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Quadratic Equations · Gujarat Board (GSEB) STD 11 Science (GSEB - English Medium). 15 questions.

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Question 13 Marks
Show the following quadratic equation by factorization method: $\sqrt{5}\text{x}^2+\text{x}+\sqrt{5}=0$
Answer
$\sqrt{5}\text{x}^2+\text{x}+\sqrt{5}=0$ We will apply discriminant rule, $\text{x}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}\ ...(\text{A})$
$D = b^2 - 4ac =1^2-4.\sqrt{5}.\sqrt{5}$ = 1 - 20 = -19 From (A) $\text{x}=\frac{-1\pm\sqrt{-19}}{2.\sqrt{5}}$ $=\frac{-1\pm\sqrt{-19}\text{ i}}{2\sqrt{5}}$
​​​​​​​Thus, $\therefore\text{x}=\frac{-1\pm\sqrt{19}\text{ i}}{2\sqrt{5}}$
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Question 23 Marks
Show the following quadratic equation by factorization method: $-x^2 + x - 2 = 0$
Answer
$-x^2 + x - 2 = 0$ We will apply discriminant rule, $\text{x}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}\ ...(\text{A})$ Where $D = b^2 - 4ac = 1^2 - 4.(-1). (-2) = 1 - 8 = -7$ From (A) $\text{x}=\frac{-1\pm\sqrt{-7}}{2.\sqrt{-1}}$
$=\frac{-1\pm\sqrt{-7}\text{ i}}{-2}$ Thus, $\therefore\text{x}=\frac{-1\pm\sqrt{7}\text{ i}}{-2}$
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Question 33 Marks
Show the following quadratic equation by factorization method: $8x^2 - 9x + 3 = 0$
Answer
$8x^2 - 9x + 3 = 0$ We will apply discriminant rule, $\text{x}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}\ ...(\text{A})$
Where $D = b^2 - 4ac = (-9)^2 - 4.8.3 = 81 - 96 = -15$ From (A) $\text{x}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-(-9)\pm\sqrt{-15}}{2.8}$ $=\frac{-9\pm\sqrt{15}\text{ i}}{16}$ $\therefore\text{x}=\frac{9\pm\sqrt{15}\text{ i}}{16}$
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Question 43 Marks
Show the following quadratic equation by factorization method: $\sqrt{2}\text{x}^2+\text{x}+\sqrt{2}=0$
Answer
$\sqrt{2}\text{x}^2+\text{x}+\sqrt{2}=0$ We will apply discriminant rule, $\text{x}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}\ ...(\text{A})$
Where $D = b^2 - 4ac =1^2-4.\sqrt{2}.\sqrt{2}$ = 1 - 8 = -7 From (A) $\text{x}=\frac{-1\pm\sqrt{-7}}{2.\sqrt{2}}$ $=\frac{{-1}\pm\sqrt{7}\text{ i}}{2\sqrt{2}}$
$\therefore\text{x}=\frac{-{1}\pm\sqrt{7}\text{ i}}{2\sqrt{2}}$
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Question 53 Marks
Show the following quadratic equation: $ix^2 - x + 12i = 0$
Answer
We will apply discriminate rule on $ax^2 + bx + c = 0 \text{x}=\frac{-\text{b}\pm\sqrt{\text{b}^2-4\text{ac}}}{2\text{a}}$ Now, $ix^2 - x + 12i = 0 \text{x}=\frac{-(-1)\pm\sqrt{(-1)^2-4(\text{i)}(12\text{i}})}{2\text{i}}$
$=\frac{1\pm\sqrt{1+48}}{2\text{i}}$
$=\frac{1\pm\sqrt{49}}{2\text{i}}$
$=\frac{1\pm7}{2\text{i}}$
$=\frac{8}{2\text{i}},\frac{-6}{2\text{i}}$
$=\frac{4}{\text{i}},-\frac{3}{\text{i}}$
$=-4\text{i},3\text{i}$
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Question 63 Marks
Show the following quadratic equation by factorization method: $\text{x}^2 +2\text{ x} +\frac{3}{2}=0$
