Question

Three test Poinst: x²+2x-8</0​

Answer 1

Hello Buddy!!

Required Solution:-

${x}^{2} + 2x - 8 \leqslant 0 \\ {x}^{2} + 4x - 2x - 8 \leqslant 0 \\ x(x + 4) - 2(x - 8) \leqslant 0 \\ (x + 4)(x - 2) \leqslant 0$

• Value of x is -4 (or) 2

By Substituting x = -4

${x}^{2} + 2x - 8 \leqslant 0 \\ {( - 4)}^{2} + 2( - 4) - 8 \leqslant 0 \\ 16 - 8 - 8 \leqslant 0 \\ 16 - 16 \leqslant 0 \\ 0 = 0$

Hence, Satisfied!!

By Substituting x = 2

${x}^{2} + 2x - 8 \leqslant 0 \\ {(2)}^{2} + 2(2) - 8 \leqslant 0 \\ 4 + 4 - 8 \leqslant 0 \\ 0 = 0$

Hence, Satisfied!!

• $\bold{\red{x²+2x-8 = 0}}$

@MrMonarque❤️

Hope It Helps You!!