A quadratic equation in the variable x is an equation of the form ax2+bx+c=0, where a,b,c are real numbers, a≠0 . For example, 2x2+x−300=0 is a quadratic equation ax2+bx+c=0,a≠0 is called the standard form of a quadratic equation.
Method of Solving a Quadratic Equation:
Factorization Method
Factorize the quadratic equation by splitting the middle term.
After splitting the middle term, convert the equation into linear factors by taking common terms out.
Then on equating each factor to zero the roots are determined.
For example:
⇒2x2−5x+3 (Split the middle term)
⇒2x2−2x−3x+3 (Take out common terms to determine linear factors)
⇒2x(x−1)−3(x−1)
⇒(x−1)(2x−3) (Equate to zero)
⇒(x−1)(2x−3)=0
When (x−1)=0 , x=1
When (2x−3)=0 , x=32
So, the roots of 2x2−5x+3 are 1 and 32
Method of completing the square
The solution of a quadratic equation can be found by converting any quadratic equation to perfect square of the form (x+a)2−b2=0.
To convert quadratic equation x2+ax+b=0 to perfect square equate b i.e., the constant term to the right side of equal sign then add square of half of a i.e., square of half of coefficient of x both sides.
To convert quadratic equation of form ax2+bx+c=0,a≠0 to perfect square first divide the equation by a i.e., the coefficient of x2 then follow the above-mentioned steps.
For example:
⇒x2+4x−5=0 (Equate constant term 5 to the right of equal sign)
⇒x2+4x=5 (Add square of half of 4 both sides)
⇒x2+4x+(42)2=5+(42)2
⇒x2+4x+4=9
⇒(x+2)2=9
⇒(x+2)2−(3)2=0
It is of the form (x+a)2−b2=0
Now,
⇒(x+2)2−(3)2=0
⇒(x+2)2=9
⇒(x+2)=±3
⇒x=1
and x=−5
So, the roots of x2+4x−5=0are 1 and −5
By using the quadratic formula
The root of a quadratic equation ax2+bx+c=0 is given by formula
x=−b±b2−4ac−−−−−−−√2a , where b2−4ac−−−−−−−√ is known as discriminant.
If b2−4ac−−−−−−−√≥0 then only the root of quadratic equation is given by
x=−b±b2−4ac−−−−−−−√2a
For example:
⇒x2+4x+3
On using quadratic formula, we get
⇒x=−4±(4)2−4×1×3−−−−−−−−−−−−−√2×1
⇒x=−4±16−12−−−−−−√2
⇒x=−4±4-√2
⇒x=−4±22
⇒x=−4+22 , x=−1
⇒x=−4−22 ,x=−3
So, the roots of x2+4x+3=0 are −1 and −3
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