RS Aggarwal Test: Quadratic Equations - Class 10 MCQ

2x² + 3 + 2√6x + x² = 3x² - 5x
3 + 2√6x + 5x= 0

Which is not a quadratic equation.

(a) The given equation is x² - 4x + 3√2 = 0.
On comparing with ax² + bx + c = 0, we get
a = 1, b = -4 and c = 3√2
The discriminant of x² - 4x + 3√2 = 0 is
D = b² - 4ac
= (-4)² - 4(1)(3√2) = 16 - 12√2 = 16 - 12 x (1.41)
= 16 - 16.92 = -0.92
⇒ b² - 4ac < 0
(b) The given equation is x² + 4x - 3√2 = 0
On comparing the equation with ax² + bx + c = 0, we get
a = 1, b = 4 and c = -3√2
Then, D = b² - 4ac = (-4)² - 4(1)(-3√2)
= 16 + 12√2 > 0
Hence, the equation has real roots.
(c) Given equation is x² - 4x - 3√2 = 0
On comparing the equation with ax² + bx + c = 0, we get
a = 1, b = -4 and c = -3√2
Then, D = b² - 4ac = (-4)² - 4(1) (-3√2)
= 16 + 12√2 > 0
Hence, the equation has real roots.
(d) Given equation is 3x^​2 + 4√3x + 4 = 0.
On comparing the equation with ax² + bx + c = 0, we get
a = 3, b = 4√3 and c = 4
Then, D = b² - 4ac = (4√3)² - 4(3)(4) = 48 - 48 = 0
Hence, the equation has real roots.
Hence, x² - 4x + 3√2 = 0 has no real roots.

(a) Given that, 2x² - 3x + 6 = 0
On comparing with ax² + bx + c = 0, we get
a = 2,  b = -3 and c = 6
∴ Sum of the roots = 
So, sum of the roots of the quadratic equation 2x² - 3x + = 0 is not 3, so it is not the answer
(b) Given that, -x² + 3x - 3 =0
On compare with ax² + bx + c = 0, we get
a = -1, b = 3 and c = -3
∴ Sum of the roots = 
So, sum of the roots of the quadratic equation -x² + 3x - 3 = 0 is 3, so it is the answer.
(c) Given that, 
⇒ 2x² - 3x + √2 = 0
On comparing with ax² + bx + c = 0, we get
a = 2, b = -3 and c = √2
∴ Sum of thee roots = 
So, sum of the roots of the quadratic equation  is not 3, so it is not the answer.
(d) Given that, 3x² - 3x + 3 = 0
⇒ x² - x + 1 = 0
On comparing with ax² + bx + c = 0, we get
a = 1, b = -1 and c = 1
∴ Sum of the roots = 
So, sum of the roots of the quadratic equation 3x² - 3x + 3 = 0 is not 3, so it is not the answer.

Given equation is (x² + 1)² - x² = 0
⇒ x⁴ + 1 + 2x² - x² = 0 [∵ (a + b)² = a² + b² + 2ab]
⇒ x⁴ + x² + 1 = 0
Let x² = y
∴ (x²)² + x² + 1 = 0
y² + y + 1 = 0
On comparing with ay² + by + c = 0, we get
a = 1, b = 1 and c = 1
Discriminant, D = b² - 4ac
= (1)² - 4(1)(1)
= 1 - 4 = -3
Since, D < 0
∴ y² + y + 1 = 0 i.e., x⁴ + x² + 1 = 0 or (x² + 1)² - x² = 0 has no real roots.

We have a quadratic equation: 

If we have standard equation ax² + bx + c  then D = b² - 4ac

a= 2, b= -√5, c= 1

D = (-√5)² - (4x2x1)

D= 5 - 8

D = -3

As the value of D<0 so there is no real root

x² - 0.09 = 0
x² = 0.09
x = √0.09
x = 0.3

As 1/2 is a root then it will satisfy the given equation.

Put x = 1/2

1/4 +(k×1/2) - 5/4 =0
k×1/2 =5/4 -1/4
k =2

3x² - 2x - 5 = 0
3x² + 3x - 5x - 5 = 0
3x(x+1) - 5(x+1) = 0
(x+1)(3x-5) = 0
x = -1 and 5/3.

Hence roots of given polynomial are -1 and 5/3.

As, discriminant (b² - 4ac) of the equation is negative. Therefore, no real roots.