Given:
Different equations
To find:
the equation whose solutions have -4 and 4
To determine the equations with solutions -4 and 4, we will solve each of th given equation
[tex]\begin{gathered} a)\text{ x}^2\text{ = 8} \\ x\text{ = }\pm\sqrt{8}\text{ = }\pm\sqrt{4\times2} \\ x\text{ = }\pm\text{2}\sqrt{2}\text{ \lparen not a solution of -4 and 4\rparen} \end{gathered}[/tex][tex]\begin{gathered} b)\text{ x}^2\text{ + 16 = 0} \\ x^2=\text{ -16} \\ x\text{ = }\pm\sqrt{-16} \\ root\text{ of -16 gives a complex number. Hence, no solution of -4 and 4} \end{gathered}[/tex][tex]\begin{gathered} c)\text{ 2x}^2\text{ = 32} \\ divide\text{ both sides by 2:} \\ x^2\text{ = }\frac{32}{2} \\ x^2\text{ = 16} \\ x\text{ = }\pm\sqrt{16}\text{ = }\pm4 \\ x\text{ = 4 and -4} \end{gathered}[/tex][tex]\begin{gathered} d)\text{ -3x}^2\text{ = -48} \\ divide\text{ both sides by -1:} \\ division\text{ of same signs give positive sign} \\ 3x^2\text{ = 48} \\ x^2\text{ = 48/3} \\ x^2\text{ = 16} \\ x\text{ = }\pm\sqrt{16}\text{ = }\pm4 \\ x\text{ = 4 and - 4} \end{gathered}[/tex][tex]\begin{gathered} e)\text{ }6x^2\text{ + 56 = -40} \\ 6x^2\text{ = -40 - 56} \\ 6x^2\text{ = -96} \\ x^2\text{ = -96/6} \\ x^2\text{ = -16} \\ x\text{ = }\pm\sqrt{-16} \\ root\text{ of a negative number gives a complex number.} \\ Hence,\text{ no solution of -4 and 4} \end{gathered}[/tex][tex]\begin{gathered} f)\text{ 27 - 5x}^2\text{ = -53} \\ add\text{ 5x}^2\text{ }to\text{ both sides:} \\ 27\text{ - 5x}^2+\text{ 5x}^2\text{ = -53 + 5x}^2 \\ 27\text{ = -53 + 5x}^2 \\ \\ add\text{ 53 to btoh sides:} \\ 27\text{ + 53 = 5x}^2 \\ 80\text{ = 5x}^2 \\ divide\text{ both sides by 5:} \\ \frac{80}{5}=\text{ x}^2 \\ x^2\text{ = 16} \\ x\text{ = }\pm\sqrt{16}\text{ = }\pm4 \\ x\text{ = 4 and -4} \end{gathered}[/tex]