Let , be the roots of the equation x^2-x+p=0 and , be the roots the equation x^2-4x+q=0; p…

Question

Let $\alpha ,\beta $ be the roots of the equation ${x}^{2}-x+p=0$ and $\gamma ,\delta $ be the roots the equation ${x}^{2}-4x+q=0$; $p,q\in Z$. If $\alpha ,\beta ,\gamma ,\delta $ are in G.P., then $|p+q|$ equals :

Accepted Answer

Let the four roots $\alpha ,\beta ,\gamma ,\delta $, which are in G.P., be represented as $a,ar,a{r}^{2}$, and $a{r}^{3}$ respectively. Here, $a$ is the first term and $r$ is the common ratio.

For the first quadratic equation ${x}^{2}-x+p=0$, the roots are $\alpha $ and $\beta $.

Using the sum and product of the roots, we get:

Sum of roots: $\alpha +\beta =-\frac{x}{{x}^{2}}$

$a+ar=1\, \implies \, a(1+r)=1\,$

Product of roots: $\alpha \beta =\frac{}{{x}^{2}}$

$a(ar)=p\, \implies \, {a}^{2}r=p\,$

For the second quadratic equation ${x}^{2}-4x+q=0$, the roots are $\gamma $ and $\delta $.

Similarly:

Sum of roots: $\gamma +\delta =-\frac{-4}{1}$

$a{r}^{2}+a{r}^{3}=4\, \implies \, a{r}^{2}(1+r)=4\,$

Product of roots: $\gamma \delta =\frac{q}{1}$

$(a{r}^{2})(a{r}^{3})=q\, \implies \, {a}^{2}{r}^{5}=q\,$

Now, we can solve for $a$ and $r$ by dividing Equation 3 by Equation 1:

$\frac{a{r}^{2}(1+r)}{a(1+r)}=\frac{4}{1}$

${r}^{2}=4$

$r=2\, \, r=-2$

Let us check both cases using the condition given in the problem that $p,q\in Z$ (they are integers).

Case 1: When $r=2$

Substitute $r=2$ into Equation 1:

$a(1+2)=1\, \implies \, 3a=1\, \implies \, a=\frac{1}{3}$

Now, finding the value of $p$ using Equation 2:

$p={a}^{2}r={(\frac{1}{3})}^{2}(2)=\frac{2}{9}$

Since $p$ must be an integer ($p\in Z$), $\frac{2}{9}$ is not valid. Thus, we reject $r=2$.

Case 2: When $r=-2$

Substitute $r=-2$ into Equation 1:

$a(1-2)=1\, \implies \, -a=1\, \implies \, a=-1$

Now, finding the value of $p$ using Equation 2:

$p={a}^{2}r=(-1{)}^{2}(-2)=1	imes (-2)=-2$

Since $-2$ is an integer, this is a valid value for $p$.

Next, we find the value of $q$ using Equation 4:

$q={a}^{2}{r}^{5}$

$q=(-1{)}^{2}(-2{)}^{5}$

$q=1	imes (-32)=-32$

Both $p$ and $q$ are integers, which satisfies the given condition.

Finally, we need to calculate the value of $|p+q|$:

$|p+q|=|-2+(-32)|$

$|p+q|=|-34|=34$

Correct Answer:

Option C

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