A Quadratic equation is an equation with the highest degree two. The standard form is written as ax2+bx+c=0 where a is not equal to zero and a,b are the coefficients of x2and x respectively; c is a constant. Some of the examples are, 2x2+3x+5; x2-4x+5; 4x2-16 etc. Let us now learn how to solve Quadratic Equations. The following are the different methods used in solving Quadratic Equations,
Completing the square method
Using the Quadratic Formula
Any equation of the form ax2+bx+c=0 and ax2+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 x2+4x-5=0 for x
Solution: Given equation x2+4x-5=0
To factor the equation we find the product of the coefficient of x2 and the constant term which is,
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
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 x2+6x-7=0
Solution: Given equation x2+6x-7=0
The constant term is taken to the other side of the equal to sign
x2+6x = 7
Now consider the coefficient of the x term which is 6, half the value and square it which gives 6/2=32=9
Converting the left hand side to a squared form by adding 9 on both sides
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 ax2+bx+c, the solutions using the formula is given as, [-bsqrt(b2-4ac)]/2a.
Here the given equation is compared to the standard form, the values substituted appropriately and finally solved for the variable
For example, solve 3x2+5x+1=0 for x
Solution: Given equation 3x2+5x+1=0
Comparing the terms with the standard equation ax2+bx+c=0
a=3, b=5 and c=1
Substituting the values of a,b and c in the formula and simplifying gives
Solutions are, x= [-5+sqrt(13)]/6 and x= [-5 - sqrt(13)]/6
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