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# Quadratic Equation Tricks & Tips

Any equation having in the form of ax2+bx+c=0, is a quadratic equation in where x represents an unknown. a, b & c represents the real numbers such that a is not equal to 0.

In this Equation:

a is the coefficient of x2

b is the coefficient of x

c is the constant term

How to solve Single Variable Quadratic Equation ?

In single variable quadratic equation we will be given one quadratic equation in the form of  ax2+bx+c=0 & we have to find the value of that unknown variable x.

In quadratic Equations We have the following options to choose from :-

1. X >Y
2. Y >X

iii. X >= Y

1. Y>= X
2. X = Y or relationship cannot be established.

(i) Now, Suppose on solving the quadratic equation we get X = -1, 2 And Y = -2, -3

On putting these values on number line we will see that X lies on the right of Y. Hence, We can say that X>Y or Y<X.

(ii) Now, Suppose if after solving the equation we get X = -1 , 2 and Y= 3,4.

On putting these values on number line we see that Y lies to the right of X without touching X at any point. Hence, We can say that Y>X or X<Y.

(iii) If X= 3,4 and Y= 2,3. Putting this on number line we get X to the right of Y but touching Y at one single point i.e., 3. Hence, X>=Y or Y<=X.

(iv) If X=3,4 and Y=4,5. Putting these values on number line we will see that Y lies to the right of X but touching X at 4. Hence, Y>=X or X<=Y.

(v) Now, Suppose X= 2,4 and Y= 3,5. Putting these values on number line we will see that X touches Y at all the points between 3 and 4 i.e., at 3, 3.1, 3.2, 3.3 etc. upto 4 hence, in this case we cannot relate X and Y and Hence, Will answer as Relationship cannot be established.

Also suppose if y=7 and X= 4,8 then X<y also X>y . Hence, in this case also no relationship can be established.

Example:

x2+13x+40=0 —-(1)

y2+7y+12=0 —– (2)

First Solve the Equation (1)

Therefore, +13 = (+5) + (+8)

Also, +40 = (+5) * (+8)

Now Change the signs & Divide them with the coefficient of x2 i.e with +1

Therefore, we get

-5/1= -5 & -8/1= -8

Thus, -5 & -8 are the required values of x.

Now similarly Solve the 2nd Equation,

+7 = (+3) + (+4)

Also +12 = (+3) * (+4)

Now Change the signs & Divide them with the coefficient of y2 i.e with +1

-3/1=-3 & -4/1=-4

Thus, -3 & -4 are the required values of y.

Therefore Now we have x= -5, -8

y= -3, -4

Therefore, by comparing both the values we can say that x < y

Short Tricks to Solve the Quadratic Equation :

 +      + –          – –       + +        + +       – –         + –        – +        –

Remember this diagram friends to solve the Questions of quadratic equation shortly in the exam but before exam you will need to practice more and more question to get speed and also to get familiar with the tricks.

1. 1. I.5×2– 87x + 378 = 0                 3y2– 49y + 200 = 0

I.5×2 – 45x – 42x + 378 = 0

or,5x(x – 9) – 42(x – 9) = 0

or,(5x – 42) (x – 9) = 0

x = 9,5

II.3y2 – 24y – 25y + 200 = 0

or,3y(y – 8) –25(y – 8) = 0

or,(y – 8) (3y – 25) = 0

y = 8,3

So , Answers : x > y

Now for this question we can solve this in less than thirty second how lets see,

Step-1. First take the Equation-I means equation with  variable x.

It is in the form of   ax2  +   (-)b x + c.

Step-2. See the Sign of b & c means sign of coefficient of x variable and sign of last one,

It is first negative(-) and second positive(+)

Step -3.From above table we can say that both the roots of x will be positive.

Step-4.Make Calculation 378*5=1890 first divide by 2 then 3 and then by 5 and follow above rule, so for this it will be 45 and 42 and now you don’t need to write another two to three lines of taking common and then making in the forms of roots. This is your answer because we know from table that both root have positive sign.

Step-5 For those question in which coefficient of X2 is not one like in this question you will need to divide both the roots by (a).

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Example 1:

(i) 6X^2 +11X + 3 = 0

(ii) 6X^2 + 10X +4= 0

(i) Shortcut tricks :
This equation +6 is coefficient of x2 .
+ 11 is coefficient of x
+3 is constant term.
Step 1: we multiply (+6) x (+3) = +18 = 2x3x3
Step 2: we break 2x3x3 in two parts such that addition between them is 11.
So, 2x3x3 = 2 x 9. Also,  9 +2 = 11

So , +9 and +2 = Sum of is +11 .

Step 3: Change the sign of both the factors , So +9 = -9 and +2 = -2 .
and divide by coefficient of x2 , So we get -9 / 6 = -3 / 2 and -2 / 6 = – 1 / 3 .

Therefore, X = -3/2 , -1/3

Similarly Solving (ii),

6×4= 24 = 2x2x2x3

Break 2x2x2x3 into two parts such that their sum becomes 10.

2×2 and 2×3 are two parts of 24 whose sum is 10.

Now, Change the sign of both the factors  and divide by coefficient of X^2.

So, +4 = > -4/6

And +6 = > -6/6

Y = –2/3 , -1

So, in all X= -1.5, -0.33

And, Y = -0.667, -1

Now, Putting these values on Number line we get to knw that from -0.667 to -1

The values of X and Y coincides .

