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x28x+16=0x^2-8x+16=0

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Q. x28x+16=0x^2-8x+16=0
  1. Identify Structure: Identify the structure of the quadratic equation.\newlineThe given equation is in the standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0, where a=1a = 1, b=8b = -8, and c=16c = 16.
  2. Check Perfect Square: Check if the quadratic is a perfect square trinomial.\newlineA perfect square trinomial is of the form (ax)2±2abx+b2(ax)^2 \pm 2abx + b^2, which factors into (ax±b)2(ax \pm b)^2.\newlineComparing x28x+16x^2 - 8x + 16 with the perfect square trinomial structure, we see that (x)22(4)x+(4)2(x)^2 - 2(4)x + (4)^2 matches, suggesting that the equation might factor into (x4)2(x - 4)^2.
  3. Factor Equation: Factor the quadratic equation.\newlineSince x28x+16x^2 - 8x + 16 is a perfect square trinomial, it factors as (x4)(x4)(x - 4)(x - 4) or (x4)2(x - 4)^2.
  4. Solve for x: Solve the factored equation for xx.\newlineSet the factored form equal to zero: (x4)2=0(x - 4)^2 = 0.\newlineTo find the value of xx, take the square root of both sides: x4=0x - 4 = 0.
  5. Find Solution: Find the solution to the equation.\newlineAdd 44 to both sides of the equation: x=4x = 4.