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MATHS 3 Marks Question

Quadratic Equations · Rajasthan Board (RBSE) STD 11 Science (Rajasthan - 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 = b2 - 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:
-x2 + x - 2 = 0
Answer
-x2 + x - 2 = 0
We will apply discriminant rule,
$\text{x}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}\ ...(\text{A})$
Where D = b2 - 4ac
= 12 - 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:
8x2 - 9x + 3 = 0
Answer
8x2 - 9x + 3 = 0
We will apply discriminant rule,
$\text{x}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}\ ...(\text{A})$
Where D = b2 - 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 = b2 - 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:
ix2 - x + 12i = 0
Answer
We will apply discriminate rule on ax2 + bx + c = 0
$\text{x}=\frac{-\text{b}\pm\sqrt{\text{b}^2-4\text{ac}}}{2\text{a}}$
Now,
ix2 - 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:
$\sqrt{3}\text{x}^2-\sqrt{2 }\text{x}+2\sqrt{3}=0$
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 = b2 - 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 73 Marks
Show the following quadratic equation by factorization method:
21x2 - 28x + 10 = 0
Answer
21x2 - 28x + 10 = 0
We will apply discriminant rule,
$\text{x}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}\ ...(\text{A})$
Where D = b2 - 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 83 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 = b2 - 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 93 Marks
Show the following quadratic equation by factorization method:
17x2 - 28x + 12 = 0
Answer
17x2 + 28x + 12 = 0
We will apply discriminant rule,
$\text{x}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}\ ...(\text{A})$
Where D = b2 - 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 103 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 = b2 - 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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Question 113 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 = b2 - 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 123 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
ax2 + 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 $\sqrt{-15+8\text{i}}=\text{a}+\text{bi}$
⇒ -15 + 8i = (a + bi)2
⇒ -15 + 8i = a2 - b2 + 2abi
⇒ a2 - b2 = -15 and 2abi = 8i
Now (a2 + b2) = (a2 - b2) + 4a2b2
⇒ (a2 + b2) = (15)2 + 64 = 289
⇒ a2 + b2 = 17
Solving a2 - b2 = -15 and a2 + b2 = 17, we get
 a2 = 1 and b2 = 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 133 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 143 Marks
Show the following quadratic equation:
2x2 − (3 + 7i) x + (9i − 3) = 0
Answer
2x2 − (3 + 7i) x + (9i − 3) = 0
2x2 - 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 153 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 = b2 - 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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