5.Find the zeroes of the quadratic polynomial P(x) = x2 + x-12 andverify the relationship between the zeroes and the coefficients. About the author Ximena
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. Reply
Answer: Given :– [tex] \bf \: {x}^{2} + x – 12[/tex] To Find :– Relationship the zeroes and the coefficients. Solution :– We know that • Sum of Zeroes [tex] \sf \: \alpha + \beta = \dfrac{ – b}{a} [/tex] [tex] \sf \: \alpha + \beta = \dfrac{ – ( – 1)}{1} [/tex] [tex] \sf \: \alpha + \beta = \dfrac{1}{1} [/tex] [tex] \sf \blue{ \alpha + \beta = 1}[/tex] • Product of Zeroes [tex] \sf \: \alpha \beta = \dfrac{c}{a} [/tex] [tex] \sf \: \alpha \beta = \dfrac{ – 12}{1} [/tex] [tex] \alpha \beta = – 12[/tex] Lets factorise [tex] \sf \: {x}^{2} + x – 12[/tex] [tex] \sf \: {x}^{2} + (4x – 3x) – 12[/tex] [tex] \sf \: {x}^{2} + 4x – 3x + 12 = 0[/tex] [tex] \sf \: x(x + 4) \: or \: 3(x + 4)[/tex] [tex] \sf \: (x – 3) \: or \: (x + 4)[/tex] Either [tex] \bf \: {x} – 3 = 0 [/tex] [tex] \bf \: x \: = 0 + 3[/tex] [tex] \bf \: x = 3[/tex] Or, [tex] \bf \: x + 4 = 0[/tex] [tex] \bf \: x = 0 – 4[/tex] [tex] \bf \: x = – 4[/tex] By putting the value 3 + (-4) 3 – 4 = -1 And, 3(-4) -12 Reply
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:
Given :–
[tex] \bf \: {x}^{2} + x – 12[/tex]
To Find :–
Relationship the zeroes and the coefficients.
Solution :–
We know that
• Sum of Zeroes
[tex] \sf \: \alpha + \beta = \dfrac{ – b}{a} [/tex]
[tex] \sf \: \alpha + \beta = \dfrac{ – ( – 1)}{1} [/tex]
[tex] \sf \: \alpha + \beta = \dfrac{1}{1} [/tex]
[tex] \sf \blue{ \alpha + \beta = 1}[/tex]
• Product of Zeroes
[tex] \sf \: \alpha \beta = \dfrac{c}{a} [/tex]
[tex] \sf \: \alpha \beta = \dfrac{ – 12}{1} [/tex]
[tex] \alpha \beta = – 12[/tex]
Lets factorise
[tex] \sf \: {x}^{2} + x – 12[/tex]
[tex] \sf \: {x}^{2} + (4x – 3x) – 12[/tex]
[tex] \sf \: {x}^{2} + 4x – 3x + 12 = 0[/tex]
[tex] \sf \: x(x + 4) \: or \: 3(x + 4)[/tex]
[tex] \sf \: (x – 3) \: or \: (x + 4)[/tex]
Either
[tex] \bf \: {x} – 3 = 0 [/tex]
[tex] \bf \: x \: = 0 + 3[/tex]
[tex] \bf \: x = 3[/tex]
Or,
[tex] \bf \: x + 4 = 0[/tex]
[tex] \bf \: x = 0 – 4[/tex]
[tex] \bf \: x = – 4[/tex]
By putting the value
3 + (-4)
3 – 4
= -1
And,
3(-4)
-12