# CAT Algebra Quadratic Equations

## The three sections of CAT, Quantitative Aptitude, Logical Reasoning, and Data Interpretation, and Verbal Ability, each comprise questions and sets to present criteria for evaluating the applicant’s potential in being future managers and entrepreneurs.

The results are accorded in terms of percentiles, that is, by comparing, allotting, and equalizing results across ranks.

The scrutiny, as a result, would be to determine the candidate’s skills such as confidence, in attempting questions they are assured of, as the wrong attempts are awarded negative scores, their proficiency in concluding the test within the time allotted, their ability in prioritizing the sets, which to attempt first, which later and which to leave, and of course, their potential in performing under stress and pressure.

Another point is that while CAT is an important step for admission to one’s dream institute, the process does not end with it.

CAT can be called the preliminary stage for the admission procedure, it is followed by a call for interview, group discussion, written ability test, and this would also be determined by the applicant’s profile, academic performance, work experience, co-curricular, and so forth. The applicants, therefore, need to adopt a multitudinous focus, whereby they can be a rung closer to their aim.

The applicants should begin their preparation by getting in touch with the purview of the examination, brushing up on current affairs and general knowledge, and work on their pace and time management.

The general pattern of the Common Admission Test comprises a hundred questions, divided into three sections, as specified, and has a time frame of three hours, an hour for each section.

The Quantitative Aptitude section, comprising questions based on mathematics and entailing a variety of subjects, broadly, number system, algebra, mensuration, time and distance, and modern maths.

Algebra is a branch of mathematics, which uses symbols and letters in place of unknown quantities, also known as variables, as their value varies as per the conditions.

The algebraic expressions, made up of these variables, represent the given situation with the required mathematical operations. An important concept for algebra is equations, which as the name suggests, represent equality between two expressions, signifying that one state of affairs is the same as the other and under what conditions.

**These equations, vary as per**

- Terms, that are distinct parts of the equation; the number of terms determines the expression, an expression with one term is a monomial, two is a binomial, three is a trinomial, also an expression with two or more than two terms is referred to as a polynomial.
- Power or degree, that is the highest exponent assumed in the given expression.

**Equations**– Equations as specified above, evince equality, thereby can be used to determine the value of a variable that results in the agreement of both sides.

An equation with a degree or highest power 1, is called a linear equation, it can be in one or more variables, for example, 2x+1=5, 3x+2y=4, 7a+10b=110, and so forth.

**Quadratic Equations**

An equation of degree 2, that is, the highest order of the variable is two is called a quadratic equation. The general form for a quadratic equation is ax²+bx+c=0, where x is the variable and a and b are real numbers.

The roots of a quadratic equation are the values of x or the variable, which satisfy the equation. Quadratic equations do not have more than two distinct solutions, or roots, or zeroes.

Framing quadratic equations: For framing a quadratic equation, the given conditions and situations should be followed and translated into requisite mathematical operations.

- For example, the difference between two positive integers is 3 and the sum of their squares is 117.

Assume one of the numbers to be x, it has been given that their difference is 3, which implies the other number is 3+x, as x-y=3 or y=3+x

Also, sum of their squares is 117

→x²+(3+x)²=117,

→x²+x²+6x+(3)²=117

→2x²+6x=117-9

→2x²+6x=108, which is the required quadratic equation

**Solving a quadratic equation will be taken up further.**

- Forming quadratic equations where roots are given:

The general form of a quadratic equation is ax^2+bx+c, roots are the solutions which satisfy the equation,

Sum of roots = (-b)/a, product of roots=C/a,

Example when roots are -3, 1/2,

Sum of roots = -3+ 1/2 →(-b)/a=(-6+1)/2,

→(-b)/a=-5/2

→-b=-5,or b=5,a=2

→C/a=-3×1/2=-3/2,

→c=-3, a=2

The equation thus is,

ax²+bx+c=2x²+5x-3

Conversely, if the sum of product of roots is given, sum=1, product=6, -b/a=1,c/a=6

→b=-1, c=6, a=1

→The equation would be ax²+bx+c=x²-x+6

**Solving Quadratic Equations:**

- General principles followed while solving an equation are,
- Start by opening all the brackets, it reveals the further steps to be taken, for example, x²+2(x²+1)=3, x²+2x²+2=3, here the bracket signifies multiplication, care should be taken in cases of the negative sign before and after the brackets, as it changes other signs as well.
- Regroup the terms so as to put like terms together, as the operations can be used only when terms are similar, for example (x²+2)+(2x²+x)=7, arranging like terms together, x²+2x²+x+2=7
- Both sides of the equation should agree with each other, in other words, equality should be maintained, thus any operation on the side should be used on the other side as well

For example, x²-x+6=0, to eliminate 6, subtract 6 from both sides, x²-x=-6 - The next step is to simplify the equation and determine the solution.
- The final step is to verify the result obtained if it satisfies the given equality or not.

