Math, asked by tanveerkaur5689, 3 months ago

Q. Find the zeroes of the quadratic polynomial : x2 – 2x – 8 and verify the relationship between the zeroes and the coefficients.​

Answers

Answered by SuitableBoy
118

Answer:

 \\

\large{\underbrace{\underline{\bf\bigstar\:Required\:Solution:-}}}

 \\

» In this question, first we are supposed to find the zeroes of the quadratic equation given.

» To find the zeroes of the polynomial,we may use the quadratic formula or middle term splitting.

» After finding the zeroes, we will verify the relationship between the zeroes and the coefficients of the polynomial.

 \\

Finding the zeroes of the polynomial :

 \\

» Here, we would us middle term splitting to find the zeroes.

 \colon \implies \:  \bf {x}^{2}  - 2x - 8 = 0 \\  \\

 \colon \implies \bf \:  {x}^{2}  - 4x + 2x - 8 = 0 \\  \\

 \colon \implies \bf \: x(x - 4) + 2(x - 4) = 0   \\  \\

 \colon \implies \bf \: (x + 2)(x - 4) = 0 \\  \\

So,

Either -

 \mapsto \sf \: x + 2 = 0 \\  \\   \dashrightarrow \boxed { \pink{ \bf{x =  - 2}}}

Or -

 \mapsto \sf \: x - 4 = 0 \\  \\  \dashrightarrow \boxed{ \bf { \pink{x = 4}}}

 \\

\therefore\;{\underline{\sf Zeroes ~of~the~Polynomial~are \bf{-2}~\sf and~\bf{4.}}}\\

 \\

_____________________________

 \\

Verifying the Relationship between Zeroes and Coefficients :

 \\

» Standard form of a Quadratic Equation is -

 \odot \:   \boxed{ \tt a {x}^{2}   + bx + c}

Compare it with given equation -

 \odot \: \boxed{ \tt  {x}^{2} - 2x - 8  }

So,

  • a = 1
  • b = -2
  • c = -8

We know :

  • Zeroes = -2 & 4

Now,

 \colon \displaystyle \leadsto \sf \: sum \: of \: zeroes =   \bf\frac{ - b}{a}

LHS :

 \colon \rarr \sf \: sum \: of \: zeroes =  \bf - 2 + 4 \\  \\  \implies  \boxed{ \bf \purple2}

RHS :

 \displaystyle \colon \rarr \sf \:  \frac{ - b}{a}   =  \bf  \frac{ - ( - 2)}{1}  \\  \\ \implies \boxed{ \bf \purple2}

So,

LHS = RHS

Hence Verified.

 \\

And

 \displaystyle \colon \leadsto \sf \: product \: of \: zeroes =  \bf \frac{c}{a}

LHS :

 \colon \rarr \sf \: product \: of \: zeroes =   \bf - 2 \times 4 \\  \\  \implies \boxed{ \bf \blue{ - 8}}

RHS :

 \displaystyle \colon \rarr \sf \:  \frac{c}{a}  =   \bf\frac{ - 8}{1}  \\  \\  \implies \boxed{ \bf \blue{ - 8}}

So,

LHS = RHS

Hence Verified.

 \\

_____________________________

Similar questions