Let a,b be two non-zero real numbers. If p and r are the roots of the equation x% - 8ax + 2a = 0 and q and s are the roots of the equation x? + 12bx + 6b = 0, El Ea cen -1_p-1j such that Sos arein AP, then a b~" is equal to -

1. Root relations for a quadratic

For a quadratic in the form
$x^2 + Cx + D = 0$ we always have
• Sum of roots = $-C$
• Product of roots = $D$

2. Data from the two given quadratics

1. Equation-1: $x^2 - 8ax + 2a = 0$
   Roots: $p , r$
   • $p + r = 8a$
   • $pr = 2a$

2. Equation-2: $x^2 + 12bx + 6b = 0$
   Roots: $q , s$
   • $q + s = -12b$
   • $qs = 6b$

3. Turn roots upside-down (reciprocals)

We often need $\frac1p + \frac1r$, so we note: $\frac1p + \frac1r = \frac{p + r}{pr} = \frac{8a}{2a} = 4$ Likewise $\frac1q + \frac1s = \frac{q + s}{qs} = \frac{-12b}{6b} = -2$ Call the reciprocals $x = \frac1p,\; y = \frac1q,\; z = \frac1r,\; w = \frac1s$ so we already know two sums: $x + z = 4 \quad\text{and}\quad y + w = -2$

4. A.P. conditions for four consecutive terms

Because $x,\,y,\,z,\,w$ are in an Arithmetic Progression,

1. $2y = x + z$ (the second term is the mean of its neighbours)
2. $2z = y + w$ (the third term is the mean of its neighbours)

5. Solve step-by-step

• From (1):
$2y = x + z = 4 \implies y = 2$

• Plug $y$ into (2):
$2z = y + w = 2 + w \implies w = 2z - 2$

But we also have $y + w = -2$: $2 + w = -2 \;\Rightarrow\; w = -4$ Insert this $w$ back:
$w = 2z - 2 \implies -4 = 2z - 2 \implies z = -1$

Finally use $x + z = 4$: $x + (-1) = 4 \implies x = 5$

So $\frac1p = 5,\; \frac1q = 2,\; \frac1r = -1,\; \frac1s = -4$

Hence $p = \frac15,\; q = \frac12,\; r = -1,\; s = -\frac14$

6. Extract a and b from their respective equations

For a: Root r = -1 satisfies $x^2 - 8ax + 2a = 0$. Plug $x = -1$: $1 + 8a + 2a = 0 \;\Rightarrow\; 10a = -1 \;\Rightarrow\; a = -\frac1{10}$

For b: Root q = \frac12 satisfies $x^2 + 12bx + 6b = 0$. Plug $x = \tfrac12$: $\left(\tfrac12\right)^2 + 12b\left(\tfrac12\right) + 6b = 0$ $\tfrac14 + 6b + 6b = 0 \;\Rightarrow\; \tfrac14 + 12b = 0 \;\Rightarrow\; b = -\frac1{48}$

7. Required ratio

$\frac{a}{b} = \frac{-\dfrac1{10}}{-\dfrac1{48}} = \frac{48}{10} = \frac{24}{5}$

Hence, $a\,/\,b = 24/5$.

🧒 Imagine two Magic Machines (Quadratic Equations)

1. First Machine pops out two numbers p and r when you feed it a special "key" called a.
2. Second Machine pops out two numbers q and s when you feed it another key called b.

Now you flip every number upside-down (take its reciprocal). So you look at 1/p , 1/q , 1/r , 1/s.

The cool rule in the question says these four flipped numbers sit nicely on a straight number line at equal gaps (they form an Arithmetic Progression, A.P.).

Because the gaps are equal, the middle of the first three is also the average of its neighbours, and the middle of the last three is the average of its neighbours. Those two simple "equal gap" facts are all you need.

Using the usual A.P. shortcut and the well-known sum and product tricks for quadratic roots, you can quickly find the exact values of a and b and then their ratio a/b.

In the end the magic ratio comes out to 24/5.

Step-by-Step Solution

1. For $x^2 - 8ax + 2a = 0$ (roots $p,r$): $p + r = 8a, \quad pr = 2a$
2. For $x^2 + 12bx + 6b = 0$ (roots $q,s$): $q + s = -12b, \quad qs = 6b$
3. Define $x = 1/p,\; y = 1/q,\; z = 1/r,\; w = 1/s$. Then $x + z = 4 \quad\text{and}\quad y + w = -2$
4. Because $x,y,z,w$ are in A.P.: $2y = x + z \implies y = 2$ $2z = y + w \implies w = 2z - 2$ Using $y + w = -2$ gives $w = -4$ and hence $z = -1$. Finally $x = 4 - z = 5$.
5. Therefore $p = \tfrac15,\; q = \tfrac12,\; r = -1,\; s = -\tfrac14$
6. Substitute $r = -1$ into the first quadratic: $1 + 8a + 2a = 0 \;\Rightarrow\; a = -\tfrac1{10}$
7. Substitute $q = \tfrac12$ into the second quadratic: $\tfrac14 + 12b\cdot\tfrac12 + 6b = 0 \;\Rightarrow\; b = -\tfrac1{48}$
8. Compute the ratio: $\frac{a}{b} = \frac{-\tfrac1{10}}{-\tfrac1{48}} = \frac{48}{10} = \boxed{\tfrac{24}{5}}$