The value of [tex]x[/tex] is -1 and 3/2
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I assume we are asked to solve for [tex] x[/tex] in the equation:
[tex]\displaystyle \frac{6}{x}+2=4x[/tex]
To make the equation simpler, multiply both sides by [tex]x[/tex]
[tex]\displaystyle x\bigg(\frac{6}{x}+2\bigg)=x(4x)[/tex]
[tex]6+2x=4x^2[/tex]
Move the terms to the right side to get the quadratic equation:
[tex]4x^2-2x-6=0 [/tex]
In order to solve for [tex]x[/tex] in the quadratic equation, we use the quadratic formula:
[tex]\boxed{\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}} [/tex]
Set [tex] a=4[/tex], [tex] b =-2[/tex] and [tex]c=-6[/tex]
[tex]\displaystyle x=\frac{-(-2)\pm\sqrt{(-2)^2-4(4)(-6)}}{2(4)}[/tex]
[tex]\displaystyle x=\frac{2\pm\sqrt{4+96}}{8}[/tex]
[tex]\displaystyle x=\frac{2\pm\sqrt{100}}{8}[/tex]
[tex]\displaystyle x=\frac{2\pm10}{8}[/tex]
Solving,
[tex]\begin{array}{cccc} x=\dfrac{2+10}{8}& & &x=\dfrac{2-10}{8} \\ \\ x=\dfrac{12}{8} & & & x=\dfrac{-8}{8} \\ \\ \boxed{x=\dfrac{3}{2}}& & & \boxed{x=-1 }\end{array} [/tex]
Therefore, the roots are -1 and 3/2.
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Hope it helps.
