Question

find the zeroes of 2x^2 - 3x - 2 and verify the relation between zero and coefficient​

Answer 1

$2 {x}^{2} - 3x - 2 \\ \\ = > 2 {x}^{2} - 4x + x - 2 \\ \\ = > 2x(x - 2) + 1(x - 2) \\ \\ = > (x - 2)(2x + 1)$

Then, zeroes will be 2 and -1/2 ✔✔

Answer 2

☯GIVEN :

• $\sf {2x^2 - 3x - 2}$

☯TO DO :

• Find the zeros and verify the relation between zero and coefficient.

➲SOLUTION :

$\implies \sf{2x^2 - 3x - 2 = 0}$

$\implies \sf{2x^2 - 4x + x- 2 = 0}$

$\implies \sf{2x \big(x - 2 \big)+ 1 \big(x- 2 \big) = 0}$

$\implies \sf{\big(x - 2 \big) \big(2x + 1 \big) = 0}$

$\implies \sf{\big(x - 2 \big) = 0 \: , \: \big(2x + 1 \big) = 0}$

$\implies\sf{x = 0 + 2\: , \: 2x = 0 - 1}$

$\implies\sf{x = 2\: , \: 2x = - 1}$

$\implies \underline {\boxed{ \red{\bf{x = 2\: , \: x = \dfrac{- 1}{2}}}}}$

Verifying the relation between zero and coefficient :

• Here , a =2 , b=-3 and c = -2

Sum of zeros :

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

$\implies \sf {2 + \dfrac{ - 1}{2} = \dfrac{ -( - 3 )}{2} }$

$\implies \sf {\dfrac{ 4 + (- 1)}{2} = \dfrac{ 3}{2} }$

$\implies \sf {\dfrac{ 4- 1}{2} = \dfrac{ 3}{2} }$

$\implies { \boxed{\bf{ \pink{\dfrac{ 3}{2} = \dfrac{ 3}{2} }}}}$

Product of zeros :

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

$\implies \sf{ \cancel{2 }\times \dfrac{ - 1}{ \cancel{2}} = \cancel{\dfrac{ - 2}{ 2}}}$

$\implies{ \underline {\boxed{ \pink{\bf{ - 1= - 1}}}}}$

$\huge {\green{\therefore }}$ Verified.