Question

Use square roots for the problem. Which equation(s) have -4 and 4 as solutions? Select all that apply

Answer 1

Answer:
`\begin{gathered} 2x^2\text{ = 32 \lparen option C\rparen} \\ -3x^2\text{ = -48 \lparen option D\rparen} \\ 27\text{ - 5x}^2\text{ = -53 \lparen option F\rparen} \end{gathered}`

Step-by-step explanation:

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

`\begin{gathered} a)\text{ x}^2\text{ = 8} \\ x\text{ = }\pm√(8)\text{ = }\pm√(4*2) \\ x\text{ = }\pm\text{2}√(2)\text{ \lparen not a solution of -4 and 4\rparen} \end{gathered}`
`\begin{gathered} b)\text{ x}^2\text{ + 16 = 0} \\ x^2=\text{ -16} \\ x\text{ = }\pm√(-16) \\ root\text{ of -16 gives a complex number. Hence, no solution of -4 and 4} \end{gathered}`
`\begin{gathered} c)\text{ 2x}^2\text{ = 32} \\ divide\text{ both sides by 2:} \\ x^2\text{ = }(32)/(2) \\ x^2\text{ = 16} \\ x\text{ = }\pm√(16)\text{ = }\pm4 \\ x\text{ = 4 and -4} \end{gathered}`
`\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√(16)\text{ = }\pm4 \\ x\text{ = 4 and - 4} \end{gathered}`
`\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√(-16) \\ root\text{ of a negative number gives a complex number.} \\ Hence,\text{ no solution of -4 and 4} \end{gathered}`
`\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:} \\ (80)/(5)=\text{ x}^2 \\ x^2\text{ = 16} \\ x\text{ = }\pm√(16)\text{ = }\pm4 \\ x\text{ = 4 and -4} \end{gathered}`