Question

find the roots of given quadratic equation X square + 4 x minus 12 is equals to zero​

Answer 1

Given: x² + 4x - 12 = 0.

To find: The roots of the equation.

Answer:

FIRST METHOD: Splitting the middle term.

x² + 4x - 12 = 0

x² + 6x - 2x - 12 = 0

x(x + 6) - 2(x + 6) = 0

(x - 2)(x + 6) = 0

x - 2 = 0 or x + 6 = 0

x = 2; x = -6

SECOND METHOD: Quadratic formula.

Formula:

$\tt x\ =\ \dfrac{-b\ \pm\ \sqrt{b^2\ -\ 4ac}}{2a}$

From the equation,

• a = 1
• b = 4
• c = -12

Using them in the formula,

$\tt x\ =\ \dfrac{-4\ \pm\ \sqrt{(4)^2\ -\ 4\ \times\ 1\ \times\ -12}}{2\ \times\ a}\\\\\\x\ =\ \dfrac{-4\ \pm\ \sqrt{16\ +\ 48}}{2}\\\\\\x\ =\ \dfrac{-4\ \pm\ \sqrt{64}}{2}\\\\\\x\ =\ \dfrac{-4\ \pm\ 8}{2}\\\\\\x\ =\ \dfrac{-4\ +\ 8}{2}\ \ or\ x\ =\ \dfrac{-4\ -\ 8}{2}\\\\\\\bf x\ =\ 2\ \tt and\ \bf x\ =\ -6$

Therefore, the zeros of the quadratic equation x² + 4x - 12 = 0 are 2 and -6.

Answer 2

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$\huge\tt{GIVEN:}$

• x²+4x-12=0

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$\huge\tt{TO~FIND:}$

• Roots of the equation

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$\huge\tt{SOLUTION:}$

Splitting the terms,

↪x²+4x-12=0

↪x²+6x-2x-12=0

↪x(x+6)-2(x+6)=0

↪(x-2)(x+6)=0

↪x-2=0 and, x+6=0

↪x = 2 ; x = -6

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Henceforth,

$\tt{The~zeros~of~the~quadratic~equation~}$$\tt{x^2~ + ~4x ~-~ 12~ = 0 ~are ~2 ~and ~-6.}$

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