## Quadratic equations

Shortcut Tricks are very important things in competitive

And, before starting anything just do a math practice set. First of all, write down twenty math problems related to this topic on a page. And, do first ten maths using basic formula of this math topic. You also need to keep track of the time. After finish write down total time taken by you to solve those ten maths. Now, read our examples on quadratic equations shortcut tricks and practice few questions. And, after this do remaining ten questions and apply shortcut formula on those math problems. Again keep track of Timing. Now, you will surely see the improvement in your timing this time. But, this is not all you need. You need more practice to improve your timing more.

### Few Important things to Remember

Math section in a competitive exam is the most important part of the exam. And, it doesn’t mean that other topics are not so important. You can get a good score only if you get a good score in math section. And, you can get good score only by practicing more and more. So, all you need to do is to do math problems correctly within time, and this can be achieved only by using shortcut tricks.

Again it does not mean that you can’t do maths without using shortcut tricks. You may have that potential to do maths within time without using any shortcut tricks. But other peoples may not do the same. So Quadratic equations shortcut tricks here for those people. We try our level best to put together all types of shortcut methods here. But if you see any tricks are missing from the list then please inform us. Your little help will help others.

### Quadratic Equation

So, in a mathematical calculation, a quadratic equations is came from the Latin word that is quadrature’s which is called square is a structure. In maths exam papers there are two or three question are given from this chapter.

And, this type of problem are given in Quantitative Aptitude which is a very essential paper in banking exam. So, under below given some more example for your better practice.

### Example:

**ax ^{2} + bx + c = 0**

It is equal to 0 and the a, b, c are the constant value and we can say that x represent as unknown.

The a, b, c are the constant and quadratic coefficient or linear coefficient. Quadratic equation hold the only power of x which is also non negative integer.

### Quadratic Equations: Example #1

6x^{2} +11x + 3 = 0

- 3/2 and 1/3
- 2/3 and 3/2
- -3/2 and -1/3
- -2/3 and -3/2

Show Answer Show How to Solve Open Rough Workspace

**Answer:**Option (C)

**How to Solve**

In this equation +6 is coefficient of x

^{2}.

+ 11 is coefficient of x.

+3 is constant term.

Now, we multiply (+6) x (+3) = +18

Then, we break +18 in two parts such that addition between them is 11.

+18 = 9 + 2 = 11, and product of both factors is 18.

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

Change the sign of both the factors , So +9 = -9 and +2 = -2.

Now, divide by coefficient of x^{2},

So, we get,

-9 / 6 = -3 / 2, and

-2 / 6 = -1 / 3.

**Rough Workspace**

### Quadratic Equations: Example #2

4y^{2} + 12y + 8 = 0

- -2 and -1
- -2 and -3
- -4 and -2
- -1 and -3

Show Answer Show How to Solve Open Rough Workspace

**Answer:**Option (A)

**How to Solve**

4 x 8 = +32

We break +32 in two parts such that addition between them is 12.

+32 = (+8) + (+ 4) = +12.

Change sign of both factor and divide by coefficient of y^{2},

So, we get,

– 8 / 4 = -2.

– 4 / 4 = -1.

**Rough Workspace**

### Quadratic Equations: Example #3

x^{2} + 10x + 4 = 0

- 2 and 3/2
- 1 and 2/3
- -2 and -3/2
- -1 and -2/3

Show Answer Show How to Solve Open Rough Workspace

**Answer:**Option (D)

**How to Solve**

6 x 4 = +24

We break + 24 in two parts such that addition between them is 10.

+24 = (+6) + (+ 4) = +10.

Change sign of both factor and divide by coefficient of x^{2},

So, we get,

– 6 / 6 = -1.

– 4 / 6 = -2 / 3.

**Rough Workspace**

### Quadratic Equations: Example #4

x^{2} + 9x + 20 = 0

- -3 and -4
- -5 and -4
- -4 and -2
- -2 and -5

Show Answer Show How to Solve Open Rough Workspace

**Answer:**Option (B)

**How to Solve**

5 x 4 = +20

We break + 20 in two parts such that addition between them is 9.

+20 = (+5) + (+ 4) = +9.

Change sign of both factor and divide by coefficient of x^{2},

So, we get,

– 5 / 1 = -5.

– 4 / 1 = -4.

**Rough Workspace**

### Quadratic Equations: Example #5

3y^{2} + 19y + 28 = 0

- -3 and -5/7
- -4 and -5/3
- -4 and -7/3
- -6 and -3/7

Show Answer Show How to Solve Open Rough Workspace

**Answer:**Option (C)

**How to Solve**

3 x 28 = +84

We break + 84 in two parts such that difference between them is 19.

+84 = (+12) + (+ 7) = +19 .

Change sign of both factor and divide by coefficient of y^{2}, that is 3.

So, we get,

y = -12 / 3 = -4.

y = -7 / 3 = – 7 / 3.

