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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.

$$\begin{array}{l}px^{2}+qx+r=0\end{array}$$
with real coefficients p, q, r and
$$\begin{array}{l}p\neq 0\end{array}$$
. Now, let us assume that the discriminant d < 0 i.e.
$$\begin{array}{l}b^{2}-4ac< 0\end{array}$$
.

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

 $$\begin{array}{l}x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}=\frac{-b\pm i\sqrt{4ac-b^{2}}}{2a}\end{array}$$

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.

## Quadratic Equations Class 11 Examples

1. Find the roots of equation

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

Solution: Give,

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

i.e.

$$\begin{array}{l}x^{2} = -2\end{array}$$
or x =
$$\begin{array}{l}\pm \sqrt{2}i\end{array}$$

2. Solve

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

Solution: Given

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

Therefore, discriminant D =

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

Therefore, the solution of given quadratic equation =

$$\begin{array}{l}\frac{-1\pm \sqrt{-19}}{2\sqrt{5}}=\frac{-1\pm 19i}{2\sqrt{5}}\end{array}$$

3. Solve

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

Solution: Given

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

Therefore, discriminant D =

$$\begin{array}{l}b^{2}-4ac=1-4=-3\end{array}$$

Therefore, the solution of given quadratic equation =

$$\begin{array}{l}\frac{-1\pm \sqrt{-3}}{2}=\frac{-1\pm 3i}{2}\end{array}$$