We have already discussed about Polynomial in Algebra. A polynomial equation having degree of two is known as Quadratic equation. “Quad” means four but there Quadratic means ‘to make square.’
Standard form of Quadratic equation:
The standard form of Quadratic equation is
\(ax^{2}+bx+c = 0\)
Where, x is an unknown and a,b,c are the constants such that \(a \neq 0\)
If a becomes 0 then the equation will become a linear equation rather than quadratic equation.
Example: Determine all the quadratic equation given below- (i) \(3x^{2}+ 5x + 1 = 0\) (ii) \( 13x + 7 = 0\) (iii) \(3x^{5}+ 32x^{2} + 7 = 0\) (iv) \(9x^{2} – 36 = 0\) Solution: (i) and (iv) forms the quadratic equation. |
Solution for Quadratic equation:
As quadratic equation has degree of 2, hence the number of solution that will satisfy the given equation will also be equal to 2.
There can be more than one method for finding the solution for quadratic equation:
- Factoring the Quadratic equation
- Completing the square method
- Sridhara’s Formula
Formula for Quadratic Equation:
Also known as Sridhara’s Method, which is mostly common for finding the roots of quadratic equation (\(ax^{2}+bx+c = 0\)), which is given as:
\(x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\)
here \(\pm\) represent both the values (one positive and one negative), hence there are two possible roots for the equation.
Discriminant (D):
The term \(b^{2}-4ac\) is known as discriminant (D), as it differentiate two roots one being positive and other being negative.
So the formula becomes:
\(\frac{-b \pm \sqrt{D}}{2a}\), (where \(D = b^{2}-4ac\))
Nature of Roots of a Quadratic Equation:
Depending upon the value of discriminant, the quadratic equation can have 3 different types of roots, namely:
S.no |
Value of Discriminant (D) |
Nature of Roots |
1 |
\(b^{2}-4ac > 0\) |
Two distinct Real roots |
2 |
\(b^{2}-4ac = 0\) |
Two equal roots |
3 |
\(b^{2}-4ac < 0\) |
Two distinct Complex Roots |
Example: Find the roots of the equation \(2x^{2} + 7x + 3 = 0\) Solution: Given: \(2x^{2} + 7x + 3 = 0\) Here a = 2, b = 7 and c = 3 Finding the value of D, \(D = b^{2}-4ac = 7^{2}- 4\times 2 \times 3 = 49 – 24\) \(\Rightarrow D = 25\) As \(D >0\), the equation has two distinct real roots. \(x = \frac{-b \pm \sqrt{D}}{2a}\) \(\Rightarrow x = \frac{-7 \pm \sqrt{25}}{2 \times 2}\) \(\Rightarrow x = \frac{-7 \pm 5}{4}\) \(\Rightarrow x = \frac{-7 – 5}{4}\; or \; \frac{-7 + 5}{4}\) \(\Rightarrow x = -3 \; or \; \frac{-1}{2}\) |
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Related Links:
Quadratics Quadratic Formula Solving Quadratic Equation Standard form of Quadratic equation |