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

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.

2. Quadratic Formula

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 = b² - 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  = x² - 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

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 :

Answer : (e)

Q2. (i) q² - 11q + 30 = 0 (ii) 2p² - 7p + 6=0

Sol:

Answer : (b)
Q3. (i) 6q² + 1/2 = 7/2 q (ii) 12p² + 2 = 10p
Sol:
(i) 6q² + 1/2 = ⁷/₂ q => 12q² - 7q + 1 = 0

(ii) 12p² + 2 = 10p => 6p² - 5p + 1 = 0

Answer : (e)

Q4. x - 1/x = y, then x² + 1 / x² = ?
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
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 x² + 1 / x² = 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 = x² + ( x+ 1 )²

4x² + 1 - 4x = x^2 + x² + 1 + 2x

4x² + 1 - 4x = 2x² + 2x +1

2x² - 6x = 0

x = 3,

H = 2(3) - 1 = 5

P = 3 + 1 = 4

Q12. The solution of √25-x² = x-1 are

Sol:

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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