If x, y and z are real numbers such that (x - 4)2 + (y - 5)2 + (z - 6)2 = 0 then (x + y + z) is equal to
1). -15
2). 3
3). 7
4). 15
We know that if the sum of squares of real numbers = 0 then each number is individually 0.
Here given that, (x - 4)2 + (y - 5)2 + (z - 6)2 = 0
From above information we can say,
x – 4 = 0 ⇒ x = 4;
y – 5 = 0 ⇒ y = 5;
z – 6 = 0 ⇒ z = 6;
∴ (x + y + z) = (4 + 5+ 6) = 152. If x = 3 + 2√2, find the value of \({\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)^2}\)
3. Let ABC be an equilateral triangle and AD perpendicular to BC. Then $AB^{2}+BC^{2}+CA^{2}$ =
4. If $a^{3} + b^{3}$ = 28 and a + b = 4, then what is the value of ab?
7. If $ p^{2}+q^{2}=7 pq$ , then the value of $\frac{a}{b}+\frac{b}{a}$ is equal to?