An equation in which the highest power of the variable is 2 is called a quadratic equation.For example, the equation of the type axThere are many ways for solving a quadratic equations but while in exams we need a quick answer^{2}+bx+c=0 denotes a quadratic equation.

so there is a shortcut method which used for solving a quadratic equations in less time but before let me tell you the most basic and accurate method which is used for it.

### Solving Quadratic Equations

When we solved a quadratic expression or equation, it will always gives two values of variable. these values are called

**roots of the equation or expression**.- By taking square roots or Factor Method
- By Quadratic Formula

#### 1. Factorization of Quadratic Equation

**Condition for Factorization of a quadratic Equation**

- If b
^{2}- 4ac > 0, then the quadratic equation can be factorized. - If b
^{2}- 4ac < 0, then the quadratic equation cannot be factorized.

#### 2. Quadratic Formula

Discriminant of ax

^{2}+bx+c = 0 is D = b^{2}- 4ac and the two values of x obtained from a quadratic equation are called roots of the equation which denoted by α and β sign.
For a quadratic equations ax

If α,β are the roots of ax^{2}+bx+c = 0^{2}+ bx + c = 0 then

α = ( - b + √ b

^{2}- 4ac ) /2a
β = ( - b -√ b

^{2}- 4ac ) /2a .- When D = 0 ,Both the roots will be real an equal ( a = b ) and rational.
- When D > 0 but not perfect square then the roots will be irrational, unequal and real .
- When D > 0 and perfect square then the roots will be rational, unequal and real .
- When D < 0 imaginary roots

**Example:**

x

D = b^{2}- 4x + 3 = 0 ; a = 1, b= -4, c = 3^{2 }- 4ac = (-4)2 -4×1×3 = 16-12 = 4 > 0 and also a perfect square. So, the roots will be distinct rational numbers.

### Formulating a Quadratic Equation

when you have the roots and you need to make a quadratic equation from that roots then formula that is used = x

Now let you have roots 2 and 3 then the quadratic equation will be using formula( x

^{2 }- x (sum of the roots)+product of the roots = 0Now let you have roots 2 and 3 then the quadratic equation will be using formula( x

^{2}-x(sum of the roots)+product of the roots = 0 ) is : x^{2}- (2+3)×x + (3×2) = 0 ; x^{2}-5x+6 = 0 .**Note:**

- A quadratic equation ax
^{2}+ bx + c = 0 will have reciprocal roots, if a =c . - When a quadratic equation ax
^{2}+ bx + c = 0 has one root equal to zero, then c = 0. - When both the roots are equal to zero, b = 0 and c =0.
- When the roots of the quadratic equation ax
^{2}+bx = c are negative reciprocals of each other, then c = -a . - If they have both the roots common, then a/a
_{1}+ b/b_{1}+ c/c_{1}. - The square root of any negative number will be an imaginary number like √-25 = √(-1)× 25 = √-1×√ 25 = 5i

**Example :**

Forming a quadratic equation whose roots are 3 and 5 and verifying them.

the equation will be

x

^{2}- sum of the roots × x + products of the roots = 0

x

^{2}- { 3+ (-5)}×x + 3×(-5) = 0

x

^{2}- (-2)x+(-15) = 0 or x

^{2}+ 2x - 15 = 0

Verification : D = 4+ 60 = 64

### Solving Quadratic Equation using shortcut techniques

#### Direction (1-3) : In each question, one or more equations are provided. On the basis of these, you have to find out relation between p and q.

Give answer (a) if p = q , Give answer (b) if p > q , Give answer (c) if q > p , Give answer (d) if p ≥ q and give answer (e) if q ≥ p .

**Q1. (i) 4p**^{2}- 5p +1 = 0 (ii) q^{2}-2q+1 = 0
Sol :

Answer : (e)

Q2. (i) q

Sol: ^{2}- 11q + 30 = 0 (ii) 2p^{2}- 7p + 6=0Answer : (b)

**Q3. (i) 6q**

Sol:

^{2}+ 1/2 = 7/2 q (ii) 12p^{2}+ 2 = 10pSol:

(i) 6q

^{2}+ 1/2 =

^{7}/

_{2}q => 12q

^{2 }- 7q + 1 = 0

**Q4. x - 1/x = y, then x**

^{2 }+ 1 / x^{2 }= ?**Sol:**

**Q5. If (x-1) and (x+1) be the sides (in cm) of a right triangle, then the value of x is ?**

**Sol:**

(H)

^{2}= (P)

^{2}+ B)

^{2}

(x+1)

^{2}= (x)

^{2}+ (x-1)

^{2}

x

^{2}+1+2x =

^{ }x

^{2 }+ x +1 - 2x

-x

^{2}+ 4x = 0

x = 4

**Q6. Two consecutive positive even integers whose squares have the sum 164 are ?**

**Sol:**

(x)

^{2}+ (x+2)

^{2}= 164

x

^{2}+x

^{2}+4+4x = 164

2x

^{2}+4x+4-164 = 0

2(x

^{2}+2x-80) = 0

x

^{2}+2x-80 = 0

(x

^{2}+10-8x-80 = 0

(x-8) (x+10) = 0

x = 8 , x = -10

hence numbers 8 & 10.

**Q7. The sum of two numbers is 15 and the sum of their reciprocals is 3/10 then the numbers are ?**

**Sol:**

x+y = 15

1/x + 1/y = 3/10

x = 15 - y

1/15-y + 1/y = 3/10

150 = 45y - 3y

3y

3(y

(y-10) (y-5) = y = 10 , y = 5.

Let Breadth = x, Length = x

Area = x

x

-4a + 36 = 0

a = 36/ 4 = 9.

^{2}3y

^{2}- 45y + 150 = 03(y

^{2}- 15y +50 ) = 0(y-10) (y-5) = y = 10 , y = 5.

**Q8. If the length of a rectangle is square of the breadth and area is 64 find the length ?****Sol:**Let Breadth = x, Length = x

^{2}Area = x

^{3}x

^{3}= 64, x = 4 , B = 4, L = 16**Q9. If x**^{2 }+ 1 / x^{2 }= 102 , then the value of x - 1/x =?**Sol :****Q10. The sum of a number and its reciprocal is 17/4, then the number is ?****Sol:****Q11. The hypotenus of a right angled triangle is 1 m less than 2 times the smallest side. If the third side is 1 m more than the smallest then sides of the triangle are ?**

**Sol :**

B = x

H = 2x -1

P = x + 1

( 2x - 1 )

^{2}= x^{2}+ ( x+ 1 )^{2}
4x

^{2 }+ 1 - 4x = x^{2 + }x^{2 }+ 1 + 2x
4x

^{2 }+ 1 - 4x = 2x^{2 }+ 2x +1
2x

^{2}- 6x = 0
x = 3,

H = 2(3) - 1 = 5

P = 3 + 1 = 4

**Q12. The solution of √25-x**

^{2 }= x-1 are**Sol:****Q13. If x - ax + 20 is exactly divisible by x-4 then a = ?**

**Sol :**

x

x- 4 ; x =4^{2}- ax + 20
(4)

16 - 4a + 20 = 0^{2}- a (4) + 20 = 0-4a + 36 = 0

a = 36/ 4 = 9.