A quadratic equation in the variable \(x\) is an equation of the form \(ax^2+bx+c =0\), where \(a, b,c\) are real numbers, \(a \neq 0\). The degree of the quadratic equation is \(2\).
Example:
\(x^2 + 9x +15 =0\).
It is of the form \(ax^2+bx+c =0\).
Therefore, the given equation is a quadratic equation.
Procedure to solve the quadratic equation by the factorisation method:
Step 1: Write the given equation in standard form.
Step 2: Express the middle term as the sum of two terms such that the sum satisfies the middle term, and the product should satisfy the extreme product.
Step 3: Group the expression into two linear factors by taking the common term outside.
Step 4: Now, solve for \(x\) by equating each linear factor to zero. The obtained values of \(x\) are called the roots or zeroes of the equation.
Step 2: Express the middle term as the sum of two terms such that the sum satisfies the middle term, and the product should satisfy the extreme product.
Step 3: Group the expression into two linear factors by taking the common term outside.
Step 4: Now, solve for \(x\) by equating each linear factor to zero. The obtained values of \(x\) are called the roots or zeroes of the equation.
Example:
Find the roots of the quadratic equation \(2x^2 + 4 = 9x\). 2x 2 +4=9
Solution:
The given equation is \(2x^2 + 4 = 9x\)
.
.
Let us first write the given equation in standard form.
\(2x^2 -9x +4 = 0\)
Now, split the middle term by the above procedure.
\(2x^2 -x -8x +4 = 0\)
Group the expression into two linear factors by taking the common term outside.
\(x(2x -1) -4(2x -1) =0\)
\((2x -1)(x -4) =0\)
\\(2x -1(1)(x−4)=0
Now, solve for \(x\) by equating each linear factor to zero.
\(2x−1=0\) or \(x−4=0\)
\(x= \frac{1}{2}\) or \(x=4\)
Therefore, the roots of \(2x^2 + 4 = 9x\) are \(\frac{1}{2}\) and \(4\).
Important!
Any quadratic equation can have at most two roots.