(x+3 i)(x+7 i)=0 \therefore x=-3 i,-7 i$
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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) $\text{x}=-\frac{-(-6)\pm\sqrt{-4}}{2.5}$ $=\frac{6\pm2\text{i}}{10}$ $=\frac{3\pm\text{i}}{5}$ $\therefore\text{x}=\frac{3}{5}+\frac{\text{i}}{5},\frac{3}{5}-\frac{\text{i}}{5}$
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Question 92 Marks
Show the following quadratic equation by factorization method: $x^2 + 4x + 7 = 0$
Answer
$x^2+4 x+7=0$ We will apply discriminant rule, $x=\frac{-b \pm \sqrt{D}}{2 a} \ldots(A)$ Where $D=b^2-4 a c=(-4)^2-4 \cdot 1 \cdot 7=-12$ From
(A) $\text{x}=-\frac{(-4)\pm\sqrt{-12}}{2}$
$=\frac{4\pm2\sqrt{3}\text{i}}{2}$
$=2\pm\sqrt{3}\text{i}$ $\therefore\text{x}=2+\sqrt{3}\text{i},2-\sqrt{3}\text{i}$
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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) $\text{x}=-\frac{(-2)\pm\sqrt{-4}}{2}$
$=\frac{-2\pm2\text{i}}{2}$ $=-1\pm\text{i}$ $\therefore\text{x}=-1+\text{i},-1-\text{i}$
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Question 112 Marks
Show the following quadratic equation by factorization method: $4 x^2+1=0$
Answer
$4 x^2+1=0 \Rightarrow(2 x)^2-i^2=0\left[\because i^2=-1\right] \Rightarrow(2 x+i)(2 x-i)=0 \Rightarrow$ Either $2 x+i=0$ or $2 x-i=0 \Rightarrow x=\frac{-i}{2}$ or $x=\frac{i}{2}$ $\therefore \mathrm{x}=\frac{-\mathrm{i}}{2}, \frac{\mathrm{i}}{2}$
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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) $\text{x}=\frac{-1\pm\sqrt{-3}}{2}$
$=\frac{-1\pm\sqrt{3\text{i}}}{2}$ $\therefore\text{x}=\frac{-1}{2}+\frac{\sqrt{3}}{2}\ \text{i},\frac{-1}{2}-\frac{\sqrt{3}}{2}\ \text{i}$
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Question 132 Marks
Show the following quadratic equation: $i x^2-4 x-4 i=0$
Answer
$i x^2-4 x-4 i=0 \Rightarrow i x^2+4 i^2 x+4 i^3=0 \Rightarrow x^2+4 i x+4 i^2=0 \Rightarrow x^2+2 i x+2 i x+4 i^2=0 \Rightarrow x(x+2 i)+2 i(x+2 i)=0 \Rightarrow(x$
$+2 i)(x+2 i) \therefore x=-2 i,-2 i$
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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) $\text{x}=\frac{-7\pm\sqrt{-3}}{2.13}$
 $=\frac{-7\pm2\sqrt{3}\text{ i}}{26}$ $\therefore\text{x}=\frac{-7}{26}\pm\frac{\sqrt{3}}{26}\ \text{i}$
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Question 152 Marks
Show the following quadratic equation: $x^2-(5-i) x+(18+i)=0$
Answer
$\mathrm{x}^2-(5-\mathrm{i}) \mathrm{x}+(18+\mathrm{i})=0 \Rightarrow \mathrm{x}^2-5 \mathrm{x}-\mathrm{i} \mathrm{x}+18+\mathrm{i}=0 \Rightarrow \mathrm{x}^2-(3-4 \mathrm{i}) \mathrm{x}-(2+3 \mathrm{i}) \mathrm{x}+(18+\mathrm{i})=0 \Rightarrow \mathrm{x}(\mathrm{x}-(3-4 \mathrm{i}))-(2+$
3i) $(x-(3-4 i))=0 \Rightarrow(x-(2+3 i))(x-(3-4 i))=0 \Rightarrow x=2+3 i$ or $3-4 i$
