Question

If the equation $x^2 - (p+q)x + 4 = 0; x^2 - 2qx + 8 = 0$ and $x^2 - (2p+q)x + 12 = 0$ have exactly

Answer 1

6

Answer 2

The problem statement is incomplete, ending with "have exactly". Based on the typical structure of such problems in competitive exams like JEE/NEET, it is highly probable that the question implies "have exactly one common root" and asks for a specific value related to the parameters $p$ and $q$, or the common root itself. We will proceed by finding the conditions for these three quadratic equations to have a common root.

Let the common root be $\alpha$. Substituting $\alpha$ into each equation, we get:

1. $\alpha^2 - (p+q)\alpha + 4 = 0 \quad (I)$
2. $\alpha^2 - 2q\alpha + 8 = 0 \quad (II)$
3. $\alpha^2 - (2p+q)\alpha + 12 = 0 \quad (III)$

Subtract equation $(I)$ from equation $(II)$: $(\alpha^2 - 2q\alpha + 8) - (\alpha^2 - (p+q)\alpha + 4) = 0$ $\alpha^2 - 2q\alpha + 8 - \alpha^2 + p\alpha + q\alpha - 4 = 0$ $p\alpha - q\alpha + 4 = 0$ $(p-q)\alpha = -4 \quad (IV)$

Subtract equation $(II)$ from equation $(III)$: $(\alpha^2 - (2p+q)\alpha + 12) - (\alpha^2 - 2q\alpha + 8) = 0$ $\alpha^2 - 2p\alpha - q\alpha + 12 - \alpha^2 + 2q\alpha - 8 = 0$ $-2p\alpha + q\alpha + 4 = 0$ $(-2p+q)\alpha = -4 \quad (V)$

From equations $(IV)$ and $(V)$, we have: $(p-q)\alpha = -4$ $(-2p+q)\alpha = -4$

Since the right-hand sides are equal, the left-hand sides must be equal: $(p-q)\alpha = (-2p+q)\alpha$

If $\alpha = 0$, then from $(IV)$ we get $0 = -4$, which is a contradiction. Therefore, $\alpha \neq 0$. Since $\alpha \neq 0$, we can divide both sides by $\alpha$: $p-q = -2p+q$ $3p = 2q$ $q = \frac{3}{2}p$

Now substitute $q = \frac{3}{2}p$ into equation $(IV)$: $(p - \frac{3}{2}p)\alpha = -4$ $(-\frac{1}{2}p)\alpha = -4$ $p\alpha = 8$

From $p\alpha = 8$, we can express $p$ as $p = \frac{8}{\alpha}$. Then, $q = \frac{3}{2}p = \frac{3}{2} \left(\frac{8}{\alpha}\right) = \frac{12}{\alpha}$.

Substitute these expressions for $p$ and $q$ in terms of $\alpha$ into any of the original equations. Let's use equation $(I)$: $\alpha^2 - (p+q)\alpha + 4 = 0$ $\alpha^2 - \left(\frac{8}{\alpha} + \frac{12}{\alpha}\right)\alpha + 4 = 0$ $\alpha^2 - \left(\frac{20}{\alpha}\right)\alpha + 4 = 0$ $\alpha^2 - 20 + 4 = 0$ $\alpha^2 - 16 = 0$ $\alpha^2 = 16$ $\alpha = \pm 4$

Now we have two possible values for the common root $\alpha$:

Case 1: $\alpha = 4$ Using $p = \frac{8}{\alpha}$ and $q = \frac{12}{\alpha}$: $p = \frac{8}{4} = 2$ $q = \frac{12}{4} = 3$ In this case, the product $pq = 2 \times 3 = 6$.

The equations become:

1. $x^2 - (2+3)x + 4 = 0 \implies x^2 - 5x + 4 = 0 \implies (x-1)(x-4)=0$. Roots are $1, 4$.
2. $x^2 - 2(3)x + 8 = 0 \implies x^2 - 6x + 8 = 0 \implies (x-2)(x-4)=0$. Roots are $2, 4$.
3. $x^2 - (2(2)+3)x + 12 = 0 \implies x^2 - 7x + 12 = 0 \implies (x-3)(x-4)=0$. Roots are $3, 4$. All three equations have exactly one common root, which is $4$.

Case 2: $\alpha = -4$ Using $p = \frac{8}{\alpha}$ and $q = \frac{12}{\alpha}$: $p = \frac{8}{-4} = -2$ $q = \frac{12}{-4} = -3$ In this case, the product $pq = (-2) \times (-3) = 6$.

The equations become:

1. $x^2 - (-2-3)x + 4 = 0 \implies x^2 + 5x + 4 = 0 \implies (x+1)(x+4)=0$. Roots are $-1, -4$.
2. $x^2 - 2(-3)x + 8 = 0 \implies x^2 + 6x + 8 = 0 \implies (x+2)(x+4)=0$. Roots are $-2, -4$.
3. $x^2 - (2(-2)+(-3))x + 12 = 0 \implies x^2 + 7x + 12 = 0 \implies (x+3)(x+4)=0$. Roots are $-3, -4$. All three equations have exactly one common root, which is $-4$.

In both valid scenarios, the product $pq$ is $6$. This makes $pq=6$ a unique value, which is often the expected answer for such problems when the common root itself is not unique.