Question

Find the quadratic polynomial whose zeros are 1 and -3. Verify the relation between the coefficients and the zeroes off the polynomial. Plz do answer a correct and appropriate answer...

Answer 1

SolutioN :

Let,

• Zero be α and β.
• α = 1.
• β = - 3.

✎ Sum of Zeros.

→ α + β = 1 + ( - 3 )

→ α + β = 1 - 3.

→α + β = - 2.

$\rule{90}1$

✎ Product Of Zero.

→ α * β = 1 * ( - 3 )

→ α * β = - 3.

$\rule{90}1$

Now,

→ K [ x² - Sx + P ]

Where as,

• K Constant term.
• S Sum of Zero.
• P Product of Zero.

→ K [ x² - ( -2 )x + ( -3 ) ]

→ K [ x² + 2x - 3 ]

$\rule{90}1$

» Let's Verify relations between Zero and coefficient.

We have, Polynomial.

→ x² + 2x - 3.

☛ Compare With General Expression.

ax² + bx + c.

Where as,

• a = 1.
• b = 2.
• c = - 3.

✎ Sum of Zeros.

→ α + β = - b / a

→ 1 + ( -3 ) = - 2 / 1

→ - 2 = - 2.

$\rule{90}1$

✎ Product Of Zero.

→ α * β = c / a.

→ 1 * - 3 = - 3.

→ - 3 = - 3.

✡ Hence Verify.

Answer 2

Answer:

$\underline{\bigstar\:\textsf{Required Quadratic Polynomial :}}$

$:\implies\sf p(x)=(x-1)(x-(-\:3))\\\\\\:\implies\sf p(x)=(x-1)(x+3)\\\\\\:\implies\sf p(x)=x(x+3)-1(x+3)\\\\\\:\implies\sf p(x)=x^2+3x-x-3\\\\\\:\implies\sf p(x)=x^2+2x-3$

⠀⠀⠀$\rule{160}{1.5}$

$\boxed{\begin{minipage}{5.5 cm} { \bigstar\: \textsf{For a Quadratic Polynomial :}}\\\\ {\qquad\sf p(x) = ax ^\sf2 \sf + bx + c}\\\sf with zeroes \alpha\:\sf and\:\beta \\\\\\ {\textcircled{\footnotesize1}} \:\:\alpha +\beta= \dfrac{ - \:b}{a}\:\:\bigg\lgroup\bf Sum\:of\:Zeroes\bigg\rgroup \\\\\\{\textcircled{\footnotesize2}} \: \:\alpha \beta= \sf\dfrac{c}{a}\:\:\bigg\lgroup\bf Product\:of\:Zeroes\bigg\rgroup\end{minipage}}$

⠀

☯ $\underline{\textsf{Relation b/w coefficients \& zeroes :}}$

${\qquad\maltese\:\:\textsf{Sum of Zeroes :}} \\\\\dashrightarrow\sf\:\:\alpha +\beta= \dfrac{ - \:b}{a}\\\\\\\dashrightarrow\sf\:\:1+(-\:3)=\dfrac{-\:2}{1}\\\\\\\dashrightarrow\:\:\underline{\boxed{\sf-\:2=-\:2}}\\\\\\{\qquad\maltese\:\:\textsf{Product of Zeroes :}}\\\\\dashrightarrow\sf\:\:\alpha\beta=\dfrac{c}{a}\\\\\\\dashrightarrow\sf\:\:1-\:3=\dfrac{-\:3}{1}\\\\\\\dashrightarrow\:\:\underline{\boxed{\sf -\:3=-\:3}}$