Br ME Cue NN RVI Wo Ca TPN]. 1h RL ax wn ECE 1S BARE AEE Ee Te RE oy Rl PANS. 141 Let ond b be the roots of the equation x? =10cx - 11d = 0 and those of x? =10ax - 11b = 0 are c, d, then find the value of a + b + c + d, assuming that a, b, c, d are distinct.

Key ideas you must know

1. Vieta’s Relations For a quadratic $x^2+px+q=0$ having roots $r_1,\,r_2$: $r_1+r_2 = -p \quad\text{and}\quad r_1 r_2 = q$
2. Symmetric manipulation Often, instead of finding every root separately, you can set up equations that link their sums and products. That is exactly what Vieta gives us.

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Step-by-step logical chain

1. Write Vieta for the first quadratic
   • Sum of its roots: $a+b = 10c$
   • Product of its roots: $ab = 11d$
2. Write Vieta for the second quadratic
   • Sum of its roots: $c+d = 10a$
   • Product of its roots: $cd = 11b$
3. Try to eliminate $d$ and $b$ first. From $ab = 11d$ we get $d = \dfrac{ab}{11}$. From $cd = 11b$ we get $b = \dfrac{cd}{11}$.
4. Get an a–c relation. Substituting $b$ into $cd = 11b$ and simplifying quickly gives $ac = 121$
5. Reduce everything to a single variable (here we keep only $a$):
   • $c = \dfrac{121}{a}$
   • $b = \dfrac{1210}{a} - a$ (after using $a+b=10c$)
   • $d = \dfrac{ab}{11}$ (plug the above $b$)
6. Use $c+d = 10a$ to obtain one cubic equation in $a$ only. A neat factorisation shows that the ‘easy’ value $a=11$ is not allowed (it would make $a$ and $c$ equal). The remaining choices of $a$ are both solutions of $a^2 + 121a + 121 = 0$
7. Magic cancellation inside the total. The sum $S = a+b+c+d$ turns into $S = 10\Bigl(a + \frac{121}{a}\Bigr)$ and, because every acceptable $a$ satisfies $a + \dfrac{121}{a} = -121$, we finally get $S = -1210$
8. Distinctness check. For each of the two admissible $a$ values the corresponding $b,\,c,\,d$ all turn out different, so the answer is valid.

Hence, regardless of which admissible set of roots you take, $a+b+c+d = -1210$.

What does the question say?

We have two quadratic (degree-2) equations:

1. $x^2 - 10c\,x + 11d = 0$ – its two answers (roots) are $a$ and $b$.
2. $x^2 - 10a\,x + 11b = 0$ – its two answers (roots) are $c$ and $d$.

We are told that all four letters $a,\,b,\,c,\,d$ are different, and we must find the sum $a+b+c+d$.

How to think about it (kid-friendly)

Imagine two magic fruit trees:

• Tree-1 produces fruits labelled $a$ and $b$. The recipe for this tree says “add the fruit labels together and you get $10c$, multiply the labels and you get $11d$.”
• Tree-2 produces fruits labelled $c$ and $d$. Its recipe says “add the fruit labels and you get $10a$, multiply them and you get $11b$.”

By only adding and multiplying the labels (no solving scary equations yet!), we can keep replacing one letter by another and finally write everything in just one letter – say $a$. After a little tidy-up, we discover a fixed value for the grand total $a+b+c+d$ … no matter which actual numbers make the letters.

That final total turns out to be $-1210$.

Detailed Solution

Using Vieta’s relations:

1. For $x^2 - 10c\,x + 11d = 0$ (roots $a,\,b$)

   $$$$

a+b &= 10c \tag{1}\ ab &= 11d \tag{2} \end{aligned}$$

2. For $x^2 - 10a\,x + 11b = 0$ (roots $c,\,d$)

   $$$$

c+d &= 10a \tag{3}\ cd &= 11b \tag{4} \end{aligned}$$

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Step 1 Express $d$ and $b$

From (2): $d = \dfrac{ab}{11}$.

From (4): $b = \dfrac{cd}{11}$.

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Step 2 Get a clean $a$–$c$ link

Plug $b$ from (4) into (2):

$ab = 11d \;\Longrightarrow\; a\Bigl(\frac{cd}{11}\Bigr) = 11d \;\Longrightarrow\; ac = 121$

Therefore c = \frac{121}{a}. \tag{5}

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Step 3 Find $b$

Using (1):

b = 10c - a = \frac{1210}{a} - a. \tag{6}

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Step 4 Insert into (3) to get an equation in $a$

Equation (3): $c + d = 10a$ Substitute $c$ from (5) and $d=\dfrac{ab}{11}$ with $ab$ from (6):

$\frac{121}{a} + \frac{ab}{11} = 10a$

But $ab = 1210 - a^2$ (multiply $a$ with (6)). Hence

$\frac{121}{a} + \frac{1210 - a^2}{11} = 10a$

Multiply every term by $11a$:

$1331 + a(1210 - a^2) = 110a^2$

Re-arrange:

a^3 + 110a^2 - 1210a - 1331 = 0. \tag{7}

Factorising, we get

$\bigl(a-11\bigr)\bigl(a^2 + 121a + 121\bigr) = 0.$

Because the roots must be distinct, $a\neq11$ (that would make $a=c$). Thus $a$ satisfies

a^2 + 121a + 121 = 0. \tag{8}

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Step 5 Compute the required sum

We need $S = a + b + c + d.$ From (1): $a+b = 10c$, so $S = 10c + c + d = 11c + d.$ Insert $c$ from (5) and $d = \dfrac{ab}{11}$:

$S = 11\Bigl(\frac{121}{a}\Bigr) + \frac{ab}{11}.$ Since $ab = 1210 - a^2$ (above),

$S = \frac{1331}{a} + \frac{1210 - a^2}{11} = \frac{110\,(a^2+121)}{11a} = 10\Bigl(a + \frac{121}{a}\Bigr).$

From (8), divide by $a$ to get $a + 121 + \frac{121}{a} = 0 \;\Longrightarrow\; a + \frac{121}{a} = -121.$

Hence

$\boxed{S = -1210}.$

Both legitimate values of $a$ obtained from (8) give the same total, and the corresponding $b,\,c,\,d$ are indeed all different, so the answer is confirmed.