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: px² +qx + r = 0 with real coefficients p, q, r and p≠0. Now, let us assume that the discriminant d < 0 i.e., b²-4ac< 0.

The solution of above quadratic equation will be in the form of complex numbers given by:

Important Notes: 

1. A polynomial equation has at least one root
2. A polynomial equation of degree n has n roots
3. The values of a variable, that satisfy the given equation are called roots of a quadratic equation
4. The solution to quadratic equations can also be calculated using the factorisation method
5. If α and β are the roots of a quadratic equation, then the equation is x² – (α + β) x + αβ = 0
6. 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.

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Quadratic Equations Class 11 Examples

$\begin{array}{l}\text{1. Find the roots of equation }{x}^2+2=0\end{array}$

Solution:

$\begin{array}{l}\text{Given, }{x}^2+2=0\end{array}$

$\begin{array}{l}i.e.,x^2=-2orx=\pm \sqrt{2}i\end{array}$

$\begin{array}{l}\text{2. Solve }\sqrt{5}{x}^2+x+\sqrt{5}=0\end{array}$

Solution:

$\begin{array}{l}\text{Given }\sqrt{5}{x}^2+x+\sqrt{5}=0\end{array}$

$\begin{array}{l}\text{Therefore, discriminant D = }{b}^2-4ac=1-4(\sqrt{5}\times \sqrt{5})=-19\end{array}$

$\begin{array}{l}\text{Therefore, the solution of given quadratic equation = }\frac{-1\pm \sqrt{-19}}{2\sqrt{5}}=\frac{-1\pm 19i}{2\sqrt{5}}\end{array}$

$\begin{array}{l}\text{3. Solve }{x}^2+x+1=0\end{array}$

Solution:

$\begin{array}{l}\text{Given, }{x}^2+x+1=0\end{array}$

$\begin{array}{l}\text{Therefore, discriminant D = }{b}^2-4ac=1-4=-3\end{array}$

$\begin{array}{l}\text{Therefore, the solution of given quadratic equation = }\frac{-1\pm \sqrt{-3}}{2}=\frac{-1\pm 3i}{2}\end{array}$

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