KPK Board 10th Class Math Ch 1 Quadratic Equation Short Questions Answers
| Class: | 10th Class | Subject: | Math |
| Chapter: | All | Board: | KPK Boards |
KPK Board 10th Class Math Ch 1 Quadratic Equation Short Questions Answers
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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 the Mathematics 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 10th classes short questions Answer
Mardan Board 10th classes short questions Answer
Peshawar Board 10th classes short questions Answer
Swat Board 10th classes short questions Answer
Dera Ismail Khan Board 10th classes short questions Answer
Kohat Board 10th classes short questions Answer
Abbottabad Board 10th classes short questions Answer
Bannu Board 10th classes short questions Answer
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Solve 2x<sup>4</sup> – 5w<sup>2</sup> + 2 = 0
Solution: 2x4 – 5w2 + 2 = 0……………..(i)
Let w2= x ……….(ii) Then w4 w2…………. (iii)
Put (ii) , (iii) , in (i)
2x2 – 5w + 2 = 0
=> 2x2 – 4x – x + 2 = 0
=> 2x ( x – 2 ) – 1 ( x-2) = 0
=> ( x-2 ) (2x – 1) = 0
=> x- 2 = 0 or 2x – 1 = 0
=> x= 2 or 2x = 1
=> x = 1/2
Put x = 2 in (ii) => w2 = 2 take square root
=> √w2 = +√2 => w = +√2
Put x = ½ in (ii) => w2 = ½ take square root
=> √w2 = +√½ => w = +√½ = + 1/√2
Solution set = {+√2 = + 1/√2}
Find the constant a and b such that x = -1 and x = 1 are both solutions to the equation</p> <p>ax<sup>2 </sup>+ bx + 2 = 0
Solution ax2 + bx + 2 = 0 ……..….(i)
Put x = -1 in (i) => a(-1)2+ b ( -1 ) + 2 = 0
=> a – b + 2 = 0 ………………(iii)
Put x = 1 in (i) => a(1)2+ b (1)+ 2=0
=> a + b + 2 = 0 (iv)
Adding (iii) and (iv)
a – b + 2 = 0
(a+b+2=0)/(2a+4=0)
=> 2a = – 4 => a = – 4/2
=> 2a = -4 => a = -4/2
=> a = -2 put a = -2 in (iv)
=> – 2 + b + 2 = 0 => b = 0
Thus a = – 2 , b = 0 Ans.
Find all values of x such that x<sup>2</sup>+ 5x + 6 = 0 and x<sup>2</sup>+ 19x + 34 = 0 are equal.
Solution: Since x2+ 5x+6=0 and x2+ 19x+34 =0 are equal.
Then x2+ 5x+6 = x2+ 19x+34
=> x2+x2+ 5x-19=34- 6=0
=> – 14 x = 28 => x= -28/14
=> x = -2 Ans.
Challenge: Find the solution to the equation.</p> <p>49x<sup>^2</sup>- 316+123=0
Solution: 49x2– 316+123=0
=> 49x2– 294x-22x+123=0
=> 49x (x-6)-22(x – 6 )=0
=> (x-6) (49x -22)=0
=> (x-6) =0 or 49x -22=0
=> x = 6 or 49x = 22
=> x= 22/49
=> Thus solution set = { 6 , 22/49 }
How to Write Perfect Short Answers?
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