Equation Formula
Quadratic equation is one of the fundamental concept of algebra. By using formula, this quadratic form can be solved and solutions can be verified too. In this case there are two solutions. The formula for equation is given below.
Let ax2+ bx + c = 0 is the quadratic equation, the solution will be,
Equation Problems
Let us discuss the solutions of quadratic equations.
Solved Examples
Question 1: Find out the roots of the given equation x2 + 2x – 3 = 0.
The given quadratic equation is,
x2 + 2x – 3 = 0
Here a = 1, b = 2 and c = -3
The formula for the solution,
Solution:
The given quadratic equation is,
x2 + 2x – 3 = 0
Here a = 1, b = 2 and c = -3
The formula for the solution,
x =
\(\begin{array}{l}\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\end{array} \)
x =
\(\begin{array}{l}\frac{-2 \pm \sqrt{2^{2}-4\times1\times-3}}{2\times1}\end{array} \)
x =
\(\begin{array}{l}\frac{-2 \pm \sqrt{4+12}}{2}\end{array} \)
x =
\(\begin{array}{l}\frac{-2+4}{2}\end{array} \)
= 1 or x = \(\begin{array}{l}\frac{-2-4}{2}\end{array} \)
= -3 Question 2: Find out the roots of 2x2 – 3x – 5 = 0.
The given quadratic equation is,
2x2 – 3x – 5 = 0
Here a = 2, b = -3 and c = -5
The formula for the solution,
x =
x =
Solution:
The given quadratic equation is,
2x2 – 3x – 5 = 0
Here a = 2, b = -3 and c = -5
The formula for the solution,
x =
\(\begin{array}{l}\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\end{array} \)
x =
\(\begin{array}{l}\frac{3\pm \sqrt{3^{2}-4\times2\times-5}}{2\times2}\end{array} \)
x =
\(\begin{array}{l}\frac{3\pm \sqrt{49}}{4}\end{array} \)
x =
\(\begin{array}{l}\frac{3+7}{4}\end{array} \)
= \(\begin{array}{l}\frac{5}{2}\end{array} \)
or x = \(\begin{array}{l}\frac{3-7}{4}\end{array} \)
= -1
| More topics in Equation Formula | |
| Linear Equations Formula | |
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