KPK Board 10th Class Math Ch 2 Theory of Quadratic Equation Short Questions Answers
KPK Board 10th Class Math Ch 2 Theory of Quadratic Equation Short Questions Answers
We are providing all Students from 5th class to master level all exams preparation in free of cost. Now we have another milestone achieved, providing all School level students to the point and exam oriented preparation question answers for all science and arts students.
After Online tests for all subjects now it’s time to prepare the next level for KPK board students to prepare their short question section here. We have a complete collection of all classes subject wise and chapter wise thousands questions with the right answer for easy understanding.
We provided 10th class Mathematics short questions answers on this page of all KPK boards. Students of KPK boards can prepare theMathematics subject for annual examinations.
In this List we have included all KPK boards and both Arts and Science students. These Boards students can prepare their exam easily with these short question answer section
Malakand Board 12th classes short questions Answer
Mardan Board 12th classes short questions Answer
Peshawar Board 12th classes short questions Answer
Swat Board 12th classes short questions Answer
Dera Ismail Khan Board 12th classes short questions Answer
Kohat Board 12th classes short questions Answer
Abbottabad Board 12th classes short questions Answer
Bannu Board 12th classes short questions Answer
All above mention KPK Boards students prepare their annual and classes test from these online test and Short question answer series. In coming days we have many other plans to provide all kinds of other preparation on our Gotest website.
How to Prepare KPK Board Classes Short Question Answer at Gotest
- Just Open the desired Class and subject which you want to prepare.
- You have Green bars which are Questions of that subject Chapter. Just click on Bar, it slides down and you can get the right answer to those questions.
3x2- 5x+k=0 are equal.
Here a = 3 , b = -5 , c = k
Since roots are equal then discriminate is equal to zero. i.e.
B2– 4ac=0 ………..(ii)put values in (ii)
=> (-5)2 – 4 (3)(k) = 0 => 25 = 12k
=> k = 25/12
Evaluate ( -1+ √(-3) )7+ ( -1- √(-3) )7.
Solution: ( -1+ √(-3) )7+ ( -1- √(-3) )7
As (-1+i√3)/2 = w and (-1+i√3)/2 =w2
=> -1+i √3 = 2w and -1-i√3 = 2w2
Then
( -1+ √(-3) )7+ ( -1- √(-3) )7 = (2w)7 + (2w2 )7
=> 27 w7+27 w14
=> 27 (w2.w+w12.w2)
=> 27 [(w3 )2.w+(w3 )4.w2]
As w3 = 1
=> 2^7 ((1)3.w+(14.w2)
=> 27 (w+w2)
=> 128 (-1) AS w+ w2 = -1
=> -128
(i) 4x2- 1=0 (ii) 3x2+ 4x =0
Solution: (i) 4x2+ 0.x – 1 =0
Here a = 4 , b = 0 , c = -1
As sum of the roots S = – b/a……………(i)
Put values (i).
S = – 0/4 => S = 0
Now product of the roots = p = S = c/a …………….. (ii) put values in ……(i)
=> P = – 1/4
(ii) 3x2+ 4x+0 = 0
Here a = 3 , b = 4 , c = 0
As sum of the roots S = – b/a …………(i)
Put values in (i).
S = – 4/3
Product of the roots = S = – c/a put values
=> P = – 0/3 => P = 0
Solution: 3x2+ (2K+1 )x+k-5 = 0……………(i)
Here a = 3 , b = 2k + 1 , c k – 5
As sum of the roots S = – b/a …………..(i)
Put values in (i).
=> S = – (2k+1)/3 …………………. (ii)
As product of the roots = P = – c/a ………………… (iii) put values in (iii)
P = (k-5)/3…………….. (iv)
AS Sum of the roots = product of the roots
Then from (ii) and (iv)
– (2k+1)/3 = (k – 5)/3 => -(2k + 1 ) = k – 5
=> – 2k – 2 = k – 5 => – 2 + 5 = k + 2k
=> 3k = 3 => K = 3/3 = 1
Solution: x2– 3x+k+1=0……………(i)
Let a and β are the roots of equation ……(i)
α-β = 1 ……………….. (i)
Here a = 1 , b = -3 , c = k + 1
Sum of the roots α+β= -b/a
α+β-(-3)/1 => α+β=3 ……………………….(ii)
Product of the roots αβ= c/a
αβ= (k+1)/1 => αβ= k+1…………….(iii)
Adding (i) and (ii)
α-β= 1
= (α+β=3)/(2a=4)
=> α=4/2 => α = 2
Put α=2 in (ii)
2 + β = 3 => β =3-2
=> β = 1
Now put α=2 and β = 1 in ( iii)
(2)(1) = k + 1 => 2 – 1 = k
=> k = 1
Thus k = 1