Answer
$\text{x}^2 +2\text{ x} +\frac{3}{2}=0$ We will apply discriminant rule, $\text{x}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}\ ...(\text{A})$ Where $D = b^2 - 4ac =(-2)^2-4(1)\Big(\frac{3}{2}\Big)$ = 4 - 6 = -2 From (A) $\text{x}=\frac{-(-2)\pm\sqrt{-2}}{2(1)}$ $=\frac{2\pm\text{i}\sqrt{2}}{2}$$=1\pm\frac{\text{i}}{\sqrt{2}}$
Thus, $\therefore\text{x}=1\pm\frac{\text{i}}{\sqrt{2}}$
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Question 73 Marks
Show the following quadratic equation: $2\text{x}^2+\sqrt{15}\ \text{ix}-\text{i}=0$
Answer
$2\text{x}^2+\sqrt{15}\ \text{ix}-\text{i}=0$ Comparing the given Equation with the general form $ax^2 + bx + c = 0$, we get a = 2, $\text{b}=\sqrt{15}\text{i},\text{c}=-\text{i}$ Substituting a and in. $\alpha=\frac{-\text{b}+\sqrt{\text{b}^2-4\text{ac}}}{2\text{a}}$ and $\beta=\frac{-\text{b}-\sqrt{\text{b}^2-4\text{ac}}}{2\text{a}}$ $\text{a}=\frac{-\sqrt{15}\text{ i}+\sqrt{-15+8}\text{ i}}{4}$ and $\beta=\frac{-\sqrt{15}\text{ i}-\sqrt{-15+8}\text{ i}}{4}$ Let $\sqrt{-15+8\text{i}}=\text{a}+\text{bi}$
$\Rightarrow-15+8 i=(a+b i)^2 \Rightarrow-15+8 i=a^2-b^2+2 a b i \Rightarrow a^2-b^2=-15 \text { and } 2 a b i=8 i $
$\text { Now }\left(a^2+b^2\right)=\left(a^2-b^2\right)+4 a^2 b^2=\left(a^2+b^2\right)=(15)^2+$$64=289 \Rightarrow a^2+b^2=17 \text { Solving } a^2-b^2=-15 \text { and } a^2+b^2=17 \text {, we get } a^2=1 \text { and } b^2=16$
$\Rightarrow\text{a}=\pm1$ and $\text{b}=\pm4$ ⇒ a = 1, b = 4 or a = -1, b = -4 $\therefore\sqrt{-15+8\text{i }}=1+4\text{i},-1-4\text{i}$ when $\sqrt{-15+8}\text{i}=1+4\text{i}$ $\alpha=\frac{-\sqrt{15}\text{i}+1+4\text{i}}{4}=\frac{1+(4-\sqrt{15})\text{i}}{4}$ and $\beta=\frac{-\sqrt{15}\text{i}-(1+4\text{i})}{4}=\frac{-1-(4+\sqrt{15})\text{i}}{4}$ When $\sqrt{-15+8\text{i}}=-1-4\text{i}$ $\alpha=\frac{-\sqrt{15}\text{i}-1-4\text{i}}{4}=\frac{-1-(4+\sqrt{15})}{4}$ and $\beta=\frac{-\sqrt{15}\text{i}-(-1-4\text{i})}{4}=\frac{1+(4-\sqrt{5})\text{i}}{4}$
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Question 83 Marks
Show the following quadratic equation by factorization method: $\text{x}^2+\frac{\text{x}}{\sqrt{2}}+1=0$
Answer
$\text{x}^2+\frac{\text{x}}{\sqrt{2}}+1=0$ $\Rightarrow\sqrt{2\text{x}}^2+\text{x}+\sqrt{2}=0$ We will apply discriminant rule, $\text{x}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}\ ...(\text{A}) D = b^2 - 4ac =1^2-4.\sqrt{2}.\sqrt{2}$ = 1 - 8 = -7 From (A) $\text{x}=\frac{-1\pm\sqrt{-7}}{2\sqrt{2}}$ $=\frac{-1\pm\sqrt{-7}}{2\sqrt{2}}$ Thus, $\therefore\text{x}=\frac{-1\pm\sqrt{7}\text{ i}}{2\sqrt{2}}$
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Question 93 Marks
Show the following quadratic equation: $2 x^2-(3+7 i) x+(9 i-3)=0$
Answer
$2 x^2-(3+7 i) x+(9 i-3)=02 x^2-3 x-7 i x+(9 i-3)=0(2 x-3-i)(x-3 i)=0\left(x-\frac{3+i}{2}\right)(x-3 i)=0 x=\frac{3+i}{2}, 3 i$
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Question 103 Marks
Show the following quadratic equation by factorization method: $\sqrt{3}\text{x}^2-\sqrt{2 }\text{x}+2\sqrt{3}=0$
