Question

Compute the zeroes of the polynomial 4x^2 – 4x – 8. Also, establish a relationship between the zeroes and coefficients.

Answer 1

Qᴜᴇsᴛɪᴏɴ :-

Compute the zeroes of the polynomial 4x^2 – 4x – 8. Also, establish a relationship between the zeroes and coefficients. ?

Sᴏʟᴜᴛɪᴏɴ :-

Put The Given Quadratic Polynomial Equals to 0 .

→ 4x² - 4x - 8 = 0

→ 4(x² - x - 2) = 0

→ x² - x - 2 = 0

→ x² - 2x + x - 2 = 0

→ x(x - 2) + 1(x - 2) = 0

→ (x - 2)(x + 1) = 0

→ x = 2 & (-1) .

____________________

Now, First Relation is :-

→ Sum of Zeros = - (coefficient of x) /(coefficient of x²)

Putting both values ,

→ 2 + (-1) = -(-4)/4

→ 1 = 1 ✪✪ Hence Verified. ✪✪

Second Relation :-

→ Product Of Zeros = Constant Term / (coefficient of x²)

Putting both Values ,

→ 2 * (-1) = (-8) / (4)

→ (-2) = (-2) ✪✪ Hence Verified. ✪✪

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Answer 2

${\huge{\bf{\red{\underline{Solution:}}}}}$

${\bf{\blue{\underline{Given:}}}}$

${\star{\sf{ \: 4 {x}^{2} - 4x - 8 }}}$

${\bf{\blue{\underline{Now:}}}}$

Divide by 4,

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

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

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

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

Take,

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

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

So, the value of 4x²-4x-8 is zero when x=-1,2

Therefore,the zeros of 4x²-4x-8 are -1 and 2.

Now,

$\star\boxed{\sf{ \purple {Sum \: of \: zeros \: = \frac{ - (Coefficient \: of \: x)}{Coefficient \: of \: {x}^{2} } }}}$

${ : \implies{\sf{ - 1 + 2 = \frac{ - ( - 4)}{ 4} }}}$

${ : \implies{\sf{ 1 = \frac{ - ( - 1)}{ 1} }}}$

${ : \implies \boxed{\sf{ 1 = 1 }}}$

$\star\boxed{\sf{ \purple {Product \: of \: zeros \: = \frac{ constant \: term}{coefficient \: of \: {x}^{2} } }}}$

${ : \implies {\sf{ ( - 1 )(2) = \frac{ - 8}{4} }}}$

$: \implies\boxed {\sf{ - 2 = - 2 }}$

Hence Verified.