what are the possible values of x if (4x-5)^2=49? check all that apply. -4/5, -1/2, 3, 5, 7
To solve an equation with a square yo would take the square root of both sides.
(4x - 5)^2 = 49
sqrt((4x - 5)^2) = sqrt49
4x - 5 = + - 7
4x - 5 = 7 and 4x - 5 = -7
4x = 12 4x = -2
x = 3 x = -1/2
The possible value of x is 3. So option 3 is the correct answer.
"A linear equation is an equation of a straight line, written in one variable. The only power of the variable is 1. Linear equations in one variable may take the form ax + b = 0".
For the given situation,
The linear equation is given as \((4x-5)^2=49\)
Taking square root on both sides,
⇒ \(4x- 5=\sqrt{49}\)
⇒ \(4x-5=7\)
Substitute, \(x=-\frac{4}{5}\)
⇒\(4(\frac{-4}{5} )-5=7\)
⇒\(\frac{-41}{5} = 7\)
This is not equal. So x does not satisfy -4/5.
Substitute, \(x=\frac{-1}{2}\)
⇒ \(4(\frac{-1}{2} )-5=7\)
⇒ \(-7=7\)
This is not equal. So x does not satisfy -1/2.
Substitute, \(x=3\)
⇒ \(4(3)-5=7\)
⇒ \(7=7\)
This is equal. So x satisfies 3.
Substitute, \(x=5\)
⇒ \(4(5)-5=7\)
⇒ \(15=7\)
This is not equal. So x does satisfy 5.
Substitute, \(x=7\)
⇒ \(4(7)-5=7\)
⇒ \(23=7\)
This is not equal. So x does not satisfy 7.
Hence we can conclude that the possible value of x is 3. So option 3 is the correct answer.
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