Quadratic Equation Easy Questions and Answers

On comparing it with ax² + bx + c = 0 , we get a = 1, b = p, c = q
The roots of the equation x² + px + q = 0 are equal if
b² - 4ac = 0
⇒ p² - 4q = 0 ⇒ p² = 4q.

Correct Option: B

On comparing it with ax² + bx + c = 0 , we get a = 1, b = p, c = q
The roots of the equation x² + px + q = 0 are equal if
b² - 4ac = 0
⇒ p² - 4q = 0 ⇒ p² = 4q.
Thus , required answer is option B .

According to question ,
Let, the two consecutive odd positive integers be 2x + 1 and 2x + 3 where x is a whole number.
Now, ( 2x + 1 )² + ( 2x + 3 )² = 290

Correct Option: A

According to question ,
Let, the two consecutive odd positive integers be 2x + 1 and 2x + 3 where x is a whole number.
Now, ( 2x + 1 )² + ( 2x + 3 )² = 290
⇒ 4x² + 4x + 1 + 4x² + 12x + 9 = 290
⇒ 8x² + 16x - 280 = 0
⇒ x² + 2x - 35 = 0
⇒ ( x + 7 ) ( x - 5 ) = 0
⇒ x = 7, - 5
But, x = −7 is not possible, since −7 is not a whole number .
∴ x = 5.
We get , 2x + 1 = 2 x 5 + 1 = 11 and 2x + 3 = 2 x 5 + 3 = 13
Thus , two consecutive positive odd integers are 11 and 13 .

As we know that ,
∴ α, β are the roots of the equation 2x² - 3x + 1 = 0

∴ α + β =	3	........... ( 1 )
	2

and αβ =	1	........... ( 2 )
	2

We are to form a quadratic equation whose roots are

α	and	β
β	and	α

Correct Option: C

As we know that ,
∴ α, β are the roots of the equation 2x² - 3x + 1 = 0

∴ α + β =	3	........... ( 1 )
	2

and αβ =	1	........... ( 2 )
	2

We are to form a quadratic equation whose roots are

α	and	β
β	and	α

S = sum of the roots =	α	+	β	=	α² + β²	=	( α + β )² - 2αβ
	β		α		αβ		αβ

S =	3	²	- 2	1
	2			2

		1
		2

∴ Using ( 1 ) and ( 2 ) , we get

S =	9	- 1
	4

	1
	2

S =	5	x	2	=	5
	4		1		2

P = Product of the roots =	α	×	β	= 1
	β		α

Hence the required quadratic equation is x² − (sum of the roots)x + (Product of the roots) = 0

⇒ x² -	5	x + 1 = 0
	2

⇒ 2x² - 5x + 2 = 0.