## Trigonometry

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:

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 $$x^{2}+2=0$$

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 = $$\frac{-1\pm \sqrt{-3}}{2}=\frac{-1\pm 3i}{2}$$