Quadratic Equation Easy Questions and Answers

As per the given above equations , we can say that
Ⅰ.
16x² + 20x + 6 = 0
⇒ 16x² + 12x + 8x + 6 = 0

Ⅱ.
10y² + 38y + 24 = 0
⇒ 10y² + 30y + 8y + 24 = 0
⇒ 10y(y + 3) + 8(y + 3) = 0

Correct Option: B

As per the given above equations , we can say that
Ⅰ. 16x² + 20x + 6 = 0
⇒ 16x² + 12x + 8x + 6 = 0
⇒ 4x(4x + 3) + 2(4x + 3) = 0
⇒ (4x + 2)(4x + 3) = 0

⇒ x = -	3	, -	2
	4		4

Ⅱ.10y² + 38y + 24 = 0
⇒ 10y² + 30y + 8y + 24 = 0
⇒ 10y(y + 3) + 8(y + 3) = 0
⇒ (10y + 8)(y + 3) = 0

⇒ y = - 3, -	4
	5

∴ x > y
From above both equations it is clear that correct answer is x > y .

From the given above equations , we have
From equation Ⅰ.
17x² + 48x − 9 = 0
⇒ 17x² + 51x − 3x − 9 = 0

From equation Ⅱ.
13y² − 32y + 12 = 0
⇒ 13y² − 26y − 6y + 12 = 0

Correct Option: A

From the given above equations , we have
From equation Ⅰ.
17x² + 48x − 9 = 0
⇒ 17x² + 51x − 3x − 9 = 0
⇒ 17x(x + 3) − 3(x + 3) = 0
⇒ (17x − 3)(x + 3) = 0

⇒ x = - 3,	3
	17

From equation Ⅱ.
13y² − 32y + 12 = 0
⇒ 13y² − 26y − 6y + 12 = 0
⇒ 13y(y − 2) − 6(y − 2) = 0
⇒ (13y − 6)(y − 2) = 0

⇒ y = 2,	6
	13

The given quadratic equation is
a²x² - 3abx + 2b² = 0 &⇒ a² x² - 3abx + 2b² = 0
⇒ ax ( ax - 2b ) - b ( ax - 2b ) = 0
⇒ ( ax - 2b ) ( ax - b ) = 0
⇒ ax - 2b = 0 or ax - b = 0

⇒ x =	2b	or x =	b
	a		a

Correct Option: B

The given quadratic equation is
a² x² - 3abx + 2b² = 0 &⇒ a² x² - 3abx + 2b² = 0
⇒ ax ( ax - 2b ) - b ( ax - 2b ) = 0
⇒ ( ax - 2b ) ( ax - b ) = 0
⇒ ax - 2b = 0 or ax - b = 0

⇒ x =	2b	or x =	b
	a		a

Thus, the two roots of the given quadratic equation are

2b	and	b
a	and	a

According to question , we can say that
Since α and β are the roots of the quadratic equation
ax²+ bx + c = 0

∴ α + β = −	b	and αβ =	c
	a		a

Correct Option: B

According to question , we can say that
Since α, β are the roots of the quadratic equation
ax²+ bx + c = 0

∴ α + β = −	b	and αβ =	c
	a		a

Now, α³ + β³ = ( α + β )³ - 3αβ( α + β )

=		-	b		³	- 3 x	c	x	- b
			a				a		a

α³ + β³ =	- b³	+	3bc	=	- b² + 3abc
	a³		a²		a³

Hence , α³ + β³ =	b ( 3ac - b² )
	a³

We have , the given quadratic eq. 3z ² + 5z + k
Comparing it with ax² + bx + c = 0, we get a = 3, b = 5, c = k
Now , D = b² - 4ac = 25 - 12k
For equal linear factors to exist, D ≥ 0
⇒ 25 - 12k ≥ 0
⇒ 25 ≥ 12k

Correct Option: B

⇒ k ≤	25
	12

Therefore, the set of real numbers ≤	25	gives the set of value of k for which the given
	12

quadratic polynomial can be factored into the product of real linear factors.