Question

Let alpha and beta be the roots of ax^2+bx+c=0 and alpha1 and -beta be the roots so a1x^2+b1x+c1=0, then find a quadratic equation whose roots are alpha and alpha1 .

Answer 1

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Answer 2

Answer:

The required quadratic equation is found to be:

$(aa_1)x^2+(a_1b-ab_1)x-\frac{cc_1}{\beta^2}=0$

Step-by-step explanation:

We know that the relation between the roots and coefficients of the first quadratic equation is given by:

$\alpha \beta=\frac{c}{a}$                         ...(i)

$\alpha + \beta=-\frac{b}{a}$                  ...(ii)

Similarly, the relation between the roots and coefficients of the second expression is given by:

$\alpha_1-\beta=-\frac{b_1}{a_1}$                ...(iii)

$-\alpha_1 \beta=\frac{c_1}{a_1}$                      ...(iv)

by (ii) and (iii), we get:

$\alpha + \frac{b}{a}=-\alpha_1+\frac{b_1}{a_1}$

$\alpha +\alpha_1=\frac{b_1}{a_1}- \frac{b}{a}$

$\alpha +\alpha_1=\frac{ab_1-a_1b}{aa_1}$              ...(v)

We can write the multiplication of the alphas as:

$\alpha \alpha_1=-\frac{cc_1}{\beta^2aa_1}$                   ...(vi)

Now, finding the coefficients, we get:

$\alpha+\alpha_1=-\frac{b_n}{a_n}=\frac{a_1b-ab_1}{aa_1}$

and

$\alpha \alpha_1=\frac{c_n}{a_n}=-\frac{cc_1}{\beta^2aa_1}$

by these, we can say:

$a_n=aa_1$, $b=a_1b-ab_1$, $c = -\frac{cc_1}{\beta^2}$

Thus, the required equation will be:

$(aa_1)x^2+(a_1b-ab_1)x-\frac{cc_1}{\beta^2}=0$

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