Question

(a) 12-31-4=0 (b) 2x2-7x+3=0
(c) x2-3x=10
(a) (2,3) and (6,6)
(c) (-8, 15)
(b) (-6, 8)
(a) (3, 4)
Points (2, 1) and (7, 8) in the ratio 3:1 (3 Marks)
Find the area of triangle when the vertices are given (3 Marks)
ACTIVITY-02 PASSING PACKAGE-02
7) Find the Mean, Median and Mode of the following data (3 Marks)
C.I 15-25
25-35
35-45 45-55
F
6
11
55-68
5.
1) Using Quadratic formula find the roots of the equations (2 Mars
=) Find the distance between the following points (2 Marks)
(b) (2,5) and (-3,-7) (c) (3.4) and (0,0)
=) Find the distance between Origin and the given point (2 Marks)
4) Find the Co-ordinates of the point which divides the line segment joining the
Find the ratio in which the line segment joining the points (-3,10) and (6.-) is divided by
(a) (1, 1), (2, 3) and (4, 5) (b) (1,-1), (-4, 6) and (-3,-5) (C) (2, 3), (-1,0) and (2.4)​

Answer 1

Answer:

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Answer 2

Answer:

The roots of quadratic equation are the values of the variable that satisfy the equation. They are also known as the "solutions" or "zeros" of the quadratic equation. For example, the roots of the quadratic equation x2 - 7x + 10 = 0 are x = 2 and x = 5 because they satisfy the equation. i.e.,

when x = 2, 22 - 7(2) + 10 = 4 - 14 + 10 = 0.

when x = 5, 52 - 7(5) + 10 = 25 - 35 + 10 = 0.

But how to find the roots of a general quadratic equation ax2 + bx + c = 0? Let us try to solve it for x by completing the square.

ax2 + bx = - c

Dividing both sides by 'a',

x2 + (b/a) x = - c/a

Here, the coefficient of x is b/a. Half of it is b/(2a). Its square is b2/4a2. Adding b2/4a2 on both sides,

x2 + (b/a) x + b2/4a2 = (b2/4a2) - (c/a)

[ x + (b/2a) ]2 = (b2 - 4ac) / 4a2 (using (a + b)² formula)

Taking square root on both sides,

x + (b/2a) = ±√ (b² - 4ac) / 4a²

x + (b/2a) = ±√ (b² - 4ac) / 2a

Subtracting b/2a from both sides,

x = (-b/2a) ±√ (b² - 4ac) / 2a (or)

x = (-b ± √ (b² - 4ac) )/2a

This is known as the quadratic formula and it can be used to find any type of roots of a quadratic equation.