Question

the roots of the quadratic equation 6x^2-x-2=0 are​

Answer 1

Answer :-

2/3 , -1/2

Given :-

6x² -x -2 = 0

To find :-

Roots of the Quadratic equation

SOLUTION :-

Method - 1 :-

By factorisation method splitting the midlle term

6x² -x -2 = 0

• Here sum of two numbers should be -1xand product of two numbers should be -12x²

So, the numbers are

3x × -4x = -12x²

3x+(-4x) = 3x-4x

= -1x

So, split the middle term as 3x -4x

6x² -x -2 = 0

6x² + 3x -4x -2 = 0

3x(2x +1) -2(2x +1) = 0

(2x +1) (3x -2) = 0

2x +1 =0

2x = -1

x = -1/2

3x -2 =0

3x = 2

x = 2/3

So, the roots are -1/2, 2/3

Method -2 :-

Solving by the Quadratic formula

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

6x² -x - 2=0

Comparing with general form of Quadratic equation ax² + bx + c = 0

• a = 6
• b = -1
• c = -2

Substituting the values in formula

$x = \dfrac{ - ( - 1) \pm \sqrt{( - 1) {}^{2} - 4(6)( - 2)}}{2(6)}$

$x = \dfrac{ 1\pm \sqrt{1 + 48}}{12}$

$x = \dfrac{ 1\pm \sqrt{49}}{12}$

$x = \dfrac{ 1\pm 7}{12}$

$x = \dfrac{1 + 7}{12} , \cfrac{1 - 7}{12}$

$x = \dfrac{8}{12} ,\cfrac{ - 6}{12}$

$x = \dfrac{4}{6} ,\cfrac{ - 1}{2}$

$x = \dfrac{2}{3} , \cfrac{ - 1}{2}$

So, the value of x are 2/3 , -1/2

So, the roots of Quadratic equation is 2/3 , -1/2

Answer 2

Answer:

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \LARGE{\color{pink}{\textsf{\textbf{Answer :-}}}}$

$\sf6x² - x - 2 = 0$

$\sf⇒ 6x² - 4x + 3x - 2 = 0$

$\sf⇒ 2x(3x - 2) +1(3x - 2) = 0$

$\sf⇒ (2x+1)(3x - 2) = 0$

$\sf \: so \: either \:(3x - 2) = 0 \: or  \:  (2x+1) = 0$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \bf{x = 2/3 \: \: \: or \: -1/2 }$

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