CAT 2009 Results Declared, Check Now !
Quantitative Aptitude > Equations - Simple, Special and Quadratic
Practice Tests
Add Comment Bookmark share + Refresher MaterialLinear equation Free Online Preparation for CAT 2009 with Minglebox e-CAT Prep. Cover basic concepts of Equations-Simple, Special and Quadratic under Quantitative Aptitude for MBA Entrance Exam Preparation with Study material, solved examples and tests prepared by CAT coaching experts.
Linear equation in one variable A linear equation in one variable is respected as ax + b = 0, where a,b are real constants and in one variable is x = - any linear equation in one variable always has a solution.
Linear equation in 2 variable
Linear equation in 2 variable – Ab linear equation in 2 variable is represented as ax + by = c where a,b,c are real constants, where either a or b = 0 but not both simultaneously equation ax + by = c has infinite number of solution for real x and y but for suitable restrictions equation may have finite number of solutions.
Simultaneous linear equations
Simultaneously linear equations in two variable A simultaneously linear equation in 2 variable is represented as and have none, one or infinite number of solutions depending on the relationship between the constants. If If If
Equation in one variable
Equation in one variable but of higher degree. The degree of equation is the highest power of variable that exists in the equation for example
An equation of second degree in one variable is known as quadratic equation.
Quadratic equation
A equation of the form quadratic equation . Roots of a quadratic equation A value of 'x' say Therefore if there exists a number equation . In general , any quadratic will have roots either real or imaginary . Relationship between roots of the equation and co-efficients of the equation. If and method to solve a quadratic equation. Completing the square method at Step1 : add and subtract the quantity
Step 2 :- Rearrange the terms as follows
Discriminant the expression
Factorization method Before attempting this method , check the discriminant of the equation . If the discriminant D is zero or a perfect square , then this method will easily given the roots of the equation . We illustrated this by an example.
Example Question Find the roots of Solution
Step i :- Find the discriminant of the equation her by compression we get a = 1, b= -5, and c = 6 the discriminant D =
Step ii :- Consider the constant term C, and factories it into 2 factors whose product is C. here C = 6 Therefore 2 factor products of 6 are
For each pair try to see in which case the sum of factors is equal to b(the coefficient of x) here b= -5, we see that
Step iii :- Write the coefficient of x as sum of the 2 factors obtained in step So take out the common terms for first 2 terms and do the same for last 2 terms x (x-2) -3 (x-2) (x-2) is common in both the terms, so pull it out we get (x-2) (x-3) = 0
Step iv :- Equation each term equal to '0' to get the root x-2 =0 or x-3 =0 x-2 or x-3
Add CommentComments
|
|













rahulsikka826 : but for reviving basic concepts site is good
ashok7278 : let us know new shortcuts..much better..
atul2777 : some clarity is lacking in problems!
irfan2387 : its very easy now