Question : If $a+b+c=0$ and $a^2+b^2+c^2=40$, then what is the value of $a b+b c+c a$?
Option 1: –30
Option 2: –20
Option 3: –25
Option 4: –40
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: –20
Solution : Given: $a+b+c=0$ and $a^2+b^2+c^2=40$ we know that, $(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ca)$ Putting the values, we get: ⇒ $0 = 40 + 2(ab+bc+ca)$ ⇒ $ab+bc+ca = -\frac{40}{2} = -20$ Hence, the correct answer is –20.
Candidates can download this e-book to give a boost to thier preparation.
Result | Eligibility | Application | Admit Card | Answer Key | Preparation Tips | Cutoff
Question : The value of $(x^{b+c})^{b–c}(x^{c+a})^{c–a}(x^{a+b})^{a–b}$, where $(x\neq 0)$ is:
Question : If $a^2+b^2+c^2=2(a-b-c)-3$, then the value of $(a+b+c)$ is:
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, b, c are real and $a^{2}+b^{2}+c^{2}=2(a-b-c)-3, $ then the value of $2a-3b+4c$ is:
Question : If the 4-digit number 48ab is divisible by 2, 5, and 7, then what is the value of (10a – b)?
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile