A Quadratic equation is an equation with the highest degree two. The standard form is written as ax^{2}+bx+c=0 where a is not equal to zero and a,b are the coefficients of x^{2}and x respectively; c is a constant. Some of the examples are, 2x^{2}+3x+5; x^{2}-4x+5; 4x^{2}-16 etc. Let us now learn how to solve Quadratic Equations. The following are the different methods used in solving Quadratic Equations,
Factoring method
Completing the square method
Using the Quadratic Formula
Any equation of the form ax^{2}+bx+c=0 and ax^{2}+bx=0 where c is zero can be solved using the factoring method by finding the factors of the equation and solving for x.
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Examples On Solving Quadartic Equations
For example, solve x^{2}+4x-5=0 for x
Solution: Given equation x^{2}+4x-5=0
To factor the equation we find the product of the coefficient of x^{2} and the constant term which is,
Product= (1)(-5)=-5
List out the factors of -5 and choose the appropriate factors such that their sum would be equal to the coefficient of x which is 4
Factors of- 5 are -1 & 5; the sum (5-1) gives -4 and the product (5)(-1)= -5
Now the middle term 4x is split into the factors as -x and 5x
x^{2}-x+5x-5x
Re-grouping the terms, x(x-1)+ 5(x-1)
Taking (x-1) common gives (x-1)(x+5)
Solving for x, x-1=0; x+5=0
Solutions are x=1 and x=-5
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Let us solve Quadratic Equations using completing the square method. For example, solve x^{2}+6x-7=0
Solution: Given equation x^{2}+6x-7=0
The constant term is taken to the other side of the equal to sign
x^{2}+6x = 7
Now consider the coefficient of the x term which is 6, half the value and square it which gives 6/2=3^{2}=9
Converting the left hand side to a squared form by adding 9 on both sides
x^{2}+6x+9= 7+9
(x+3)^{2}=16 [a^{2}+2ab+b^{2}=(a+b)^{2}]
Square root on both sides
(x+3) = 4
We get, x+3=4 and x+3=-4
Solutions are x = 1 and x =-7
Sometimes the quadratic equations cannot be factored, in such cases the Quadratic formula is used to solve the equations. For the standard form of the equation ax^{2}+bx+c, the solutions using the formula is given as, [-bsqrt(b^{2}-4ac)]/2a.
Here the given equation is compared to the standard form, the values substituted appropriately and finally solved for the variable
For example, solve 3x^{2}+5x+1=0 for x
Solution: Given equation 3x^{2}+5x+1=0
Comparing the terms with the standard equation ax^{2}+bx+c=0
a=3, b=5 and c=1
Substituting the values of a,b and c in the formula and simplifying gives
x= [-bsqrt(b^{2}-4ac)]/2a
x= [[-5sqrt(5^{2}-4(3)(1))]/2(3)
x= [-5sqrt(13)]/6
Solutions are, x= [-5+sqrt(13)]/6 and x= [-5 - sqrt(13)]/6
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