Question

find the zeroes of quadratic polynomial 6x2 - x - 12 and verify the relationship between the zeroes and coefficients​

Answer 1

EXPLANATION.

Quadratic polynomial.

⇒ 6x² - x - 12.

As we know that,

Factorizes the equation into middle term splits, we get.

⇒ 6x² - 9x + 8x - 12 = 0.

⇒ 3x(2x - 3) + 4(2x - 3) = 0.

⇒ (3x + 4)(2x - 3) = 0.

⇒ x = -4/3  and  x = 3/2.

Sum of the zeroes.

⇒ -4/3 + 3/2.

⇒ - 8 + 9/6 = 1/6.

Products of the zeroes.

⇒ (-4/3) x (3/2).

⇒ - 12/6 = - 2.

⇒ 6x² - x - 12 = 0.

As we know that,

Sum of the zeroes of the quadratic equation.

⇒ α + β = -b/a.

⇒ α + β = -(-1)/6 = 1/6.

Products of the zeroes of the quadratic equation.

⇒ αβ = c/a.

⇒ αβ = (-12)/6 = - 2.

Hence verified.

                                                                                                                         

MORE INFORMATION.

Nature of the roots of the quadratic expression.

(1) = Real and unequal, if b² - 4ac > 0.

(2) = Rational and different, if b² - 4ac is a perfect square.

(3) = Real and equal, if b² - 4ac = 0.

(4) = If D < 0 Roots are imaginary and unequal Or complex conjugate.

Answer 2

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$\underline{\pink{\dag}{\underline{\frak{Question}}\pink{\dag}}}$

• Find the zeroes of quadratic polynomial 6x² - x - 12 and verify the relationship between the zeroes and coefficients.

$\underline{\pink{\dag}{\underline{\frak{\rm{S}\frak{tep\:by\:step\:explanation}}}\pink{\dag}}}$

• Finding zeroes of given quadratic polynomial :

$\longrightarrow\qquad\sf 6x^2 - x - 12 = 0$

• Using middle splitting method :

$\longrightarrow\qquad\sf 6x^2 - 9x + 8x - 12 = 0$

$\longrightarrow\qquad\sf 3x(2x - 3) + 4(2x - 3) = 0$

$\longrightarrow\qquad\sf (3x + 4)(2x - 3) = 0$

$\longrightarrow\qquad\sf 3x + 4 = 0\:,\:2x - 3 = 0$

$\longrightarrow\qquad\sf 3x = -4\:,\:2x = 3$

$\longrightarrow\qquad\bf{x = \red{\dfrac{-4}{3}}\:,\:x = \red{\dfrac{3}{2}}}$

$\small\red{\therefore}\:{\underline{\sf{Hence,\:zeroes\:(\alpha\:and\:\beta)\:=\:\bf{\frac{-4}{3}}\:\sf{and}\:\bf{\frac{3}{2}\:\sf{respectively.}}}}}$

• Verifying the relationship between the zeroes and coefficients :

• Sum of zeroes (α + β) = -b/a
• Product of zeroes (αβ) = c/a

$\leadsto\quad\sf \alpha + \beta\:=\:\dfrac{-b}{a}\:,\:\alpha\beta\:=\:\dfrac{c}{a}$

• Values that we have :

• α and β = -4/3 and 3/2
• b = coefficient of x = 1
• c = constant term = 12
• a = coefficient of x² = 6

• Putting all values :

$\leadsto\quad\small\sf \dfrac{-4}{3} + \dfrac{3}{2}\:=\:\dfrac{-(-1)}{6}\:,\:\dfrac{-4}{3}\:\times\:\dfrac{3}{2}\:=\:\dfrac{-12}{6}$

$\leadsto\quad\small\sf \dfrac{-8 + 9}{6}\:=\:\dfrac{1}{6}\:,\:\dfrac{-\cancel{4}}{\cancel{3}}\:\times\:\dfrac{\cancel{3}}{\cancel{2}}\:=\:\dfrac{-\cancel{12}}{\cancel{6}}$

$\leadsto\quad\small{\bf{\red{\dfrac{1}{6}}\:=\:\red{\dfrac{1}{6}}\:,\:\red{-2}\:=\:\red{-2}}}$

$\quad\qquad\qquad\underline{\pink{\dag}{\underline{\frak{\rm{H}\frak{ence,\:Verified!}}}\pink{\dag}}}$

ㅤㅤㅤㅤㅤㅤ━━━━━━━━━━

$\underline{\pink{\dag}{\underline{\frak{\rm{M}\frak{ore\:to\:know}}}\pink{\dag}}}$

• Sridhara Acharya's formula :

$\qquad\red\bigstar\:{\small{\underline{\boxed{\bf{\green{x = \dfrac{-b\:\pm\:\sqrt{b^2 - 4ac}}{2a}}}}}}}$

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