Algebra · real student question

Find the real solutions of the equation (3x - 5)^2 - 8(3x - 5) + 16 = 0.

Question

Find the real solutions of the following equation:

(3x5)28(3x5)+16=0.(3x-5)^2-8(3x-5)+16=0.

Step-by-step solution

  1. Identify the repeated expression. The binomial 3x53x-5 occurs squared and to the first power, so a substitution collapses the equation to a plain quadratic.

  2. Set u = 3x - 5. The equation becomes u28u+16=0u^2-8u+16=0. Note 16=4216=4^2 and 8=248=2\cdot 4, which is exactly the pattern of a perfect square trinomial.

  3. Factor as a perfect square. u28u+16=(u4)2u^2-8u+16=(u-4)^2, so (u4)2=0(u-4)^2=0 forces u=4u=4. Because the factor is repeated, this is a double root rather than two distinct values.

  4. Solve for x. From 3x5=43x-5=4 we get 3x=93x=9, so x=3x=3. There is only one solution, and it has multiplicity two.

  5. Substitute back. With x=3x=3: 3x5=43x-5=4, so 428(4)+16=1632+16=04^2-8(4)+16=16-32+16=0. The left side is exactly zero, confirming the single repeated root.

Answer

x=3 (a double root)x=3\ \text{(a double root)}

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