# NCERT Solutions For Class 11 Maths Chapter 5 Exercise 5.3

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## Chapter 5 Ex.5.3 Question 1

Solve the equation $${x^2} + 3 = 0$$

### Solution

The given quadratic equation is $${x^2} + 3 = 0$$

On comparing the given equation with $$a{x^2} + bx + c = 0,$$

We obtain $$a = 1,\;b = 0,$$ and $$c = 3$$

Therefore, the discriminant of the given equation is

\begin{align}D&= {b^2} - 4ac\\&= {0^2} - 4 \times 1 \times 3\\&= - 12\end{align}

Therefore, the required solutions are

\begin{align} \frac{-b\pm \sqrt{D}}{2a}&=\frac{-0\pm \sqrt{-12}}{2\times 1} \\ & =\frac{\pm \sqrt{12}i}{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \because \sqrt{-1}=i \right] \\ & =\frac{\pm 2\sqrt{3}i}{2} \\ & =\pm \sqrt{3}i\end{align}

## Chapter 5 Ex.5.3 Question 2

Solve the equation $$2{x^2} + x + 1 = 0$$

### Solution

The given quadratic equation is $$2{x^2} + x + 1 = 0$$

On comparing the given equation with $$a{x^2} + bx + c = 0,$$

We obtain $$a = 2,\;b = 1,$$ and $$c = 1$$

Therefore, the discriminant of the given equation is

\begin{align}D&= {b^2} - 4ac\\&= {1^2} - 4 \times 2 \times 1\\&= - 7\end{align}

Therefore, the required solutions are

\begin{align} \frac{-b\pm \sqrt{D}}{2a}&=\frac{-1\pm \sqrt{-7}}{2\times 2} \\ & =\frac{-1\pm \sqrt{7}i}{4}\ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \because \sqrt{-1}=i \right]\end{align}

## Chapter 5 Ex.5.3 Question 3

Solve the equation $${x^2} + 3x + 9 = 0$$

### Solution

The given quadratic equation is $${x^2} + 3x + 9 = 0$$

On comparing the given equation with $$a{x^2} + bx + c = 0,$$

We obtain $$a = 1,\;b = 3,$$ and $$c = 9$$

Therefore, the discriminant of the given equation is

\begin{align}D&= {b^2} - 4ac\\&= {3^2} - 4 \times 1 \times 9\\&= - 27\end{align}

Hence, the required solutions are

\begin{align} \frac{-b\pm \sqrt{D}}{2a}&=\frac{-3\pm \sqrt{-27}}{2\times 1} \\ & =\frac{-3\pm 3\sqrt{-3}}{2} \\ & =\frac{-3\pm 3\sqrt{3}i}{2} \ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \because \sqrt{-1}=i \right]\end{align}

## Chapter 5 Ex.5.3 Question 4

Solve the equation $$- {x^2} + x - 2 = 0$$

### Solution

The given quadratic equation is $$- {x^2} + x - 2 = 0$$

On comparing the given equation with $$a{x^2} + bx + c = 0,$$

We obtain $$a = - 1,\;b = 1$$ and $$c = - 2$$

Therefore, the discriminant of the given equation is

\begin{align}D&= {b^2} - 4ac\\&= {1^2} - 4 \times \left( { - 1} \right) \times \left( { - 2} \right)\\&= - 7\end{align}

Hence, the required solutions are

\begin{align}\frac{{ - b \pm \sqrt D }}{{2a}}&= \frac{{ - 1 \pm \sqrt { - 7} }}{{2 \times \left( { - 1} \right)}}\\&= \frac{{ - 1 \pm \sqrt 7 i}}{{ - 2}}\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ \sqrt { - 1} = i \right]\end{align}

## Chapter 5 Ex.5.3 Question 5

Solve the equation $${x^2} + 3x + 5 = 0$$

### Solution

The given quadratic equation is $${x^2} + 3x + 5 = 0$$

On comparing the given equation with $$a{x^2} + bx + c = 0,$$

We obtain $$a = 1,\;b = 3,$$ and $$c = 5$$

Therefore, the discriminant of the given equation is

\begin{align}D&= {b^2} - 4ac\\&= {3^2} - 4 \times 1 \times 5\\&= - 11\end{align}

Hence, the required solutions are

\begin{align}\frac{{ - b \pm \sqrt D }}{{2a}}&= \frac{{ - 3 \pm \sqrt { - 11} }}{{2 \times 1}}\\&= \frac{{ - 3 \pm \sqrt {11} i}}{2}\;\;\;\;\;\;\;\;\;\;\;\left[ {\sqrt { - 1} = i} \right]\end{align}

## Chapter 5 Ex.5.3 Question 6

Solve the equation $${x^2} - x + 2 = 0$$

### Solution

The given quadratic equation is $${x^2} - x + 2 = 0$$

On comparing the given equation with $$a{x^2} + bx + c = 0,$$

We obtain $$a = 1,\;b = - 1,$$ and $$c = 2$$

Therefore, the discriminant of the given equation is

\begin{align}D &= {b^2} - 4ac\\ &= {\left( { - 1} \right)^2} - 4 \times 1 \times 2\\ &= - 7\end{align}