**Rough Workspace**

### Few examples of Inequality with Shortcut Tricks

- Single variable Quadratic equation
- More than one Quadratic equation
- Linear equations
- One variable linear equations
- More than one variable linear equations
- << Go back to Inequality Methods main page

So, here we provide few shortcut tricks on this topic. Please visit this page to get updates on more Math Shortcut Tricks. You can also like our facebook page to get updates.

And, if you have any question regarding this topic then please do comment on below section. You can also send us message on facebook.

good techniques are given to to solve quadratic equations but found many mistakes… pls make it correct!!!

thank u prashant for your feedback pls mention the page where u dont understand and found mistakes ….

Very useful…like it..thanks.

i did not find any mistakes in it but the problems are of elementary sort. in my country even children can solve them easily. the problems arise when they start putting up difficult questions. so it would be very kind of you if you could complete the lecture by inclusion of mind binding problems. i think by applying the differentiation method on these problems we get rid of such difficulties

I found it correct

Please explain

Example 4.probably final values are x=-5 and x=-4 ….

sir pls give some SBI po and RBI quadratic equation which are tough to solve and how to compare them

both are tough but not impossible to solve, do practice more on basics…

very helpfull…thnx alot sir

very useful …thank you

ok i have a question :

1.x^2-x-2=0 and 2.y^2+5y+6=0

now after solving i get x=+4,-3 and y=-3,-2 so there is 1 common value -3 and other value of x is greater thn y so answer should be x>=y

but answer is relation cant be established how cn it be possible is any concept is here??

Sir we need to compare both values of x with both values of y.

I mean to say +4 is greater than both values of y, but -3 is not greater than -2 ,so ans is relation can’t be established. I hope u understood.

x^2-x-2=0. y^2+5y+6=0

x^2-2x+x-2=0. y^2+6y-y+6=0

x(x-2)+1(x-2)=0. y(y+6)-1(y+6)=0

(x+1)(x-2)=0. (y-1)(y+6)=0

x=-1,2 y=1,-6.

x>y

if y ‘s value is within the value of x then no relation

first equation roots are 2,-1 not 4, 3 and 2 nd eq roots -3,-2 so there is no relation

x=-1,2 , y= -3,-2

by comparing can’t we say y<x ?

xsquare is equal to 25 means then the answer is +or-5….the SAME WAY xpower4 is equal to 625 means wat is the x value? +or-5 ya or +5 only ya????please reply me…

and what about negTive values

Nice trick

Thanks for sharing these short cut technique but it will be good if you sharing some more question where equations are in negative like x2-x-2=0

so that we won’t get confuse in sign.

NY question is x²+5x-6.

according to the above said method, a×c = 1×6=6

6 can be broken in two ways to be equal to mid term 5 as :

6×1 = 6 and 6-1= 5………………(way 1).

and also

2×3 = 6 and 2+3= 5………………(way 2).

Assuming that I follow this short trick and not solving this equation by the traditional long cut method, as there will be no time to do so & check every equation,

if I go way 1 i get x = 1, -6.

while,

if I go way 2 i get x = -3, -2.

then how do I know which way should I go.. way 1 or way 2?… as both appear to be right in the short cut method!!!

You have to make the product of -6 which is not there in your second values i.e 2×3. So only values 6 X -1= -6 & 6 -1 =5 is valid.

x^2+5x-6

=x^2+6x-x-6

=x(x+6)-1(x+6)

=(x-1)(x+6)

x=1 and x=-6

shortcut

-6/1=-6 and 1/1=1

and remember one thing sum of 2x and 3x can be 5x bt product cant be -6

thnk u vry mch sir.. this is vry helpful

excellent tricks,this lots helps to bank aspirents

Sir can you please tell me how to compare values of x and y. Especially the cases where relation between both variable can not be established.

Bestest trick I found on net.thanxxxxxxxx

thanks from bottom of my heart

welcome, and keep visiting.

11x^2-240x-44=0

plz solve this with explanation

11x^2-240x-44=0 plz solve

Is there any other way of solving these questions specially when the coefficients are very large. With this method the calculation for such questions will become so hectic. So are there any other way to deal with that kind of problems?

How to use shortcut for big values like 9*20 cums 180 . how to divide it for 27.

Sir/Madam

Plz help me in this type of question

5x^2-87x+378=0

3y^2-49y+200=0

X>y

X=y

x<=y

No any relation being established.

Plz help me..

Useful

I’m confuse here.

we break + 24 in two parts such that addition between them is 10.

+24 = (+6) + (+ 4) = +10 .

I can’t understand it. It’s how..¿¿¿

It’s just a regular technique… Don’t again say that this is a shortcut, it’s really waste of time

Superb technique in simple way , loved it simply awesome sir .

Interesting sir we highly appreciate

Buh sir am having problem with with quatic equations lke

6x^4-35x^3+6x^2-35x+6

Sir please if there’s any tricky way

Sir cubic equation and quartic

thank you for the shortcuts i really appreciate

but I have a question to ask

what if the product of the co-efficient of x^2 and the constant is much pls explain to us hw we can easily get d 2 numbers dat can b multiplied to give the product and can be added to give the co-efficient of x

Not able to understand example 3