**For solving quadratic equations, some algebraic formulae should be kept in mind**

(a+b)²=a²+2ab+b²

(a-b)²=a²-2ab+b²

(a-b)(a+b)=a²-b²

As specified, the roots of a quadratic equation are the distinct value/s (not more than 2) that satisfy the given equation.

For solving quadratic equations, the following methods may be used:

**By factorization or splitting the middle term**

To factorize a given equation ax²+bx+c=0, Determine the product of the first and last term, that is, ac

Identify factors of ac such that the sum of the two factors is the middle term, that is, b.

Split the middle term, b, as the sum of the two factors as identified above,

Arrange the terms as pairs, the first two and last two, factorize the two pairs by finding common factors,

Equate the resultant combination to 0 to determine the value of x (since the result is a product, either of the terms can result in 0)

**For Example**, The product of two consecutive odd integers is 2499, to find the bigger integer, Assume the smaller integer as x,

the other integer would be x+2, as given the product is 2499, which implies (x)(x+2)=2499,(x²+2x)=2499; x²+2x-2499=0

To split the middle term, find the factors of 1×-2499=-2499, such that their sum gives 2.

The factors, in this case, would be, -51 and 49

x²-51x+49x-2499=0; x(x-51)+49(x-51)=2499

(x+49)(x-51)=0 or x=-49,51

Therefore the bigger odd integer is 51.

**By completing the square:**

Consider the equation ax²+bx+c=0,

for completing the square, the equation should be divided by the coefficient of x, so as to make the coefficient of the first term or x², 1; (ax²+bx+c)/a=0/a, that is, (ax²)/a+bx/a+c/a=0

Subtracting third term or the constant from both sides, x²+bx/a=-c/a,

As in the identity specified above, (a+b)²=a²+2ab+b², {in some cases, (a-b)²=a²-2ab+b², can be used} to get 2ab, add square of half the coefficient of the second term to both sides, x²+bx/a+b²/4=-c/a+b²/4,

Completing the square, we get, (x+b/2)²=-c/a+b²/4, the value of the variable can be found by taking the square root of both sides.

**For Example,**

Divide 51 into two parts whose product is 608.

Assume the first number is x since the sum is 51, the second number would be 51-x,

Given, the product of the two numbers is 608, that is, (x)(51-x)=608,

51x-x²=608; x²-51x+608=0

Subtracting 608 from both sides, x²-51x=-608;

Adding square of half of -51 to both sides, x²-51x+2601/4=-608+2601/4,

(x-51/2)²=169/4; (x-51/2)²=(13)/((2))², which gives x-51/2=13/2; x=13/2+51/2=32

Therefore the two numbers are 32 and 51-32=19.

**By using Quadratic formula**

The roots of a quadratic equation can be established by using the formula x=(-b±√(b²-4ac) ))/2a, where x is the variable and the formula gives the value of the roots.

As from the general form of a quadratic equation, ax²+bx+c=0, -b in the formula represents negative of the coefficient of the second term of the equation, a represents the coefficient of the first term and c represents the constant in the equation.

The formula is a continuation of completing the square method, which is at times a quicker way than the other two methods, either when there are a number of factors or when the last term is not factorable.

**For Example,** A two-digit number is made of two consecutive digits such that the sum of their squares is 4 less than the number. Find the two-digit number.

Assume the first digit to be x since the digits are consecutive numbers, the second digit would be x+1,

Since the given number is a two-digit number, there would be a tens digit and a digit at the one’s place. The number, therefore, is 10(x)+(x+1)=10x+x+1 or 11x+1

Also, the sum of the squares is 4 less than the number, that is, the difference between the sum of squares of the digits is 4

Sum of squares of the digits= (x)²+(x+1)²=x²+x²+2x+1

The equation, thus, is 11x+1-(2x²+2x+1)=4

11x+1-2x²-2x-1=4; -2x²+9x-4=0, or

2x²-9x+4=0

According to the formula, -b=-(-9)=9, b²=(9)²=81, 4ac=4(2)(4)=32,and 2a=2(2)=4

x=(9±√(81-32))/4, that is, x=(9±√49)/4; x=(9±7)/4

Therefore, x=4,1/2 (since the variable represents a digit, the fraction cannot be the value for x)

The number, hence is = 11(4)+1=44+1=45

To check the solution, sum of squares of 4 and 5; 4²+5²=16+25=41

Difference between the number and sum of squares of the digits 45-41=4

Thus, the answer is verified.

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