Hence, as X and Y coincides at more than 1 point,

We can Say X=Y or Relationship between X and Y cannot be established.

QUESTION

(i)           X2 – 11X + 28 = 0

(ii)         Y2 – 15Y + 56 = 0

GIVEN

In equation (i)

Sum of Root (SR) = 11

Product of Root (PR) = 28

Similarly in eq. (ii)

SR= 15

PR= 56

SOLUTION:

NORMAL METHOD:

(i). X2– 11X + 28 = 0

Now SR = -11 can be written as (-7-4 = -11)

So X2 – 7X – 4X + 28 = 0

Consider the first 2 terms and take the common term outside i.e., X here

X(X – 7) – 4X + 28 = 0

Similarly consider the last 3 terms and take the common term outside i.e., -4 here

X(X – 7) – 4(X – 7) = 0

(X – 7) (X – 4) = 0

ThereforeX = 7, 4

(ii). Y2– 15Y + 56 = 0

Now SR = -15 can be written as (-7-8 = -15)

So Y2 – 7Y – 8Y + 56 = 0

Consider the first 2 terms and take the common term outside i.e., Y here

Y(Y – 7) – 8Y + 56 = 0

Similarly consider the last 3 terms and take the common term outside i.e., -8 here

Y(Y – 7) – 8(Y – 7) = 0

(Y – 7) (Y – 8) = 0

ThereforeY = 7, 8

We have calculated the values of X and Y, now we have to compare the values with each other to deduce the relation between them

TakeX = 7, compare it with both the values of Y = 7, 8

We get, X = 7 is equal to Y = 7 i.e.,X=Y

X = 7 is smaller than Y = 8 i.e.,X<Y

Similarly TakeX = 4, compare it with both the values of Y = 7, 8

We get, X = 4 is smaller than Y = 7 i.e.,X<Y

X = 4 is smaller than Y = 8 i.e.,X<Y

So the relation between X and Y is given by both X = Y and X<Y i.e.,X≤Y

ALTERNATE METHOD:

If the givenSR is–ve then consider it as+ve

If the givenSR is+ve then consider it as–ve

Split the PR into its divisible numbers such that when the numbers are added or subtracted we get the SR

Here 7 × 4 = 28 (PR)

And 7 + 4 = 11 (SR)

Here 7 × 8 = 56 (PR)

And 7 + 8 = 15 (SR)

Therefore from both the equationsX = 7, 4 andY = 7, 8

TakeX = 7, compare it with both the values of Y = 7, 8

We get, X = 7 is equal to Y = 7 i.e.,X=Y

X = 7 is smaller than Y = 8 i.e.,X<Y

Similarly TakeX = 4, compare it with both the values of Y = 7, 8

We get, X = 4 is smaller than Y = 7 i.e.,X<Y

X = 4 is smaller than Y = 8 i.e.,X<Y

So the relation between X and Y is given by both X = Y and X<Y i.e.,X≤Y

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Example 1:P2+13P+40=0

Q2+7Q+12=0

Step1: let us take equ 1. P2+13P+40=0 in this equation coefficient of p, 13 should be split into two numbers in such a way that multiplication of both numbers should be equal to constant term 40 and addition of numbers should be equal to 13

13 can be split into (1,12) (2,11) (3,10) (4,9) (5,8) (6,7)

In these combination 5 and 8 only can give 40 while multiplying, so this is the number we are searching for,and since the coefficient of P2 is 1 and there is no negative sign in the equation, we can directly write value of P by simply changing the sign

P= -5,-8

( just for reference Actual procedure is  P2+5P+8P+40=0

P(P+5)+8(P+5)=0

(P+5)(P+8)=0

P=-5,-8 )

Step 2: Now equ2. Q2+7Q+12=0 similar process applicable for this equation to find Q, here coefficient of Q should be split into two numbers and multiplication of the numbers should give 12

7 can be split up into (1,6) (2,5) (3,4)

Combination of 3 and 4 alone satisfy our need i.e. giving 7 and 12 while adding and multiplying the numbers respectively, since there is no negative sign in the equation,we can directly write value of Q by changing sign .

Q=-3, -4

P= -5,-8

Obviously P < Q

Example 2: 2p2+12p+16=0

2q2+14q+24=0

Step 1: let us take equ1 since coefficient of p2 is 2,we have to multiply 2 with constant number 16 , now as usual 12 should be split up into two numbers and multiplication of the numbers should give ( 2*16=) 32 .

4+8 =12

4*8=32

12 can be split up into 4 and 8,

now change the sign of numbers and divide it by 2 since the coefficient of p2 is 2 ,

Thus the value of  p = -2,-4

Step 2: now take equ2. And follow the same procedure multiply 2 with constant number 24 . So addition of two numbers should be 14 and multiplication of numbers should be 48 (24*2)

6+8 =14  ;  6*8=48

Numbers are 6 and 8, now divide by 2, since the coefficient of qis 2 so the value of q is -3,-4

here, one of the numbers of p and q are same but other number of p is greater than other number of q

so ans is p ≥ q

If you have doubt in finding which one is greater, use this technique.

Number which is 1st from right hand side is greater one.

Example 3 x2-x-6=0

2y2+13y+21

Step1: Here constant number and coefficient of x are negative so the combination of numbers will be positive and negative.