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1=4\left(x-2\right)^{2}
Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)^{2}.
1=4\left(x^{2}-4x+4\right)
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-2\right)^{2}.
1=4x^{2}-16x+16
Use the distributive property to multiply 4 by x^{2}-4x+4.
4x^{2}-16x+16=1
Swap sides so that all variable terms are on the left hand side.
4x^{2}-16x+16-1=0
Subtract 1 from both sides.
4x^{2}-16x+15=0
Subtract 1 from 16 to get 15.
a+b=-16 ab=4\times 15=60
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx+15. To find a and b, set up a system to be solved.
-1,-60 -2,-30 -3,-20 -4,-15 -5,-12 -6,-10
Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 60.
-1-60=-61 -2-30=-32 -3-20=-23 -4-15=-19 -5-12=-17 -6-10=-16
Calculate the sum for each pair.
a=-10 b=-6
The solution is the pair that gives sum -16.
\left(4x^{2}-10x\right)+\left(-6x+15\right)
Rewrite 4x^{2}-16x+15 as \left(4x^{2}-10x\right)+\left(-6x+15\right).
2x\left(2x-5\right)-3\left(2x-5\right)
Factor out 2x in the first and -3 in the second group.
\left(2x-5\right)\left(2x-3\right)
Factor out common term 2x-5 by using distributive property.
x=\frac{5}{2} x=\frac{3}{2}
To find equation solutions, solve 2x-5=0 and 2x-3=0.
1=4\left(x-2\right)^{2}
Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)^{2}.
1=4\left(x^{2}-4x+4\right)
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-2\right)^{2}.
1=4x^{2}-16x+16
Use the distributive property to multiply 4 by x^{2}-4x+4.
4x^{2}-16x+16=1
Swap sides so that all variable terms are on the left hand side.
4x^{2}-16x+16-1=0
Subtract 1 from both sides.
4x^{2}-16x+15=0
Subtract 1 from 16 to get 15.
x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 4\times 15}}{2\times 4}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -16 for b, and 15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-16\right)±\sqrt{256-4\times 4\times 15}}{2\times 4}
Square -16.
x=\frac{-\left(-16\right)±\sqrt{256-16\times 15}}{2\times 4}
Multiply -4 times 4.
x=\frac{-\left(-16\right)±\sqrt{256-240}}{2\times 4}
Multiply -16 times 15.
x=\frac{-\left(-16\right)±\sqrt{16}}{2\times 4}
Add 256 to -240.
x=\frac{-\left(-16\right)±4}{2\times 4}
Take the square root of 16.
x=\frac{16±4}{2\times 4}
The opposite of -16 is 16.
x=\frac{16±4}{8}
Multiply 2 times 4.
x=\frac{20}{8}
Now solve the equation x=\frac{16±4}{8} when ± is plus. Add 16 to 4.
x=\frac{5}{2}
Reduce the fraction \frac{20}{8} to lowest terms by extracting and canceling out 4.
x=\frac{12}{8}
Now solve the equation x=\frac{16±4}{8} when ± is minus. Subtract 4 from 16.
x=\frac{3}{2}
Reduce the fraction \frac{12}{8} to lowest terms by extracting and canceling out 4.
x=\frac{5}{2} x=\frac{3}{2}
The equation is now solved.
1=4\left(x-2\right)^{2}
Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)^{2}.
1=4\left(x^{2}-4x+4\right)
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-2\right)^{2}.
1=4x^{2}-16x+16
Use the distributive property to multiply 4 by x^{2}-4x+4.
4x^{2}-16x+16=1
Swap sides so that all variable terms are on the left hand side.
4x^{2}-16x=1-16
Subtract 16 from both sides.
4x^{2}-16x=-15
Subtract 16 from 1 to get -15.
\frac{4x^{2}-16x}{4}=-\frac{15}{4}
Divide both sides by 4.
x^{2}+\left(-\frac{16}{4}\right)x=-\frac{15}{4}
Dividing by 4 undoes the multiplication by 4.
x^{2}-4x=-\frac{15}{4}
Divide -16 by 4.
x^{2}-4x+\left(-2\right)^{2}=-\frac{15}{4}+\left(-2\right)^{2}
Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-4x+4=-\frac{15}{4}+4
Square -2.
x^{2}-4x+4=\frac{1}{4}
Add -\frac{15}{4} to 4.
\left(x-2\right)^{2}=\frac{1}{4}
Factor x^{2}-4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(x-2\right)^{2}}=\sqrt{\frac{1}{4}}
Take the square root of both sides of the equation.
x-2=\frac{1}{2} x-2=-\frac{1}{2}
Simplify.
x=\frac{5}{2} x=\frac{3}{2}
Add 2 to both sides of the equation.