If a+b+c=9 and ab+bc+ca=26,find a^{2}+b^{2}+c | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) We know the algebraic identity:$(a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2(ab+bc+ca)$Given that $a+b+c = 9$ and $ab+bc+ca = 26$.Substituting these values i

Vedclass · Accepted Answer

$29$

Vedclass · Answer

(B) We know the algebraic identity:$(a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2(ab+bc+ca)$Given that $a+b+c = 9$ and $ab+bc+ca = 26$.Substituting these values into the identity:$(9)^{2} = (a^{2}+b^{2}+c^{2}) + 2(26)$$81 = (a^{2}+b^{2}+c^{2}) + 52$Subtracting $52$ from both sides:$a^{2}+b^{2}+c^{2} = 81 - 52$$a^{2}+b^{2}+c^{2} = 29$

If $a+b+c=9$ and $ab+bc+ca=26$,find $a^{2}+b^{2}+c^{2}$.

Similar Questions

Without actually calculating the cubes,find the value of $48^{3}-30^{3}-18^{3}$.

Expand $(2x - y - 5)^2$.

Expand $(x+3)(x+8)$.

Expand $\left(\frac{2x}{3} + \frac{4y}{5}\right) \left(\frac{2x}{3} - \frac{4y}{5}\right)$.

Expand $\left(\frac{x}{2}+\frac{2 y}{3}-\frac{3 z}{4}\right)^{2}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module