Question

find thequadratic polynomial whose zeroes 2,-1​

Answer 1

✪AnSwEr

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

• zeroes of polynomial
• 2 and -1

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

• polynomial

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

Let the zeroes of the polynomial be$\tt\alpha{and}\beta$

Then,

$\tt\alpha{+}\beta=\frac{-b}{a}=2-1=1$

&

$\tt\alpha{\times}\beta{=}\frac{c}{a}=2\times{-1}=-2$

From this we conclude that

a=1

b=-1

C=-2

ax²+bx+c

• Making polynomial

x²-x-2

Additional Information

$\tt\alpha{+}\beta{=}\dfrac{{1}}{1}$

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

&

$\tt\alpha{\times}\beta{=}\dfrac{-2}{1}$

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

Hence,relation verified

Answer 2

Step-by-step explanation:

$\bf \underline{Question}$

• find thequadratic polynomial whose zeroes 2,-1

_______________________________

$\bf \underline{Given}$

• zeroes 2,-1

_______________________________

$\bf \underline{To\:Find}$

• find thequadratic polynomial

____________________________

$\bf \underline{Solve\to}$

Supposed :-

• A be the first zero i.e 2

• b be the second i.e -1

Then,

$\sf\to \: a+b=(2 - 1)=1 \\ </p><p></p><p> \sf \to\: ab=2 \times (-1)=-2 \\ </p><p> \sf \to\: x^2+(a+b)x+ab \\ </p><p> \sf\to \: (X-2)(X+1)=X^2-2X+X-2 \\ </p><p> \sf\to \blue{X^2-X-2=0}$

Hope it helps you