How do you solve #7x^2 - 54 = 0#?

2 Answers
EZ as pi
Apr 30, 2016

Although this is a quadratic equation, it is an easy one because there is no term in #x#. First, change the equation into the form #x² = c#. Then find both the positive and negative square root.

Explanation:

Divide both sides by 7 first to give: #x² - 54/7 = 0#

#x² = 54/7#
#x= +-sqrt(54/7)#

#x = 2.778 or x = -2.778# (3dp)
It could be given in surd form as well.

Remember that there are always two answers to a quadratic equation.

A longer method would be to use the quadratic formula with #a = 7, b=0 and c= -54#

Meave60
Apr 30, 2016

#x=(3sqrt 6)/sqrt 7#

#x=-(3sqrt 6)/sqrt 7#

Explanation:

#7x^2-54=0#

Add #54# to both sides of the equation.

#7x^2=54#

Divide both sides by #7#.

#x^2=54/7#

Take the square root of both sides.

#x=+-sqrt(54/7)#

Simplify.

#x=+-(sqrt54)/(sqrt 7)#

Determine the prime factors for #54#.

#x=+-sqrt (2xx3xxcolor(blue)(3xx3))/(sqrt7)#

#x=+-sqrt((2xx3xxcolor(blue)(3^2)))/(sqrt7)#

Apply square root rule #sqrt(a^2)=a#.

#x=+-color(blue)(3color(black)sqrt(2xx3))/(sqrt 7)#

#x+-color(blue)(3color(black)sqrt 6)/(sqrt 7)#

Solution

#x=(3sqrt 6)/sqrt 7#

#x=-(3sqrt 6)/sqrt 7#

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