Question

If $f\left(x\right)=a_1{x}^2+b_{1}x+c_1$ and $\alpha ,\beta$ are the roots of $a{x}^2+bx+c=0$. Find
$f\left(\alpha \right)f\left(\beta \right)$.

• A. $\frac{{(c_{1}a-c{a}_1)}^2-(a{b}_1-a_{1}b)(b{c}_1-b_{1}c)}{a^2}$
• B. $\frac{{(b_{1}a-b{a}_1)}^2-(a{b}_1-a_{1}b)(b{c}_1-b_{1}c)}{a^2}$
• C. $\frac{{(c_{1}a-a{a}_1)}^2-(a{b}_1-a_{1}b)(b{c}_1-b_{1}c)}{a^2}$
• D. $\frac{{(a_{1}c-a{c}_1)}^2-(a_{1}b-a{b}_1)(b{c}_1-b_{1}c)}{a^2}$

Answer 1

The correct option is A $\frac{{(c_{1}a-c{a}_1)}^2-(a{b}_1-a_{1}b)(b{c}_1-b_{1}c)}{a^2}$

$f\left(\alpha \right)f\left(\beta \right)=(a_1{\alpha }^2+b_1\alpha +c_1)(a_1{\beta }^2+b_1\beta +c_1)={a_1}^2{\left(\alpha \beta \right)}^2+a_1{b}_1\beta \alpha (\alpha +\beta )+a_1{c}_1({(\alpha +\beta )}^2-2\alpha \beta )+b_1{c}_1(\alpha +\beta )+\alpha \beta b_1^2+c_1^2={a_1}^2{\left(\frac{c}{a}\right)}^2+\frac{a_1{b}_1(-bc)}{a^2}+a_1{c}_1\frac{b^2-2ac}{a^2}+b_1{c}_1\left(\frac{-b}{a}\right)+\frac{c}{a}{b}_1^2+c_1^2\frac{(a_1^2{c}^2-2a{a}_{1}c{c}_1+a^2{c}_1^2)+a_1{c}_1{b}^2-a_1{b}_{1}bc-ab{b}_1{c}_1+ac{b}_1^2}{a^2}=\frac{{(c_{1}a-c{a}_1)}^2-(a{b}_1-a_{1}b)(b{c}_1-b_{1}c)}{a^2}$