Hence, the required solutions are

\begin{align} \frac{-b\pm \sqrt{D}}{2a}&=\frac{-\left( -1 \right)\pm \sqrt{-7}}{2\times 1} \\ & =\frac{1\pm \sqrt{7}i}{2}\ \ \ \ \ \ \ \ \ \ \ \ \left[ \because \sqrt{-1}=i \right]\end{align}

## Chapter 5 Ex.5.3 Question 7

Solve the equation $$\sqrt 2 {x^2} - x + \sqrt 2 = 0$$

### Solution

The given quadratic equation is $$\sqrt 2 {x^2} - x + \sqrt 2 = 0$$

On comparing the given equation with $$a{x^2} + bx + c = 0,$$

We obtain $$a = \sqrt 2 ,\;b = - 1,$$ and $$c = \sqrt 2$$

Therefore, the discriminant of the given equation is

\begin{align}D&= {b^2} - 4ac\\&= {\left( { - 1} \right)^2} - 4 \times \sqrt 2 \times \sqrt 2 \\&= - 7\end{align}

Hence, the required solutions are

\begin{align} \frac{-b\pm \sqrt{D}}{2a}&=\frac{-\left( -1 \right)\pm \sqrt{-7}}{2\times \sqrt{2}} \\ & =\frac{1\pm \sqrt{7}i}{2\sqrt{2}} \ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \because \sqrt{-1}=i \right]\end{align}

## Chapter 5 Ex.5.3 Question 8

Solve the equation $$\sqrt 3 {x^2} - \sqrt 2 x + 3\sqrt 3 = 0$$

### Solution

The given quadratic equation is $$\sqrt 3 {x^2} - \sqrt 2 x + 3\sqrt 3 = 0$$

On comparing the given equation with $$a{x^2} + bx + c = 0,$$

We obtain $$a = \sqrt 3 ,\;b = - \sqrt 2 ,$$ and $$c = 3\sqrt 3$$

Therefore, the discriminant of the given equation is

\begin{align}D&= {b^2} - 4ac\\&= {\left( { - \sqrt 2 } \right)^2} - 4 \times \left( {\sqrt 3 } \right) \times \left( {3\sqrt 3 } \right)\\&= - 34\end{align}

Hence, the required solutions are

\begin{align} \frac{-b\pm \sqrt{D}}{2a}&=\frac{-\left( -\sqrt{2} \right)\pm \sqrt{-34}}{2\times \sqrt{3}} \\ & =\frac{\sqrt{2}\pm \sqrt{34}i}{2\sqrt{3}}\ \ \ \ \ \ \ \ \ \ \ \left[ \because \sqrt{-1}=i \right]\end{align}

## Chapter 5 Ex.5.3 Question 9

Solve the equation $${x^2} + x + \frac{1}{{\sqrt 2 }} = 0$$

### Solution

The given quadratic equation is $${x^2} + x + \frac{1}{{\sqrt 2 }} = 0$$

This equation can also be written as $$\sqrt 2 {x^2} + \sqrt 2 x + 1 = 0$$

On comparing the given equation with $$a{x^2} + bx + c = 0,$$

We obtain$$a = \sqrt 2 ,\;b = \sqrt 2$$ and $$c = 1$$

Therefore, the discriminant of the given equation is

\begin{align}D&= {b^2} - 4ac\\&= {\left( {\sqrt 2 } \right)^2} - 4 \times \left( {\sqrt 2 } \right) \times 1\\&= 2 - 4\sqrt 2\end{align}

Hence, the required solutions are

\begin{align}\frac{-b\pm \sqrt{D}}{2a}& =\frac{-\sqrt{2}\pm \sqrt{2-4\sqrt{2}}}{2\times \sqrt{2}} \\ & =\frac{-\sqrt{2}\pm \sqrt{2\left( 1-2\sqrt{2} \right)}}{2\sqrt{2}} \\ & =\left( \frac{-\sqrt{2}\pm \sqrt{2}\left( \sqrt{2\sqrt{2}-1} \right)i}{2\sqrt{2}} \right)\ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \because \sqrt{-1}=i \right] \\ & =\frac{-1\pm \left( \sqrt{2\sqrt{2}-1} \right)i}{2}\end{align}

## Chapter 5 Ex.5.3 Question 10

Solve the equation $${x^2} + \frac{x}{{\sqrt 2 }} + 1 = 0$$

### Solution

The given quadratic equation is $${x^2} + \frac{x}{{\sqrt 2 }} + 1 = 0$$

This equation can also be written as $$\sqrt 2 {x^2} + x + \sqrt 2 = 0$$

On comparing the given equation with $$a{x^2} + bx + c = 0,$$

We obtain $$a = \sqrt 2 ,\;b = 1$$ and $$c = \sqrt 2$$

Therefore, the discriminant of the given equation is

\begin{align}D &= {b^2} - 4ac\\&= {\left( 1 \right)^2} - 4 \times \left( {\sqrt 2 } \right) \times \left( {\sqrt 2 } \right)\\&= 1 - 8\\&= - 7\end{align}

Hence, the required solutions are

\begin{align} \frac{-b\pm \sqrt{D}}{2a}&=\frac{-1\pm \sqrt{-7}}{2\times \sqrt{2}} \\ & =\frac{-1\pm \sqrt{7}i}{2\sqrt{2}}\ \ \ \ \ \ \ \ \ \ \left[ \because \sqrt{-1}=i \right]\end{align}

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