In quadratic Equations We have the following options to choose from :-
i. X >Y
ii. Y >X
iii. X >= Y
iv. Y>= X
v. X = Y or relationship cannot be established.
(i) Now, Suppose on solving the quadratic equation we get X = -1, 2 And Y = -2, -3
On putting these values on number line we will see that X lies on the right of Y. Hence, We can say that X>Y or Y<X.
(ii) Now, Suppose if after solving the equation we get X = -1 , 2 and Y= 3,4.
On putting these values on number line we see that Y lies to the right of X without touching X at any point. Hence, We can say that Y>X or X<Y.
(iii) If X= 3,4 and Y= 2,3. Putting this on number line we get X to the right of Y but touching Y at one single point i.e., 3. Hence, X>=Y or Y<=X.
(iv) If X=3,4 and Y=4,5. Putting these values on number line we will see that Y lies to the right of X but touching X at 4. Hence, Y>=X or X<=Y.
(v) Now, Suppose X= 2,4 and Y= 3,5. Putting these values on number line we will see that X touches Y at all the points between 3 and 4 i.e., at 3, 3.1, 3.2, 3.3 etc. upto 4 hence, in this case we cannot relate X and Y and Hence, Will answer as Relationship cannot be established.
Also suppose if y=7 and X= 4,8 then X<y also X>y . Hence, in this case also no relationship can be established.
Example some quadratic equation:
Example 1:
Example 1:
(i) 6X^2 +11X + 3 = 0
(ii) 6X^2 + 10X +4= 0
Answer :
(i) Shortcut tricks :
This equation +6 is coefficient of x2 .
+ 11 is coefficient of x
+3 is constant term.
Step 1: we multiply (+6) x (+3) = +18 = 2x3x3
Step 2: we break 2x3x3 in two parts such that addition between them is 11.
So, 2x3x3 = 2 x 9. Also, 9 +2 = 11
(i) Shortcut tricks :
This equation +6 is coefficient of x2 .
+ 11 is coefficient of x
+3 is constant term.
Step 1: we multiply (+6) x (+3) = +18 = 2x3x3
Step 2: we break 2x3x3 in two parts such that addition between them is 11.
So, 2x3x3 = 2 x 9. Also, 9 +2 = 11
So , +9 and +2 = Sum of is +11 .
Step 3: Change the sign of both the factors , So +9 = -9 and +2 = -2 .
and divide by coefficient of x2 , So we get -9 / 6 = -3 / 2 and -2 / 6 = – 1 / 3 .
and divide by coefficient of x2 , So we get -9 / 6 = -3 / 2 and -2 / 6 = – 1 / 3 .
Therefore, X = -3/2 , -1/3
Similarly Solving (ii),
6x4= 24 = 2x2x2x3
Break 2x2x2x3 into two parts such that their sum becomes 10.
2x2 and 2x3 are two parts of 24 whose sum is 10.
Now, Change the sign of both the factors and divide by coefficient of X^2.
So, +4 = > -4/6
And +6 = > -6/6
Y = --2/3 , -1
So, in all X= -1.5, -0.33
And, Y = -0.667, -1
Now, Putting these values on Number line we get to knw that from -0.667 to -1
The values of X and Y coincides .
Hence, as X and Y coincides at more than 1 point,
We can Say X=Y or Relationship between X and Y cannot be established.
EXERCISE
Give answer
(i). If x>y
(ii). If x>=y
(iii). If x<y
(iv). If x<=y
(v). x=y or relationship cannot be established
1) (a) 3X^2+8X+4 = 0 and (b) 4Y^2-19Y+12= 0.
2) (a) X^2 + X-20= 0 and (b) Y^2-Y-30= 0.
3) (a) X^2- 365 = 364 and (b) y- (324) ^(1/2) = (81)^(1/2).
4) (a) 225X^2-4 = 0 and (b) (225y)^(1/2) +2 = 0
5) (a) x^2 = 729 and (b) Y= (729)^(1/2)
6) (a) 2x^2 + 11x + 14 = 0 and (b) 4y^2 + 12y +9 =0
7) (a) x^2-7x+12=0 and (b) y^2+y-12=0
8) (a) x^4- 227= 398 and (b) y^2 + 321=346
9) (a) x^2-1=0 and (b) y^2+4y+3=0
10)(a) x^2-7x+12=0 and (b) y^2-12y+32=0
Try this exercise in the minimum time answer will be provided soon.
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