Question

If the roots of the equation $px^2 + qx + r = 0$, where $2p, q, 2r$ are in G.P., are of the form $\alpha^2, 4\alpha - 4$. Then the value of $2p + 4q + 7r$ is:

• A. 0
• B. 10
• C. 14
• D. 18

Answer 1

The correct answer is not among the given options. The value of the expression is 82p, which is 82 when p=1.

Answer 2

We start with the quadratic

$$px^2+qx+r=0$$

whose roots are given to be

$$x_1=\alpha^2,\qquad x_2=4\alpha-4.$$

Also, the numbers

$$2p,\; q,\; 2r$$

are in a geometric progression (G.P.). (Recall that three numbers $a, b, c$ are in G.P. if and only if $b^2=ac$.)

Step 1. Use Vieta’s formulas.

For the quadratic we have

$$x_1+x_2 = -\frac{q}{p} \quad\Longrightarrow\quad \alpha^2+(4\alpha-4)=-\frac{q}{p},$$

so that

$$q=-p\Bigl[\alpha^2+4\alpha-4\Bigr].$$

Also,

$$x_1x_2=\frac{r}{p} \quad\Longrightarrow\quad \alpha^2(4\alpha-4)=\frac{r}{p},$$

so that

$$r=p\,\alpha^2(4\alpha-4).$$

Step 2. Use the G.P. condition on $2p,\;q,\;2r$.

Since these are in G.P.,

$$q^2=(2p)(2r)=4pr.$$

Substitute the expressions for $q$ and $r$:

$$\Bigl[-p\bigl(\alpha^2+4\alpha-4\bigr)\Bigr]^2= 4p\Bigl[p\,\alpha^2(4\alpha-4)\Bigr].$$

That is,

$$p^2\Bigl(\alpha^2+4\alpha-4\Bigr)^2= 4p^2\,\alpha^2(4\alpha-4).$$

Since $p\ne0$ we cancel $p^2$ to obtain

$$\Bigl(\alpha^2+4\alpha-4\Bigr)^2= 4\alpha^2 (4\alpha-4).$$

Step 3. Solve for $\alpha$.

Notice that

$$4\alpha-4=4(\alpha-1),$$

so the equation becomes

$$\bigl(\alpha^2+4\alpha-4\bigr)^2= 16\alpha^2(\alpha-1).$$

It is not hard to check that $\alpha=2$ is a solution. (In fact one may verify that the equation factors as $(\alpha-2)^4=0$ so that $\alpha=2$ is the unique solution.)

Step 4. Find $q$ and $r$ in terms of $p$.

Taking $\alpha=2$ we have

$$\alpha^2+4\alpha-4=4+8-4=8,\qquad\text{and}\qquad 4\alpha-4=8-4=4.$$

Thus,

$$q=-p\,(8)=-8p,$$

and

$$r=p\,(2^2)(4)=p\,(4)(4)=16p.$$

Step 5. Compute $2p+4q+7r$.

Substitute $q=-8p$ and $r=16p$:

$$2p+4q+7r=2p+4(-8p)+7(16p)=2p-32p+112p=(2-32+112)p=82\,p.$$

Because the quadratic $px^2+qx+r=0$ is determined only up to a nonzero multiplicative constant, one may choose any nonzero value for $p$. (In other words, the “numerical value” of an expression like $2p+4q+7r$ — which turns out to be $82p$ — is only unique if a normalization such as “the quadratic is monic” is imposed; here that would mean taking $p=1$ so that $2p+4q+7r=82$.)

Final remark.
The options given in the problem are
  (a) 0  (b) 10  (c) 14  (d) 18.
None of these equals 82. (Also, note that if one simply writes the answer in terms of $p,$ one obtains $82p,$ which depends on the arbitrary scale factor $p.$)

Thus, under the usual convention of choosing the quadratic to be monic (i.e. $p=1$), the value comes out as $82.$ Since $82$ is not one of the options, we conclude that none of the provided options is correct.