Answer
$\sqrt{3}\text{x}^2-\sqrt{2 }\text{x}+2\sqrt{3}=0$ We will apply discriminant rule, $\text{x}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}\ ...(\text{A})$ Where $D = b^2 - 4ac =(-\sqrt{2})^2-4.\sqrt{3}.3\sqrt{3}$ = 2 - 36 = -34 From (A) $\text{x}=\frac{-(-2)\pm\sqrt{-34}}{2.\sqrt{3}}$ $=\frac{\sqrt{2}\pm\sqrt{34}\text{ i}}{2\sqrt{3}}$ $\therefore\text{x}=\frac{\sqrt{2}\pm\sqrt{34}\text{ i}}{2\sqrt{3}}$
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Question 113 Marks
Show the following quadratic equation: $\text{x}^2−(2+\text{i})\text{x}+\sqrt{2\text{i}}=0$
Answer
$\text{x}^2−(2+\text{i})\text{x}+\sqrt{2\text{i}}=0$ $\text{x}^2-\sqrt{2}\text{x}-\text{ix}+\sqrt{2}\text{i}=0$ $\text{x}(\text{x}-\sqrt{2})-\text{i}(\text{x}-\sqrt{2})=0$ $(\text{x}-\text{i})(\text{x}-\sqrt{2})=0$ $\text{x}=\text{i},\sqrt{2}$
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Question 123 Marks
Show the following quadratic equation by factorization method: $21x^2 - 28x + 10 = 0$
Answer
$21 x^2-28 x+10=0$ We will apply discriminant rule, $x=\frac{-b \pm \sqrt{D}}{2 a} \ldots(A)$
Where $D=b^2-4 a c=(-28)^2-4.21 .10=784$ $-840=-56 \operatorname{From}(A) \mathrm{x}=\frac{-28 \pm \sqrt{-56}}{2.21}=\frac{-28 \pm 2 \sqrt{14} \mathrm{i}}{42}$
$\therefore \mathrm{x}=\frac{2}{3} \pm \frac{\sqrt{14}}{21} \mathrm{i}$
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Question 133 Marks
Show the following quadratic equation by factorization method: $3\text{x}-4\text{x}+\frac{20}{3}=0$
Answer
$3\text{x}-4\text{x}+\frac{20}{3}=0$ We will apply discriminant rule, $\text{x}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}\ ...(\text{A})$ Where $D = b^2 - 4ac =(-4)^2-4(3)\Big(\frac{20}{3}\Big)$ = 16 - 80 = -64 From (A) $\text{x}=\frac{-(-4)\pm\sqrt{-64}}{2(3)}$ $=\frac{4\pm\text{i}{8}}{6}$ $=\frac{2}{3}\pm\frac{4\text{i}}{3}$ Thus, $\therefore\text{x}=\frac{2}{3}\pm\frac{4\text{i}}{3}$
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Question 143 Marks
Show the following quadratic equation by factorization method: $17x^2 - 28x + 12 = 0$
Answer
$17x^2 + 28x + 12 = 0$ We will apply discriminant rule, $\text{x}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}\ ...(\text{A})$
Where $D = b^2 - 4ac = (-28)^2 - 4.17.12 = 784 - 816 = -32$ From (A) $\text{x}=\frac{-28\pm\sqrt{-32}}{2.17}$ $=\frac{-28\pm4\sqrt{2}\text{i}}{34}$
$\therefore\text{x}=\frac{-14\pm2\sqrt{2}\text{i}}{17}$
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Question 153 Marks
Show the following quadratic equation by factorization method: $\text{x}^2+\text{x}+\frac{1}{\sqrt{2}}=0$
Answer
$\text{x}^2+\text{x}+\frac{1}{\sqrt{2}}=0$ We will apply discriminant rule, $\text{x}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}\ ...(\text{A})$ Where $D = b^2 - 4ac =1^2-4.1.\frac{1}{\sqrt{2}}$$= 1 - 2\sqrt{2}$
From (A) $\text{x}=\frac{-1\pm\sqrt{-(2\sqrt{2}-1})}{2}$ $=\frac{-1\pm\sqrt{2\sqrt{2}-1\text{ i}}}{2}$ Thus, $\therefore\text{x}=\frac{-1\pm\sqrt{2\sqrt{2}-1\text{ i}}}{2}$
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