Question

3. If the zeroes of the polynomial x3-3x2+x+1 are a-b,a,a+b,find a and b​

Answer 1

Solution :

We have cubic polynomial p(x) = x³ - 3x² + x + 1 & zeroes of the polynomial are;

$\bullet\:\sf{\alpha =a-b}\\\\\bullet\sf{\beta =a}\\\\\bullet\sf{\gamma =a+b}$

As we know that given polynomial compared with ax³ + bx² + cx + d;

• a = 1
• b = -3
• c = 1
• d = 1

Now;

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

$\longrightarrow\sf{\alpha +\beta +\gamma=\dfrac{-b}{a} =\bigg\lgroup \dfrac{Coefficient\:of\:x^{2}}{Coefficient\:of\:x^{3}} \bigg\rgroup}\\\\\\\longrightarrow\sf{a\cancel{-d}+a+a\cancel{+d}=\dfrac{-(-3)}{1} }\\\\\longrightarrow\sf{3a=3}\\\\\longrightarrow\sf{a=\cancel{3/3}}\\\\\longrightarrow\bf{a=1..............(1)}$

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

$\longrightarrow\sf{\alpha\beta +\beta\gamma +\gamma\alpha=\dfrac{c}{a} =\bigg\lgroup \dfrac{Coefficient\:of\:x}{Coefficient\:of\:x^{3}} \bigg\rgroup}\\\\\\\longrightarrow\sf{(a-b)(a) + (a)(a+b) + (a+b)(a-b) = \dfrac{1}{1} }\\\\\longrightarrow\sf{a^{2} \cancel{- ab }+ a^{2} \cancel{+ ab} + a^{2} \cancel{-ab + ab} - b^{2} = 1}\\\\\longrightarrow\sf{3a^{2} - b^{2} = 1}\\\\\longrightarrow\sf{3(1)^{2} - b^{2} = 1\:\:[from(1)]}\\\\\longrightarrow\sf{3-b^{2} = 1}\\\\\longrightarrow\sf{-b^{2} = 1-3}$

$\longrightarrow\sf{\cancel{-}b^{2} = \cancel{-}2}\\\\\longrightarrow\sf{b^{2} = 2}\\\\\longrightarrow\bf{b=\pm\sqrt{2} }$

Thus;

The value of a & b will be 1 & ±√2 .