Some Import1. Find the zeroes of the quadratic polynomialp(x) = 9x2 -

1.	Some Import1. Find the zeroes of the quadratic polynomialp(x) = 9x2 - 42. Find the quadratic polynomial whose zeroes are1 and -3Evergreen Sell-Study in Mathematics
Answer» 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; $\longrightarrow\sf{9x^{<klux>2</klux>} -4=0}\\\\\longrightarrow\sf{(3x)^{2} -(2)^{2} =0}\\\\\longrightarrow\sf{(3x+2)(3x-2)=0\:\:\:[\therefore\:using\:a^{2} -b^{2} ]}\\\\\longrightarrow\sf{3x+2=0\:\:\:Or\:\:\:3x-2=0}\\\\\longrightarrow\sf{3x=-2\:\:\:Or\:\:\:3x=2}\\\\\longrightarrow\sf{\orange{x=\dfrac{-2}{3} \:\:\:Or\:\:\:x=\dfrac{2}{3} }}$ ∴ 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; $\boxed{\bf{The\:required\:quadratic\:polynomial\::}}}}$ $\longrightarrow\sf{x^{2} -(sum\:of\:the\:zeroes)+(product\:of\:the\:zeroes)}\\\\\longrightarrow\sf{x^{2} -(-2)x+(-3)=0}\\\\\longrightarrow\sf{\pink{x^{2} +2x-3=0}}$

Some Import1. Find the zeroes of the quadratic polynomialp(x) = 9x2 - 42. Find the quadratic polynomial whose zeroes are1 and -3Evergreen Sell-Study in Mathematics

Answer»

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;

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

∴ 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;

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

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