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Maths 3 Marks Question

Quadratic Equations · CBSE STD 11 Science (CBSE - 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 $\sqrt{-15+8\text{i}}=\text{a}+\text{bi}$
$\Rightarrow -15 + 8i = (a + bi)^2$
$\Rightarrow -15 + 8i = a^2 - b^2 + 2abi$
$\Rightarrow a^2 - b^2 = -15$ and $2abi = 8i$
Now $(a^2 + b^2) = (a^2 - b^2) + 4a^2b^2$
$\Rightarrow (a^2 + b^2) = (15)^2 + 64 = 289$
$\Rightarrow a^2 + b^2 = 17$
Solving $a^2 - b^2 = -15$ and $a^2 + b^2 = 17,$ we get
$a^2 = 1$ and $b^2 = 16$
$\Rightarrow\text{a}=\pm1$ and $\text{b}=\pm4$
$\Rightarrow a = 1, b = 4$ 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:
$2x^2− (3 + 7i) x + (9i − 3) = 0$
Answer
$2x^2− (3 + 7i) x + (9i − 3) = 0$
$2x^2 - 3x - 7ix + (9i - 3) = 0$
$(2x - 3 - i) (x - 3i) = 0$
$\Big(\text{x}-\frac{3+\text{i}}{2}\Big)(\text{x}-3\text{i})=0$
$\text{x}=\frac{3+\text{i}}{2},3\text{i}$
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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
$21x^2 - 28x + 10 = 0$
We will apply discriminant rule,
$\text{x}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}\ ...(\text{A})$
Where $D = b^2 - 4ac$
$= (-28)^2 - 4.21.10$
$= 784 - 840$
$= -56$
From $(A)$
$\text{x}=\frac{-28\pm\sqrt{-56}}{2.21}$
$=\frac{-28\pm2\sqrt{14}\text{ i}}{42}$
$\therefore\text{x}=\frac{2}{3}\pm\frac{\sqrt{14}}{21}\ \text{i}$
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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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