Question

find the quadratic polynomial for the zeroes alpha, beta given in each case 1/4,-1​

Answer 1

✪AnSwEr

${\tt{\red{\underline{\large{Given}}}}}$

• Zeroes of polynomial
• 1/4 and -1

${\sf{\green{\underline{\large{To\:Find}}}}}$

• Polynomial
• Relationship between cofficient

${\sf{\pink{\underline{\Large{Explanation}}}}}$

Let the zeroes of the polynomial be$\tt\alpha=\frac{1}{4}{and}\beta={-1}$

Then,

$\rightarrow\tt\alpha{+}\beta{=}\frac{-b}{a}=\frac{1}{4}-{1}$

=>-3/4

&

$\rightarrow\tt\alpha{\times}\beta{=}\frac{c}{a}=\frac{1}{4}\times{-1}$

=>-1/4

Here,

a=4

b=-3

C=-1

New polynomial

=>ax²+bc+c

=>4x²-3x-1

Additional Information

☞

$\rightarrow\tt\alpha{+}\beta{=}\dfrac{3}{4}$

$\rightarrow\tt\alpha{+}\beta{=}\dfrac{Cofficient\:of\:X}{Cofficient\:of\:x^2}=$

&

$\rightarrow\tt\alpha{\times}\beta{=}\dfrac{-1}{4}$

$\rightarrow\tt{\large\alpha{\times}\beta{=}\dfrac{Constant\:term}{Cofficient\:of\:x^2}}$

Hence,relation verified

Answer 2

S O L U T I O N :

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

We have α and β,we get;

α = 1/4

β = -1

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

The quadratic polynomial.

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

$\underline{\mathcal{SUM\:OF\:THE\:ZEROES\::}}}$

$\mapsto\sf{\alpha +\beta =\dfrac{-b}{a} =\dfrac{Coefficient\:of\:x}{Coefficient\:of\:x^{2} } }\\\\\\\mapsto\sf{\dfrac{1}{4} +(-1)}\\\\\\\mapsto\sf{\dfrac{1}{4} -1}\\\\\\\mapsto\sf{\dfrac{1-4}{4} }\\\\\\\mapsto\sf{\dfrac{-3}{4} }$

$\underline{\mathcal{PRODUCT\:OF\:THE\:ZEROES\::}}}$

$\mapsto\sf{\alpha \times \beta =\dfrac{c}{a} =\dfrac{Constant\:term}{Coefficient\:of\:x^{2} } }\\\\\\\mapsto\sf{\dfrac{1}{4} \times (-1)}\\\\\\\mapsto\sf{\dfrac{-1}{4} }$

Thus;

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

$\mapsto\sf{x^{2} -(sum\:of\:zeroes)x+(product\:of\:zeroes)}\\\\\\\mapsto\sf{x^{2} -\bigg(-\dfrac{3}{4} \bigg)x+\bigg(-\dfrac{1}{4} \bigg)}\\\\\\\mapsto\sf{x^{2} +\dfrac{3}{4} x-\dfrac{1}{4} }\\\\\\\mapsto\bf{x^{2} +3x-1}$