Question

Some Import
1. Find the zeroes of the quadratic polynomial
p(x) = 9x2 - 4
2. Find the quadratic polynomial whose zeroes are
1 and -3
Evergreen Sell-Study in Mathematics​

Answer 1

Solution 1:

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

The quadratic polynomial p(x) = 9x² - 4.

$\bf{\red{\underline{\bf{To\:find\::}}}}$

The zeroes.

$\bf{\red{\underline{\bf{Explanation\::}}}}$

We have p(x) = 9x² - 4

Zero of the polynomial p(x) = 0

So;

$\longrightarrow\sf{9x^{2} -4=0}\\\\\longrightarrow\sf{(3x)^{2} -(2)^{2} =0}\\\\\longrightarrow\sf{(3x+2)(3x-2)=0\:\:\:[\therefore\:using\:a^{2} -b^{2} ]}\\\\\longrightarrow\sf{3x+2=0\:\:\:Or\:\:\:3x-2=0}\\\\\longrightarrow\sf{3x=-2\:\:\:Or\:\:\:3x=2}\\\\\longrightarrow\sf{\orange{x=\dfrac{-2}{3} \:\:\:Or\:\:\:x=\dfrac{2}{3} }}$

∴ The α = -2/3 and β = 2/3 are the zeroes of the polynomial.  

Solution 2:

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

We have α = 1 and β = -3.

$\bf{\red{\underline{\bf{To\:find\::}}}}$

The quadratic polynomial.

$\bf{\red{\underline{\bf{Explanation\::}}}}$

$\star\:{\green{\underline{\boldsymbol{Sum\:of\:the\:zeroes\::}}}}}$

$\longrightarrow\sf{\alpha +\beta =\dfrac{-b}{a} }\\\\\\\longrightarrow\sf{\alpha +\beta= 1+(-3)}\\\\\\\longrightarrow\sf{\alpha +\beta =1-3}\\\\\\\longrightarrow\sf{\pink{\alpha +\beta =-2}}$

$\star\:{\green{\underline{\boldsymbol{Product\:of\:the\:zeroes\::}}}}}$

$\longrightarrow\sf{\alpha \beta =\dfrac{c}{a} }\\\\\\\longrightarrow\sf{\alpha \beta =1\times -3}\\\\\\\longrightarrow\sf{\pink{\alpha \beta =-3}}$

Now;

$\boxed{\bf{The\:required\:quadratic\:polynomial\::}}}}$

$\longrightarrow\sf{x^{2} -(sum\:of\:the\:zeroes)+(product\:of\:the\:zeroes)}\\\\\longrightarrow\sf{x^{2} -(-2)x+(-3)=0}\\\\\longrightarrow\sf{\pink{x^{2} +2x-3=0}}$

Answer 2

Correct Question -

Part 1 -

Find the zeroes of the quadratic polynomial p(x) = 9x2 - 4 .

Part 2 -

Find the quadratic polynomial whose zeroes are 1 and -3 .

Solution -

Part 1 -

$\sf{ \: Find \: the \: zeroes \: of \: the \: quadratic \: polynomial } \: \\ \\ \sf{p(x) = 9x2 - 4 .} \: \\ \\ \sf{ \: = > \: p(x) = 0 \: } \\ \\ \sf{ \: 9 {x}^{2} - 4 = 0} \: \\ \\ \sf{ \: = >9 {x}^{2} = 4 } \\ \\ \sf{ \: = > {x}^{2} = \dfrac{4}{9} \: } \\ \\ \sf{ \: x = \sqrt{ \dfrac{4}{9}} } \: \\ \\ \sf{ x = \dfrac{2}{3} \: and - \dfrac{2}{3} } \: \ \\ \\ \sf{so \: the \: require d \: zeroes \: are \: \dfrac{2}{3} \: and - \dfrac{2}{3} \: } ....... [A]$

Part 2 -

$\sf{ Find \: the \: quadratic \: polynomial \: whose \: zeroes \: are \: 1 \: and \: -3 . } \: \\ \\ \sf{ \: Hence \: \alpha \: and \: \beta \: are \: 1 \: and \: - 3 \: respectively. \: } \ \\ \\ \sf{ }\: \\ \sf{ Sum \: Of \: Zeroes \: - } \\ \\ \sf{ => \alpha + \beta = 1 - 3 = -2 } \\ \\ \sf{ Product \: Of \: Zeroes \: - } \\ \\ \sf{ => \alpha \beta = 1 \times -3 = -3 } \\ \\ \sf{We \: know \: that \: \: quadratic \: polynomial \: can\: be \: written \: as \: : } \: \\ \\ \sf{ \: = > \: p(x) = {x}^2 - px + q } \\ \\ \sf{ Where \: -} \: \\ \\ \sf{ p = Sum \: Of \: Zeroes } \\ \\ \sf{ q = Product Of Zeroes } \\ \\ \sf{ Substituting \: the \: required \: values \: - } \: \\ \\ \sf{ p(x) = {x}^2 + 2x -3 } ....... [A]$