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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) $\therefore\text{x}=\frac{-(-1)\pm\sqrt{-3}}{2}$ $=\frac{1\pm\sqrt{3\text{i}}}{2}$
$\therefore\text{x}=\frac{1}{2}+\frac{\sqrt{3}}{2}\ \text{i},\frac{1}{2}-\frac{\sqrt{3}}{2}\ \text{i}$
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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 $\text{x}=\frac{-9}{42}\pm\frac{\text{i}\sqrt{3}}{42}$
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Question 182 Marks
Show the following quadratic equation: $x^2 - (2 + i) x - (1 - 7i) = 0$
Answer
$x^2-(2+i) x-(1-7 i)=0 $
$\Rightarrow x^2-(2+i) x-(1-7 i)=0$
$\Rightarrow x^2-(3-i) x+(1-2 i) x-(1-7 i)=0 $
$\Rightarrow x(x-(3-i))+(1-2 i)(x-(3-i))=0$
$\Rightarrow[x+(1-2 i)][x-(3-i)]=0 $
$\Rightarrow x=-1+2 i, 3-i$
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Question 192 Marks
Show the following quadratic equation by factorization method: $6 x^2-17 x -12=0$
Answer
$6 x^2-17 i x-12=0 \Rightarrow 6 x^2-17 i x+12 i^2=0 \Rightarrow 6 x^2-9 i x-8 i x+12 i^2=0 \Rightarrow 3 x(2 x-3 i)-4 i(2 x-3 i)=0 \Rightarrow(3 x-4 i)(2 x-3 i)$
$=0 \Rightarrow x=\frac{4}{3} i \text { or } \frac{3}{2} i$
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Question 202 Marks
Show the following quadratic equation by factorization method: $4x^2 - 12x + 25 = 0$
Answer
$4 x^2-12 x+25=0$ Now, completing the squares, we get $(2 x-3)^2+16=0$
$\Rightarrow(2 x-3)^2-4 i^2=0 $
$\Rightarrow(2 x-3+4 i)=0$ or $(2 \mathrm{x}-3-4 \mathrm{i})=0$
$\therefore \mathrm{x}=\frac{3}{2}+2 \mathrm{i}, \frac{3}{2}-2 \mathrm{i}$
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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) $\text{x}=\frac{-1\pm\sqrt{-7}}{2.2}$
$=\frac{-1\pm\sqrt{7}\text{ i}}{4}$ $\therefore\text{x}=\frac{-1}{4}\pm\frac{\sqrt{7}}{4}\ \text{i}$
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Question 222 Marks
Show the following quadratic equation: $x^2 - x + (1 + i) = 0$
Answer
$x^2-x+(1+i)=0 x^2-i x-(1-i) x+i(1-i)=0(x-i)(x-(1-i))=0 x=i, 1-i$
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Question 232 Marks
Show the following quadratic equation: $x^2+4 i x-4=0$
Answer
$x^2+4 i x-4=0 \Rightarrow x^2+4 i x+4 i^2=0 \Rightarrow x^2+2 i x+2 i x+4 i^2=0 \Rightarrow x(x+2 i)+2 i(x+2 i)=0 \Rightarrow(x+2 i)(x+2 i)=0 \Rightarrow x$
$=-2 i,-2 i$
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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) $\text{x}=\frac{-(-10)\pm\sqrt{-8}}{54}$
$=\frac{10\pm2\sqrt{2}\text{i}}{54}$ $\frac{5\pm\sqrt{2}\text{i}}{27}$
$\therefore\text{x}=\frac{5}{27}+\frac{\sqrt{2\text{i}}}{27},\frac{5}{27}-\frac{\sqrt{2}}{27}\text{i}$
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