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Looking for Exercise 5.3 Class 11 Maths Solutions? Get step-by-step answers to Class 11 Chapter 5 Exercise 5.3 based on Complex Numbers and Quadratic Equations. Our detailed guide covers every question from the NCERT Solutions for Class 11 Maths Chapter 5, ensuring clarity and accuracy. These solutions are perfect for quick revision, exam prep, and concept building. You can also explore the NCERT Exemplar Class 11 Maths problems for deeper understanding and practice.
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Exercise 5.3
1. Solve each of the following equations:
\(x^{2}+3=0\)
\(x^{2}+3=0\)
Answer
It is given in the question that,
\(x^{2}+3=0\)
On comparing the given with \( \mathrm{a}x^{2}+\mathrm{b}x+\mathrm{c} \), we obtain
\( \mathrm{A}=1, \mathrm{b}=0 \), and \( \mathrm{c},=3 \)
Therefore, the discriminant of the given equation is
\(D=b^{2}-4 a c=0^{2}-4 \times 1 \times 3=-12\)
Therefore, the required solutions are
\(\frac{-b \pm \sqrt{D}}{2 a}=\frac{ \pm \sqrt{-12}}{2 \times 1}=\frac{ \pm \sqrt{12} i}{2}\)
\(=\frac{ \pm 2 \sqrt{3} i}{2}= \pm \sqrt{3} i\)
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2. Solve each of the following equations:
\(2 x^{2}+x+1=0\)
\(2 x^{2}+x+1=0\)
Answer
It is given in the question that,
\(2 x^{2}+x+1=0\)
On comparing the given equation with \( \mathrm{a}x^{2}+\mathrm{b}x+\mathrm{c}=0 \)
\(\mathrm{a}=2, \mathrm{b}=1 \text {, and } \mathrm{c}=1\)
Therefore, the discriminant of the given equation is
\(D=b^{2}-4 a c=1^{2}-4 \times 2 \times 1=1-8=-7\)
Therefore, the required solutions are
\(\frac{-b \pm \sqrt{D}}{2 a}=\frac{-1 \pm \sqrt{-7}}{2 \times 2}=\frac{-1 \pm \sqrt{7} i}{4}\)
3. Solve each of the following equations:
\(x^{2}+3 x+9=0\)
\(x^{2}+3 x+9=0\)
Answer
It is given in the question that,
\(x^{2}+3 x+9=0\)
On comparing the given equation with \( \mathrm{a}x^{2}+\mathrm{b}x+\mathrm{c}=0 \), we obtain \( \mathrm{a}=1, \mathrm{~b}=3 \), and \( \mathrm{c}=9 \)
Therefore, the discriminant of the given equation is
\(D=b^{2}-4 a c=3^{2}-4 \times 1 \times 9=9-36=-27\)
Therefore, the required solutions are
\(\frac{-b \pm \sqrt{D}}{2 a}=\frac{-3 \pm \sqrt{-27}}{2 \times 1}=\frac{-3 \pm 3 \sqrt{-3}}{2}=\frac{-3 \pm 3 \sqrt{3} i}{2}\)
4. Solve each of the following equations:
\(-x^{2}+x-2=0\)
\(-x^{2}+x-2=0\)
Answer
It is given in the question that,
\(-x^{2}+x-2=0\)
On comparing the given equation \( \mathrm{a}x^{2}+\mathrm{b}x+\mathrm{c}=0 \), we obtain \( \mathrm{a}=-1, \mathrm{~b}=1 \) and \( \mathrm{c}=-2 \)
Therefore, the discriminant of the given equation is
\(D=b^{2}-4 a c=1^{2}-4 \times(-1) \times(-2)=1-8=-7\)
Therefore, the required solutions are
\(\frac{-b \pm \sqrt{D}}{2 a}=\frac{-1 \pm \sqrt{-7}}{2 \times(-1)}=\frac{-1 \pm \sqrt{7} i}{-2}\)
5. Solve each of the following equations:
\(x^{2}+3 x+5=0\)
\(x^{2}+3 x+5=0\)
Answer
It is given in the question that,
\(x^{2}+3 x+5=0\)
On comparing the given equation \( \mathrm{a}x^{2}+\mathrm{b}x+\mathrm{c}=0 \), we obtain \( \mathrm{a}=1, \mathrm{~b}=3 \), and \( \mathrm{c}=5 \)
Therefore, the discriminant of the given equation is
\(D=-b^{2}-4 a c=3^{2}-4 \times 1 \times 5=9-20=-11\)
Therefore, the required solutions are
