Question : If $a+b+c=16,ab+bc+ca=81$, then what is the value of $a^2+b^2+c^2$?
Option 1: 94
Option 2: 98
Option 3: 89
Option 4: 87
New: SSC CHSL tier 1 answer key 2024 out | 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: 94
Solution : Given: $a+b+c=16, ab+bc+ca=81$ Now, $a^2+b^2+c^2$ = $(a+b+c)^2-2(ab+bc+ca)$ = $16^2-2×81$ = $256-162$ = $94$ Hence, the correct answer is 94.
Candidates can download this e-book to give a boost to thier preparation.
Result | Eligibility | Application | Admit Card | Answer Key | Preparation Tips | Cutoff
Question : For real $a, b, c$ if $a^2+b^2+c^2=ab+bc+ca$, then value of $\frac{a+c}{b}$ is:
Question : If a + b + c = 1, ab + bc + ca = –1, and abc = –1, then what is the value of a3 + b3 + c3?
Question : If $a + b + c = 12$ and $ab + bc + ca = 22$, then what is the value of $a^3 + b^3 + c^3 - 3abc ?$
Question : If $a + b + c = 0$ and $ab + bc + ca = -11$, then what is the value of $a^2+b^2+c^2$?
Question : If $a^2+b^2+c^2=ab+bc+ca$, then the value of $\frac{11a^4+13b^4+15c^4}{16a^2b^2+19b^2c^2+17c^2a^2}$ is:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile