Question

Match each quadratic equation with its solution set.

Answer 1

Answer:

• 2x^2-32 = 0 ===> (-4,4)
• 4x^2 -100=0 ===> (-5,5)
• x^2 -55=9 ==>(-8, 8)
• x^2-140= -19 ===>(-11 ,11)

Step-by-step explanation: Further explanation

$2x^2-32=0\\\\\mathrm{Add\:}32\mathrm{\:to\:both\:sides}\\\\2x^2-32+32=0+32\\\\2x^2=32\\\\\frac{2x^2}{2}=\frac{32}{2}\\\\\mathrm{For\:}x^2=f\left(a\right)\\\\\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{16},\:x=-\sqrt{16}\\\\x=\sqrt{16},\:x=-\sqrt{16}\\x=4,\:x=-4$

$4x^2-100=0\\\mathrm{Add\:}100\mathrm{\:to\:both\:sides}\\4x^2-100+100=0+100\\4x^2=100\\\frac{4x^2}{4}=\frac{100}{4}\\x^2=25\\\mathrm{For\:}x^2=f\left(a\right)\\\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\x=\sqrt{25},\:x=-\sqrt{25}\\\\x=5,\:x=-5$

$x^2-140=-19\\x^2-140+140=-19+140\\x^2=121\\\mathrm{For\:}x^2=f\left(a\right)\\\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\x=\sqrt{121},\:x=-\sqrt{121}\\x=11,\:x=-11$

$x^2-55=9\\x^2-55+55=9+55\\x^2=64\\\mathrm{For\:}x^2=f\left(a\right)\\\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\x=\sqrt{64},\:x=-\sqrt{64}\\x=8,\:x=-8$

$2x^2-18=0\\2x^2-18+18=0+18\\2x^2=18\\\frac{2x^2}{2}=\frac{18}{2}\\x^2=9\\\mathrm{For\:}x^2=f\left(a\right)\\\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\x=\sqrt{9},\:x=-\sqrt{9}\\x=3,\:x=-3$

Answer 2

Answer:

2x²-32 ⇒ x²=16⇒ (-4,4)

4x²-100 ⇒x²=25 ⇒(-5,5)

x²-55=9 ⇒x²=64 ⇒(-8,8)

x²-140=-19 ⇒x²=121 ⇒(-11,11)

2x²-18=0 ⇒x²=9 ⇒(-3,3)