SUM AND PRODUCT OF THE ROOTS OF A QUADRATIC EQUATION

If a quadratic equation is given in standard form, we can find the sum and product of the roots using coefficient of x2, x and constant term. 

Let us consider the standard form of a quadratic equation,  

ax2 + bx + c = 0

(Here a, b and c are real and rational numbers)

Let α and β be the two zeros of the above quadratic equation. 

Then the formula to get sum and product of the roots of a quadratic equation are,

Sum of roots (a + ) :

= -b/a

(or)

  = -coefficient of x/coefficient of x2

Product of the roots (a) :

= c/a

(or)

= constant term/coefficient of x2

Find the sum and product of the roots for each of the following quadratic equations.

Question 1 :

x2 + 3x - 28 = 0

Solution :

Comparing x2 + 3x - 28 = 0 and ax2 + bx + c = 0, we get

a = 1, b = 3 and c = -28

Sum of the roots = -b/a

= -3/1

= -3

Product of the roots = c/a

= -28/1

= -28

Question 2 :

x2 + 3x = 0

Solution :

Comparing x2 + 3x = 0 and ax2 + bx + c = 0, we get

a = 1, b = 3 and c = 0

Sum of the roots = -b/a

= -3/1

= -3

Product of the roots (aᵦ) = c/a

= 0/1

= 0

Question 3 :

3 + (1/a) = 10/a2

Solution :

10/a- (1/a) - 3 = 0

(10 - a - 3a2)/a2 = 0

-3a2 - a + 10 = 0

In the above quadratic equation, 

coefficient of squared term a = -3,

coefficient of a term = -1

constant term = 10

Sum of the roots = -b/a

= -(-1)/(-3)

= -1/3

Product of the roots = c/a

= 10/(-3)

= -10/3

Question 4 :

3y2 - y - 4 = 0

Solution :

Comparing 3y2 - y - 4 = 0 and ax2 + bx + c = 0,

a = 3, b = -1 and c = -4

Sum of the roots = -b/a

= -(-1)/3

= 1/3

Product of the roots = c/a

= -4/3  

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