Equation Formula
Equation Formula
The quadratic equation is one of the fundamental concepts in Algebra. This quadratic form can be solved and verified using formulas. These formulas must be retained by students if they are striving to excel in the subject of Mathematics. The Extramarks learning portal offers students comprehensive reference materials to facilitate learning the Equation Formula for them. Students can also attend interactive video lectures conducted by the mentors of Extramarks for an engaging and versatile learning experience.
Equation Problems
Students can examine quadratic Equation Formula solutions on the Extramarks educational website.
Solved Examples
- Example 1 :
2×2 – 5x + 2 = 0
Solution :
Comparing 2×2 – 5x + 2 = 0 and ax2 + bx + c = 0, one gets
a = 2, b = -5 and c = 2
Then,
x = [-b ± √b2 – 4ac] / 2a
x = [-(-5) ± √(-5)2 – 4(2)(2)] / 2(2)
x = [5 ± √(25 – 16)] / 4
x = [5 ± √9] / 4
x = [5 ± 3] / 4
x = (5 + 3) /4 and x = (5 – 3)/4
x = 8/4 and x = 2/4
x = 2 and x = 1/2
Therefore, the solution is {1/2, 2}.
- Example 2:
√2f2 – 6f + 3√2 = 0
Solution:
Comparing √2f2 – 6f + 3√2 = 0 and ax2 + bx + c = 0, one gets
a = √2, b = -6 and c = 3√2
Then,
x = [-b ± √b2 – 4ac] / 2a
x = [-(-6) ± √(-6)2 – 4(√2)(3√2)] / 2(√2)
x = [6 ± √(36 – 24)] / 2√2
x = [6 ± √12] / 2√2
x = [6 ± 2√3] / 2√2
| x = [6 + 2√3] / 2√2
x = 2(3 + √3) / 2√2 x = (3 + √3) / √2 |
x = [6 – 2√3] / 2√2
x = 2(3 – √3) / 2√2 x = (3 – √3) / √2 |
Therefore, the solution is {(3 + √3)/√2, (3 – √3)/√2}.
- Example 3:
3y2 – 20y – 23 = 0
Solution:
Comparing 3y2 – 20y – 23 = 0 and ax2 + bx + c = 0, one gets
Then,
x = [-b ± √b2 – 4ac] / 2a
x = [-(-20) ± √(-20)2 – 4(3)(-23)] / 2(3)
x = [20 ± √(400 + 276)] / 6
x = [20 ± √676] / 6
x = [20 ± 26] / 6
| x = [20 + 26] / 6
x = 46 / 6 x = 23 / 3 |
x = [20 – 26] / 6
x = -6 / 6 x = -1 |
Therefore, the solution is {-1, 23/3}.
- Example 4:
36y2 – 12ay + (a2 – b2) = 0
Solution:
Comparing 36y2 – 12ay + (a2 – b2) = 0 and ax2 + bx + c = 0, one gets
a = 36, b = -12a and c = (a2 – b2)
Then,
x = [-b ± √b2 – 4ac] / 2a
x = [-(-12a) ± √(-12a)2 – 4(36)(a2 −b2)] / 2(36)
x = [12a ± √(144a2 – 144a2 + 144b2)] / 72
x = [12a ± √144b2] / 72
x = [12a ± 12b] / 72
x = 12(a ± b) / 72
x = (a ± b) / 6
FAQs (Frequently Asked Questions)
1. What is the best way to check if a quadratic Equation Formula has a correct solution?
While b and c can both be 0, a will never be 0. Quadratic equations always have two values when solved. The roots of the Equation Formula are these two values. It is always true that the roots of the equation satisfy the Equation Formula. When in doubt, one can put the values back into the equation to check the solution.
2. For quadratic equations, what is the thumb rule?
A quadratic equation cannot have a value of 0 as the value of a. In the expression, x represents the variable. The algebraic expression yields two roots when solved. Quadratic equations are solved using the quadratic Equation Formula.
3. What is the formula for a quadratic Equation Formula with x2?
Quadratic equations are written as ax2 + bx + c = 0. There is a constant a, a constant b, and a constant c, a variable x, and a constant a. x2 in the equation makes it quadratic, otherwise, it would be a linear equation.
4. In general, what is the form of a quadratic equation?
Quadratic equations are equations that have more than one term and at least one term of degree 2. The general form of the equation is ax2 + bx + c, where a,b, and c are real numbers, and a is not zero. Quadratic equations have solutions that satisfy the “x” in the Equation Formula. Equation roots are also known as equation roots.
