Question

Find the zero of the quadratic polynomials
P(x) = 6 xsquare-3 &
verife the relationship
between zeros &
its coefficent​

Answer 1

Solution :

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

We have quadratic polynomial p(x) = 6x² - 3.

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

The zeroes & verify the relationship between zeroes and it's coefficient.

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

We have p(x) = 6x² - 3

Zero of the polynomial is p(x) = 0

So;

$\mapsto\sf{6x^{2} -3=0}\\\\\mapsto\sf{6x^{2} =3}\\\\\mapsto\sf{x^{2} =\cancel{\dfrac{3}{6} }}\\\\\\\mapsto\sf{x^{2} =\dfrac{1}{2}} \\\\\\\mapsto\sf{x=\sqrt{\pm\dfrac{1}{2} } }\\\\\\\mapsto\sf{x=\pm\dfrac{1}{\sqrt{2} } }$

∴ The α = 1/√2 and β = -1/√2 are the zeroes of the polynomial.

As the given quadratic polynomial as we compared with ax² + bx + c

• a= 6
• b = 0
• c = -3

So;

$\green{\underline{\underline{\bf{Sum\:of\:the\:zeroes\::}}}}$

$\longrightarrow\sf{\alpha +\beta =\dfrac{-b}{a} =\dfrac{Coefficient\:of\:x}{Coefficient\:of\:x^{2} } }\\\\\\\longrightarrow\sf{\dfrac{1}{\sqrt{2} } +\bigg(-\dfrac{1}{\sqrt{2} } \bigg)=\cancel{\dfrac{0}{6}} }\\\\\\\longrightarrow\sf{\dfrac{1}{\sqrt{2} } -\dfrac{1}{\sqrt{2} } =0}\\\\\\\longrightarrow\sf{\red{0=0}}$

$\green{\underline{\underline{\bf{Product\:of\:the\:zeroes\::}}}}$

$\longrightarrow\sf{\alpha \times \beta =\dfrac{c}{a} =\dfrac{Constant\:term}{Coefficient\:of\:x^{2} } }\\\\\\\longrightarrow\sf{\dfrac{1}{\sqrt{2} } \times \bigg(-\dfrac{1}{\sqrt{2} } \bigg)=\cancel{\dfrac{-3}{6}} }\\\\\\\longrightarrow\sf{-\dfrac{1}{\sqrt{2}\times \sqrt{2} } } =\dfrac{-1}{2} }\\\\\\\longrightarrow\sf{\red{-\dfrac{1}{2} =-\dfrac{1}{2} }$

Thus;

Relationship between zeroes and coefficient is verified .

Answer 2

...

$\tt{\huge{\purple{ ................... }}}$

QUESTION -

Find the zero of the quadratic polynomials P(x) = 6 xsquare-3 &

verife the relationship between zeros & its coefficent.

SOLUTION :

The given polynomial is :

$f(x) = 6{x} ^ 2 - 3 \\ \\ Now \: f(x) = 0 \\ \\ => 6{x} ^ 2 - 3 = 0 \\ \\ => 6{ x } ^ 2 = 3 \\ \\ => { x }^ 2 = \dfrac{1}{2} \\ \\ => x = +- \dfrac{1} { \sqrt{2} }$

So,

The Zeroes of the given Polynomial are equal in magnitude and opposite in values.

Now, We need to verify the relationship between the Zeroes it's coefficent..

VERIFICATION -

The given Polynomial can be written as :

$p(x) = 6{x} ^ 2 + 0x - 3 \\ \\ We \: Know \: That \: : \\ \\ => Sum \: Of \: Zeroes \: = \dfrac{-b}{a} = \dfrac{0}{6} = 0$

$\alpha + \beta = 0 \: As \: They \: Cancel \: Each \: Other .$

$We \: know \: That \::- \\ \\ => Product \: Of \: Zeroes \: = \dfrac{c}{a} = \dfrac{-3}{6} = \dfrac{-1}{2} \\ \\ => \alpha \times \beta = \dfrac{1}{ \sqrt{2} } \times \dfrac{-1}{ \sqrt{2} } = \dfrac{-1}{2} \\ \\ \bold{ Hence \: Verified }$