\(\frac{-b \pm \sqrt{D}}{2 a}=\frac{-3 \pm \sqrt{-11}}{2 \times 1}=\frac{-3 \pm \sqrt{11 i}}{2}\)
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6. Solve each of the following equations:
\(x^{2}-x+2=0\)
\(x^{2}-x+2=0\)
Answer
It is given in the question that,
\(x^{2}-x+2=0\)
On comparing the given equation \( \mathrm{a}x^{2}+\mathrm{b}x+\mathrm{c}=0 \), we obtain \( \mathrm{a}=1, \mathrm{~b}=-1 \), and \( \mathrm{c}=2 \)
Therefore, the discriminant of the given equation is
\(D=b^{2}-4 a c=(-1)^{2}-4 \times 1 \times 2=1-8=-7\)
Therefore, the required solutions are
\(\frac{-b \pm \sqrt{D}}{2 a}=\frac{-(-1) \pm \sqrt{-7}}{2 \times 1}=\frac{1 \pm \sqrt{7} i}{2}\)
7. Solve each of the following equations:
\(\sqrt{2} x^{2}+x+\sqrt{2}=0\)
\(\sqrt{2} x^{2}+x+\sqrt{2}=0\)
Answer
It is given in the question that,
\(\sqrt{2} x^{2}+x+\sqrt{2}=0\)
On comparing the given equation \( \mathrm{a}x^{2}+\mathrm{b}x+\mathrm{c}=0 \), we obtain \( \mathrm{a}=\sqrt{2}, \mathrm{~b}=1 \) and \( \mathrm{c}=\sqrt{2} \)
Therefore, the discriminant od the given equation is
\(D=b^{2}-4 a c=1^{2}-4 \times \sqrt{2} \times \sqrt{2}=1-8=-7\)
Therefore, the required solutions are
\(\frac{-b \pm \sqrt{D}}{2 a}=\frac{-1 \pm \sqrt{-7}}{2 \times \sqrt{2}}=\frac{-1 \pm \sqrt{7} i}{2 \sqrt{2}}\)
8. Solve each of the following equations:
\(\sqrt{3} x^{2}-\sqrt{2} x+3 \sqrt{3}=0\)
\(\sqrt{3} x^{2}-\sqrt{2} x+3 \sqrt{3}=0\)
Answer
It is given in the question that,
\(\sqrt{3} x^{2}-\sqrt{2} x+3 \sqrt{3}\)
On comparing the given equation \( \mathrm{a}x^{2}+\mathrm{b}x+\mathrm{c}=0 \), we obtain \( \mathrm{a}=\sqrt{3}, \mathrm{~b}=-\sqrt{2} \), and \( c=3 \sqrt{3} \)
Therefore, the discriminant of the given equation is
\(\mathrm{D}=\mathrm{b}^2-4 \mathrm{ac}=(-\sqrt{2})^{2}-4(\sqrt{3})(3 \sqrt{3})=2-36=-34\)
Therefore, the required solutions are
\(\frac{-b \pm \sqrt{D}}{2 a}=\frac{-(-\sqrt{2}) \pm \sqrt{-34}}{2 \times \sqrt{3}}=\frac{\sqrt{2} \pm \sqrt{34} i}{2 \sqrt{3}} \quad[\sqrt{-1}=i]\)
9. Solve each of the following equations:
\(x^{2}+x+\frac{1}{\sqrt{2}}=0\)
\(x^{2}+x+\frac{1}{\sqrt{2}}=0\)
Answer
The given quadratic equation is
This equation can also be written as \( \sqrt{2} x^{2}+\sqrt{2} x+1=0 \)
On comparing this equation with \( \mathrm{a}x^{2}+\mathrm{b}x+\mathrm{c}=0 \), we obtain \( \mathrm{a}=\sqrt{2}, \mathrm{~b}=\sqrt{2} \), and \( \mathrm{c}=1 \)
Discriminant \( (D)=\mathrm{b}^2-4 \mathrm{ac}=(\sqrt{2})^{2}-4 \times \sqrt{2} \times 1=2-4 \sqrt{2} \)
Therefore, the required solutions are
\(\frac{-b \pm \sqrt{D}}{2 a} =\frac{-\sqrt{2} \pm \sqrt{2-4 \sqrt{2}}}{2 \times \sqrt{2}}=\frac{-\sqrt{2} \pm \sqrt{2(1-2 \sqrt{2})}}{2 \sqrt{2}}\)
\(=\left(\frac{-\sqrt{2} \pm \sqrt{2}(\sqrt{2} \sqrt{2}-1)}{2 \sqrt{2}}\right)\)
10. Solve each of the following equations:
\(x^{2}+\frac{x}{\sqrt{2}}+1=0\)
\(x^{2}+\frac{x}{\sqrt{2}}+1=0\)
Answer
The given quadratic equation is
This equation can also be written as \( \sqrt{2} x^{2}+x+\sqrt{2}=0 \)
On comparing this equation with \( \mathrm{a}x^{2}+\mathrm{b}x+\mathrm{c}=0 \), we obtain
\(a=\sqrt{2}, b=1, c=\sqrt{2}\)
\(D=b^{2}-4 a c=1^{2}-4 \times \sqrt{2} \times \sqrt{2}=1-8=-7\)
Therefore, the required solutions are
\(\frac{-b \pm \sqrt{D}}{2 a}=\frac{-1 \pm \sqrt{-7}}{2 \times \sqrt{2}}=\frac{-1 \pm \sqrt{7} i}{2 \sqrt{2}} \quad[\sqrt{-1}=i]\)
