Question

five real numbers r,a,b,c,d are such that
{(r-1)^(1/2)+2*(a-4)^(1/2)+3*(b-9)^(1/2)+4*(c-16)^(1/2)+5*(d-25)^(1/2)}=((r+a+b+c+d)/2)
Find the value of r+a+b+c+d=
a)55 b)210 c)not uniquely determined d)110

Answer 1

Answer on Question #83150 – Math – Algebra

Question

Five real numbers $r, a, b, c, d$ are such that

$$\sqrt{r - 1} + 2\sqrt{a - 4} + 3\sqrt{b - 9} + 4\sqrt{c - 16} + 5\sqrt{d - 25} = \frac{a + b + c + d + r}{2}$$

Find the value of $(r + a + b + c + d)$:

a) 55

b) 210

c) not uniquely determined

d) 110

Solution

$$\sqrt{r - 1} + 2\sqrt{a - 4} + 3\sqrt{b - 9} + 4\sqrt{c - 16} + 5\sqrt{d - 25} = \frac{a + b + c + d + r}{2}$$ $$\begin{array}{l} r - 1 = f^2, f \geq 0 \\ a - 4 = t^2, t \geq 0 \\ b - 9 = k^2, k \geq 0 \\ c - 16 = l^2, l \geq 0 \\ d - 25 = p^2, p \geq 0 \end{array}$$

Then equality can be rewritten as:

$$f + 2t + 3k + 4l + 5p = \frac{f^2 + 1 + t^2 + 4 + k^2 + 9 + l^2 + 16 + p^2 + 25}{2}$$ $$(f^2 - 2f + 1) + (t^4 - 4t + 4) + (k^2 - 6k + 9) + (l^2 - 8l + 16) + (p^2 - 10p + 25) = 0$$ $$(f - 1)^2 + (t - 2)^2 + (k - 3)^2 + (l - 4)^2 + (p - 5)^2 = 0$$

The sum of the squares of numbers is zero if each of the squares is zero, therefore:

$$f = 1; \; t = 2; \; k = 3; \; l = 4; \; p = 5$$

In this case:

$$r = 1 + f^2 = 1 + 1^2 = 2$$

Similarly:

$$a = 4 + 2^2 = 8; \; b = 9 + 3^2 = 18; \; c = 16 + 4^2 = 32; \; d = 25 + 5^2 = 50$$ $$a + b + c + d + r = 2 + 8 + 18 + 32 + 50 = 110$$

Answer: Sum is 110 (answer is option d).

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