Quadratic Equations Class 11
Quadratic Equations Class 11 Notes are available here for students. The notes are very helpful to have a quick revision before exams. Class 11 Maths Chapter 5 quadratic equations include a quadratic formula to find the solution of the given equation.
Consider the quadratic equation: \(px^{2}+qx+r=0\) with real coefficients p, q, r and \(p\neq 0\). Now, let us assume that the discriminant d < 0 i.e. \(b^{2}-4ac< 0\).
The solution of above quadratic equation will be in the form of complex numbers given by:
| \(x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}=\frac{-b\pm i\sqrt{4ac-b^{2}}}{2a}\) |
Important Notes:
- A polynomial equation has at least one root
- A polynomial equation of degree n has n roots
- The values of a variable, that satisfy the given equation are called roots of a quadratic equation
- The solution to quadratic equations can also be calculated using the factorisation method
- If α and β are the roots of a quadratic equation, then the equation is x2 – (α + β) x + αβ = 0
- The nature of roots depends on the discriminant (D) of the quadratic equation
- If D > 0, roots are real and distinct (unequal)
- If D = 0, roots are real and equal (coincident)
- If D < 0, roots are imaginary and unequal
Find solved questions based on quadratic equations using formula.
Quadratic Equations Class 11 Examples
1. Find the roots of equation \(x^{2}+2=0\)
Solution: Give, \(x^{2}+2=0\)
i.e. \(x^{2} = -2\) or x = \(\pm \sqrt{2}i\)
2. Solve \(\sqrt{5}x^{2}+x+\sqrt{5}=0\)
Solution: Given \(\sqrt{5}x^{2}+x+\sqrt{5}=0\)
Therefore, discriminant D = \(b^{2}-4ac=1-4(\sqrt{5}\times \sqrt{5})=-19\)
Therefore, the solution of given quadratic equation = \(\frac{-1\pm \sqrt{-19}}{2\sqrt{5}}=\frac{-1\pm 19i}{2\sqrt{5}}\)
3. Solve \(x^{2}+x+1=0\)
Solution: Given \(x^{2}+x+1=0\)
Therefore, discriminant D = \(b^{2}-4ac=1-4=-3\)
Therefore, the solution of given quadratic equation = \(\frac{-1\pm \sqrt{-3}}{2}=\frac{-1\pm 3i}{2}\)