Question

5.
Find the zeroes of the quadratic polynomial P(x) = x2 + x-12 and
verify the relationship between the zeroes and the coefficients.​

Answer 1

EXPLANATION.

Quadratic equation.

⇒ x² + x - 12.

As we know that,

Sum of the zeroes of the quadratic equation.

⇒ α + β = -b/a.

⇒ α + β = -(1)/1 = -1.

Products of the zeroes of the quadratic equation.

⇒ αβ = c/a.

⇒ αβ = (-12)/1 = -12.

As we know that,

Factorizes the equation into middle term splits, we get.

⇒ x² + x - 12 = 0.

⇒ x² + 4x - 3x - 12 = 0.

⇒ x(x + 4) - 3(x + 4) = 0.

⇒ (x - 3)(x + 4) = 0.

⇒ x = 3 and x = -4.

Sum of the values of x, we get.

⇒ 3 + (-4) = -1.

Products of the values of x, we get.

⇒ (3)(-4) = -12.

                                                                                                                             

MORE INFORMATION.

Conditions for common roots.

Let quadratic equation are a₁x² + b₁x + c₁ = 0 and a₂x² + b₂x + c₂ = 0.

(1) = If only one roots is common.

⇒ x = b₁c₂ - b₂c₁/a₁b₂ - a₂b₁.

⇒ y = c₁a₂ - c₂a₁/a₁b₂ - a₂b₁.

(2) = If both roots are common.

⇒ a₁/a₂ = b₁/b₂ = c₁/c₂.

Answer 2

Answer:

Given :-

$\bf \: {x}^{2} + x - 12$

To Find :-

Relationship the zeroes and the coefficients.

Solution :-

We know that

• Sum of Zeroes

$\sf \: \alpha + \beta = \dfrac{ - b}{a}$

$\sf \: \alpha + \beta = \dfrac{ - ( - 1)}{1}$

$\sf \: \alpha + \beta = \dfrac{1}{1}$

$\sf \blue{ \alpha + \beta = 1}$

• Product of Zeroes

$\sf \: \alpha \beta = \dfrac{c}{a}$

$\sf \: \alpha \beta = \dfrac{ - 12}{1}$

$\alpha \beta = - 12$

Lets factorise

$\sf \: {x}^{2} + x - 12$

$\sf \: {x}^{2} + (4x - 3x) - 12$

$\sf \: {x}^{2} + 4x - 3x + 12 = 0$

$\sf \: x(x + 4) \: or \: 3(x + 4)$

$\sf \: (x - 3) \: or \: (x + 4)$

Either

$\bf \: {x} - 3 = 0$

$\bf \: x \: = 0 + 3$

$\bf \: x = 3$

Or,

$\bf \: x + 4 = 0$

$\bf \: x = 0 - 4$

$\bf \: x = - 4$

By putting the value

3 + (-4)

3 - 4

= -1

And,

3(-4)

-12