Quadratic Equations Shortcuts with Examples

An equation in which the highest power of the variable is 2 is called a quadratic equation.For example, the equation of the type ax2+bx+c=0 denotes a quadratic equation.
There are many ways for solving a quadratic equations  but while in exams we need a quick answer
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.
  1. By taking square roots or Factor Method
  2. By Quadratic Formula

1. Factorization of Quadratic Equation

Quadratic Equations
Condition for Factorization of a quadratic Equation
  1. If b2 - 4ac  > 0, then the quadratic equation can be factorized. 
  2. If b2 - 4ac  < 0, then the quadratic equation cannot be factorized.
some basic formulas which used in while solving through factorization method.
Quadratic Equations

2. Quadratic Formula

formulae
Discriminant of ax2+bx+c = 0 is D = b2 - 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 ax2+bx+c = 0
If α,β are the roots of ax2 + bx + c = 0 then
α = ( - b + √ b2 - 4ac ) /2a
 β = ( - b -√ b2 - 4ac ) /2a .
  1. When D = 0 ,Both the roots will be real an equal ( a = b ) and rational.
  2. When D > 0 but not perfect square then the roots will be irrational, unequal and real .
  3. When D > 0 and perfect square then the roots will be rational, unequal and real .
  4. When D < 0 imaginary roots
Example:
x2 - 4x + 3 = 0  ; a = 1, b= -4, c = 3
D = b- 4ac = (-4)2 -4×1×3 = 16-12 = 4 > 0 and also a perfect square. So, the roots will be distinct rational numbers.
formulae

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- x (sum of the roots)+product of the roots = 0
Now let you have roots 2 and 3 then the quadratic equation will be using formula( x2-x(sum of the roots)+product of the roots = 0 ) is : x2 - (2+3)×x + (3×2) = 0 ; x2-5x+6 = 0 .
Note:
  • A quadratic equation ax2 + bx + c = 0 will have reciprocal roots, if a =c .
  • When a quadratic equation ax2 + 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 ax2+bx = c are negative reciprocals of each other, then c = -a .
  • If they have both the roots common, then a/a1 + b/b1 + c/c1.
  • 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
x2 - sum of the roots × x + products of the roots = 0
x2 - { 3+ (-5)}×x + 3×(-5) = 0
x2 - (-2)x+(-15) = 0 or x2 + 2x - 15 = 0
Verification : D = 4+ 60 = 64
formulae

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) 4p2 - 5p +1 = 0 (ii) q2-2q+1 = 0
Sol :
Quadratic Equations
Answer : (e)
Q2. (i) q2 - 11q + 30 = 0 (ii) 2p2 - 7p + 6=0
Sol:
Quadratic Equations
Answer : (b)
Q3. (i) 6q2 + 1/2 = 7/2 q (ii) 12p2 + 2 = 10p
Sol:

(i) 6q2 + 1/2 = 7/2 q => 12q- 7q + 1 = 0
(ii) 12p2 + 2 = 10p => 6p2 - 5p + 1 = 0


Answer : (e)

Q4. x - 1/x = y, then x+ 1 / x= ?
Sol:
Quadratic Equations questions
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
x2+1+2x = x2 + x +1 - 2x
-x2 + 4x = 0
x = 4
Q6. Two consecutive positive even integers whose squares have the sum 164 are ?
Sol:
(x)2 + (x+2)2 = 164
x2+x2+4+4x = 164
2x2+4x+4-164 = 0
2(x2+2x-80) = 0
x2+2x-80 = 0
(x2+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 - 3y2
3y2 - 45y + 150 = 0
3(y2 - 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 = x2
Area = x3
x3 = 64, x = 4 , B = 4, L = 16
Q9. If x+ 1 / x= 102 , then the value of x - 1/x =?
Sol :
Quadratic Equations questions
Q10. The sum of a number and its reciprocal is 17/4, then the number is ?
Sol:
Quadratic Equations questions
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 = x2 + ( x+ 1 )2
4x+ 1 - 4x = x2 + x+ 1 + 2x
4x+ 1 - 4x = 2x+ 2x +1
2x2 - 6x = 0
x = 3,
H = 2(3) - 1 = 5
P = 3 + 1 = 4
Q12. The solution of √25-x= x-1 are
Sol:
Quadratic Equations questions
Q13. If x - ax + 20 is exactly divisible by x-4 then a = ?
Sol :
x2- ax + 20
x- 4 ; x =4
(4)2 - a (4) + 20 = 0
16 - 4a + 20 = 0
-4a + 36 = 0
a = 36/ 4 = 9.

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