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$Find one value of x that is a solution to the equation: {:[(2x-3)^(2)=4x-6],[x=◻]:}$

Find one value of $x$ that is a solution to the equation: $\newline$ $(2 x-3)^{2}=4 x-6$ $\newline$ $x=$

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Solve a quadratic equation by factoring

Full solution

Q. Find one value of

x

that is a solution to the equation:

\newline

(2 x-3)^{2}=4 x-6

\newline

x=

Expand the squared term: Expand the squared term on the left side of the equation. $\newline$ $(2x - 3)^2 = (2x - 3)(2x - 3)$ $\newline$ $= 4x^2 - 6x - 6x + 9$ $\newline$ $= 4x^2 - 12x + 9$
Rewrite the equation: Rewrite the equation with the expanded left side. $\newline$ $4x^2 - 12x + 9 = 4x - 6$
Move all terms to one side: Move all terms to one side to set the equation to zero. $\newline$ $4x^2 - 12x + 9 - 4x + 6 = 0$ $\newline$ $4x^2 - 16x + 15 = 0$
Factor the quadratic equation: Factor the quadratic equation. $\newline$ We need to find two numbers that multiply to $4 \times 15 = 60$ and add up to $-16$ . $\newline$ The numbers $-10$ and $-6$ work because $-10 \times -6 = 60$ and $-10 + -6 = -16$ . $\newline$ So we can write the factored form as: $\newline$ $(4x - 10)(x - 6) = 0$
Solve for x: Solve for x by setting each factor equal to zero. $\newline$ $4x - 10 = 0$ or $x - 6 = 0$ $\newline$ For $4x - 10 = 0$ : $\newline$ $4x = 10$ $\newline$ $x = \frac{10}{4}$ $\newline$ $x = \frac{5}{2}$ or $2.5$ $\newline$ For $x - 6 = 0$ : $\newline$ $x = 6$ $\newline$ We have two possible solutions for x: $\frac{5}{2}$ and $x - 6 = 0$ $0$ .

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