Test: Quadratic Equations - 2 - Grade 12 MCQ

For a quadratic equation, ax² + bx + c = 0, discriminant is b² - 4ac.
Roots are . For real roots, radical must be non-negative i.e. discriminant should be greater than or equal to zero.

For a quadratic equation, ax² + bx + c = 0, discriminant is b² - 4ac.
Roots are  . For imaginary roots, radical is negative i.e. discriminant should be less than zero.

x²+1 = 0
=>x² = -1 => x = ±√−1 = ±i.

2x² + x + 1 = 0
D=1²-4*2*1 = 1-8 = -7 ≤ 0.
Since D ≤ 0, imaginary roots are there.

– x² + x – 2 = 0
=>x²-x+2 = 0
D=(-1)²-4*1*2 = 1-8 = -7 ≤ 0.
Since D ≤ 0, imaginary roots are there.

 2x² + √2x + 2 = 0
=> D = (√2)² – 4.2.2 = 2-16 = -14.
Since D ≤ 0, imaginary roots are there.

Comparing the coefficients of the above equation we get,
1/1 = b/3 = c/3
This means b = 3 and c = 3
Therefore, b * c = 9.

Given, acosθ + bsinθ = c
So, this implies acosθ = c – bsinθ
Now squaring both the sides we get,
(acosθ)² = (c – bsinθ)²
a² cos² θ = c² + b² sin² θ – 2b c sinθ
a² (1- sin² θ) = c² + b² sin² θ – 2b c sinθ
a² – a² sin² θ = c² + b² sin² θ – 2b c sinθ
Now rearranging the elements,
(a² + b²) sin² θ – 2b c sinθ +( c² – a²) = θ
So, as sum of the roots are in the form –b/a if there is a quadratic equation ax² + bx + c = 0
Now, we can conclude that
sinα + sinβ = 2bc/(a² + b²).

Given, acosθ + bsinθ = c
So, this implies acosθ = c – bsinθ
Now squaring both the sides we get,
(acosθ)² = (c – bsinθ)²
a² cos² θ = c² + b² sin² θ – 2b c sinθ
a² (1- sin² θ) = c² + b² sin² θ – 2b c sinθ
a² – a² sin² θ = c² + b² sin² θ – 2b c sinθ
Now rearranging the elements,
(a² + b²) sin² θ – 2b c sinθ +( c² – a²) = θ
So, as sum of the roots are in the formc/a if there is a quadratic equation ax² + bx + c = 0
Now , we can conclude that
sinα + sinβ = (c² + a²)/(a² + b²).

Subtracting the equation x² + ax + b = 0 to x² + bx + a = 0 by solving the equation simultaneously, we get,
(a – b)x + (b – a) = 0
So, (a – b)x = (a – b)
Therefore, x = 1
Now, putting the value of x = 3 in any one of the equation, we get,
1 + a + b = 0
Therefore, a + b = -1.

Let, y = x² + 2
Then, 2x² – 3x – 6 = 0
So, (3x)² = (2x² – 6)²
[2(y-2) – 6]² = 9(y-2)
= 4x² – 49x + 118 = 0.

Since, p and q are the roots of the equation x² + px + q =0
So, p + q = -p and pq = q
So, pq = q
And, q = 0 or p = 1
If, q = 0 then, p = 0 and if p = 1 then q = -2.

Given, (a – b)x² + (b – c)x + (c – a) = 0 and x² + px + 1 =0
So, 1 / (a – b) = p / (b – c) = 1 / (c – a)
Equating the above equation, we get,
(b – c) = p(a – b) and
(b – c) = p(c – a)
So, p(a – b) = p(c – a)
=> a – b = c – a
So, 2a = b + c which means that b, a, c are in A.P.

Comparing the coefficients of the above equation we get,
1/1 = b/3 = c/3
This means b = 3 and c = 3
Therefore, b + c = 6.

As, we know | z₁ + z₂ + …….. +z_n| ≤ |z₁| + |z₂| + ………. + |z_n|
So, |z₁ + z₂ + 3 + 4i| ≤ |z₁| + |z₂| + |3 + 4i|
Now, putting the given values in the equation, we get,
=> |z₁ + z₂ + 3 + 4i| ≤ 4 + 3 + √(9 + 16)
=> |z₁ + z₂ + 3 + 4i| ≤ 4 + 3 + 5
=> |z₁ + z₂ + 3 + 4i| ≤ 12.