PE 4 - Quadratic Equation

3x³ – 9x² + kx – 12 is divisible by x – 3

∴ f(3) = 0

∴ 3 × 3³ – 9 × 3² + 3k – 12 = 0

∴ 81 – 81 + 3k – 12 = 0

∴ 3k = 12

∴ k = 4

So, the equation will be 

3x³ – 9x² + 4x – 12 = 0

Factorising the given equation by dividing it with (x - 3), we get

(x – 3)(3x² + 4) = 0

Thus 3x² + 4 is a factor of the given equation.

Hence, option (b).

When ${\left(1-\frac{1}{a}\right)}^6$ is expanded, the the last three coefficients are:

⁶C₄, −⁶C₅ and ⁶C₆ 

⁶C₄ = 15
−⁶C₅ = −6
⁶C₆ = 1

∴ The sum of coefficients = 15 – 6 + 1 = 10

Hence, option (c).

y = x² and y = 3x + k

∴ x² = 3x + k

∴ x² – 3x – k = 0

This has two identical 

Solutions when the discriminant of the equation is 0.
∴ (−3)² – 4(–k) = 0
∴ 9 + 4k = 0
∴ k = -9/4

Hence, option (c).

Let the year of the man’s birth be y.

Then, y + x = x²

∴ y = x(x – 1)

Now looking at the options we find that only 1806 can be written in the form x(x – 1), where x = 43.

Hence, option (a).

2x² + 6x + 5y + 1 = 0
∴ 2x(x + 3) + 5y + 1 = 0   ...(1)

Also, 2x + y + 3 = 0
∴ 2x = – (y + 3)   ...(2)
∴ x = (-(y+3))/2   ...(3)

Substituting (2) & (3) in (1),
-(y + 3)$\left[\frac{-(y+3)}{2}+3\right]$ + 5y + 1 = 0

∴ $\frac{-(y+3)(-y-3+6)}{2}$ + 5y + 1 = 0

∴ –(y + 3) (–y + 3) + 10y + 2 = 0

∴ –(9 – y²) + 10y + 2 = 0

∴ y² + 10y – 7 = 0

Hence, option (c).

The sum of the roots of the equation ax² + bx + c = 0 is -b/a, where b is the coefficient of x and a is the coefficienct of x².

The given quadratic is 4x² + 5 - 8x = 0. Here, coefficient of x (b) = -8 and coefficient of x² (a) = 4.

The sum of roots of the equation 4x² + 5 - 8x = 0 is (–(-8))/4 = 2

Hence, option 4.

Substituting the values of x given in the options, we find that only x = 1 satisfies the equation.

Hence, option (a).

4x³ – 8x² – 63x – 9 = 0

Sum of roots of this equation = -((-8)/4) = 2 

Hence, option (b).