Question : If $xy=48$ and $x^2+y^2=100$, then $(x+y)$ is:
Option 1: 12
Option 2: 16
Option 3: 18
Option 4: 14
New: SSC CHSL Tier 2 answer key released | SSC CHSL 2024 Notification PDF
Recommended: How to crack SSC CHSL | SSC CHSL exam guide
Don't Miss: Month-wise Current Affairs | Upcoming government exams
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: 14
Solution : Given: $xy = 48$ and $x^{2}+y^{2} = 100$ We know the algebraic identity, $(x+y)^{2} = x^{2}+y^{2}+2xy$ By substituting the values in the above equation, we get: $⇒(x+y)^{2} = 100+2×48$ $⇒(x+y) = \sqrt{196}$ $\therefore (x+y) = 14$ Hence, the correct answer is 14.
Candidates can download this e-book to give a boost to thier preparation.
Result | Eligibility | Application | Admit Card | Answer Key | Preparation Tips | Cutoff
Question : Simplify $(3 x+2 y)^2-(3 x-2 y)^2$.
Question : If $\small x+y+z=6$ and $xy+yz+zx=10$, then the value of $x^{3}+y^{3}+z^{3}-3xyz$ is:
Question : If $x+y=4, x^{2}+y^{2}=14$ and $x>y$, then the correct value of $x$ and $y$ is:
Question : If $xy+yz+zx=0$, then $(\frac{1}{x^2–yz}+\frac{1}{y^2–zx}+\frac{1}{z^2–xy})$$(x,y,z \neq 0)$ is equal to:
Question : If $x^2+\frac{1}{x^2}=4$, then what is the value of $x^4+\frac{1}{x^4}$?
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile