Question

answer red mark question please​

Answer 1

Given polynomial :

• 2x² + 5x - 3 = 0

To Find :

• Zeroes of the polynomial.
• Verify the relationship between zeroes and coefficients.

Solution :

Let's solve and find the zeroes using the factorization method.

➣ $\mathtt{2x^2\:+5x-3=0}$

➣ $\mathtt{2x^2\:+6x\:-x\:-3=0}$

➣ $\mathtt{2x(x+3)\:-1(x+3)=0}$

➣ $\mathtt{(x+3)\:\:\:(2x-1)\:=0}$

➣ $\mathtt{(x+3)\:=0\:\:or\:(2x-1)\:=\:0}$

➣ $\mathtt{x+3=0\:\:or\:\:2x-1=0}$

➣ $\mathtt{x=\:-\:3\:\:or\:\:2x=1}$

➣ $\mathtt{x=-\:3\:or\:\:x\:=\:{\dfrac{1}{2}}}$

$\large{\boxed{\mathtt{\red{Roots\:of\:the\:equation\:are\:-3\:and\:{\dfrac{1}{2}}}}}}$

Relationship between zeroes and coefficients :

• Sum of zeroes :

Let,

• α = - 3
• β = 1/2

α + β = $\mathtt{-3\:+\:{\dfrac{1}{2}}}$

➣ $\mathtt{\dfrac{-6+1}{2}}$

➣ $\mathtt{\dfrac{-5}{2}}$

➣ $\mathtt{\dfrac{-(-5)}{2}}$

➣ $\mathtt{\dfrac{5}{2}}$

Sum of zeroes :

➣ $\mathtt{\dfrac{Coefficient\:of\:x}{Coefficient\:of\:x^2}}$

➣ $\mathtt{\dfrac{5}{2}}$

Hence, we see the relationship between zeroes and coefficients is satisfied.

Comparing the given quadratic polynomial with general form :

• a = 2 = coefficient of x²
• b = 5 = coefficient of x
• c = -3 = constant.

Product of zeroes :

αβ = $\mathtt{-3\:\times\:{\dfrac{1}{2}}}$

➣ $\mathtt{\dfrac{-3}{2}}$

Product of zeroes :

➣ $\mathtt{\dfrac{Constant}{Coefficient\:of\:x^2}}$

➣ $\mathtt{\dfrac{-3}{2}}$

Hence, relationship between product of zeroes and coefficients is satisfied.

Answer 2

AnswEr :

$\:\bullet\:\sf\ Given \: polynomial = 2x^2 - 3 + 5x \\ \\ \qquad\sf\ Or, \: it \: can \: be \: 2x^2 + 5x - 3$

$\underline{\bigstar\:\textsf{According \: to \: given \: now;}}$

$\normalsize\star{\underline{\boxed{\mathbb \red{ZEROES \: OF \: POLYNOMIAL -}}}}$

$\normalsize\ : \implies\sf\ 2x^2 + 5x - 3$

$\normalsize\ : \implies\sf\ 2x^2 + 6x - x - 3$

$\normalsize\ : \implies\sf\ 2x(x + 3) + 1(x + 3)$

$\normalsize\ : \implies\sf\ (x + 3)(2x - 1)$

$\normalsize\ : \implies$ $\begin{cases} \underline\textbf{case 1:} \\ \sf\ (x + 3) = 0 \\ \sf\ x = 0 - 3 \\ {\boxed{\sf{x = - 3}}} \end{cases}$

$\normalsize\ : \implies$$\begin{cases} \underline\textbf{case 2:} \\ \sf\ (2x -1) = 0 \\ \sf\ 2x = 1 \\ x = \frac{1}{2} \\ {\boxed{\sf{x = \frac{1}{2} }}} \end{cases}$

$\normalsize\ : \implies{\underline{\boxed{\sf \blue{Hence, \: the \: value \: of \: \alpha = - 3 \: and \: \beta = \frac{1}{2} }}}}$

$\normalsize\star{\underline{\boxed{\mathbb \red{VERIFICATION -}}}}$

$\underline{\frak{Sum \: of \: zeroes:}}$

$\normalsize\ : \implies\sf\ \alpha + \beta = \frac{-(Coefficient \: of \: x)}{Coefficient \: of \: x^2}$

$\normalsize\ : \implies\sf\ -3 + \frac{1}{2} = \frac{-b}{a}$

$\normalsize\ : \implies\sf\frac{-6 + 1}{2} = \frac{5}{2}$

$\normalsize\ : \implies\sf\frac{5}{2}= \frac{5}{2}$

$\underline{\frak{Product \: of \: zeroes:}}$

$\normalsize\ : \implies\sf\ \alpha\beta = \frac{constant \: term}{coefficient \: of \: x^2}$

$\normalsize\ : \implies\sf\ -3 \times\ \frac{1}{2} = \frac{c}{a}$

$\normalsize\ : \implies\sf\frac{-3}{2} = \frac{-3}{2}$

$\normalsize\ : \implies{\underline{\boxed{\sf \green{Hence \: Verified \:!! }}}}$