Question

If ${\left(a-5\right)}^2+{\left(b-c\right)}^2+{\left(c-d\right)}^2+{\left(b+c+d-9\right)}^2=0$, then the value of $\left(a+b+c\right)\left(b+c+d\right)$ is:

• A. $0$
• B. $11$
• C. $33$
• D. $99$

Answer 1

The correct option is D

$99$

Solution:

Step 1: The sum of all the squares is zero, so they all have to be zero:

Given:

${\left(a-5\right)}^2+{\left(b-c\right)}^2+{\left(c-d\right)}^2+{\left(b+c+d-9\right)}^2=0$.

$$\begin{gathered} a^2+b^2=0 \\ Itmeansa=b=0 \end{gathered}$$

Thus we can write the given equation as,

.$$\begin{gathered} \Rightarrow a-5=0ora=5...\left(i\right) \\ \Rightarrow b-c=0orb=c...\left(ii\right) \\ \Rightarrow c-d=0orc=d...\left(iii\right) \\ \Rightarrow b+c+d-9=0...\left(iv\right) \end{gathered}$$

Step 2: Solve for the values of $\mathit{b}\mathbf{,}\mathit{c}\mathbf{,}\mathit{d}\mathbf{.}$

$$\begin{gathered} \therefore b=c=d...\left(v\right)[Fromequation(ii)and(iii).] \\ So,byu\sin gequation\left(v\right)and\left(iv\right),weget, \\ \Rightarrow 3b=9[b+c+d-9=0] \\ \therefore b=3 \end{gathered}$$

Now we have the values $a=5,b=3,c=3andd=3.$

Step 3: Finding the value of $\left(a+b+c\right)\left(b+c+d\right)\mathbf{}$

By putting the values of $a=5,b=3,c=3andd=3$ we get,

$$\begin{gathered} =(a+b+c)(b+c+d) \\ =(5+3+3)(3+3+3) \\ =\left(11\right)\left(9\right) \\ =99. \end{gathered}$$

Final answer: Hence, correct answer is option (D).