91Ó°ÊÓ

Factoring is a key process in solving quadratic equations. It involves rewriting a quadratic equation, usually in the form of \(ax^2 + bx + c\), as a product of two binomial expressions.
In our original exercise, we are tasked with factoring the quadratics under the square roots: \(x^2 + 3x - 4\) and \(x^2 - 5x + 4\).
To factor \(x^2 + 3x - 4\), we look for two numbers that multiply to \(-4\) (the constant term \(c\)) and add up to \(3\) (the linear coefficient \(b\)). These numbers are \(4\) and \(-1\). Therefore, the expression can be factored into \((x + 4)(x - 1)\).
Similarly, for \(x^2 - 5x + 4\), we need two numbers that multiply to \(4\) and sum to \(-5\), which are \(-4\) and \(-1\). This gives us \((x - 4)(x - 1)\) as the factorization.
By factoring these quadratics, we simplify the problem, making it easier to understand and solve the equations involving square roots.

Square roots appear frequently in algebraic equations, and simplifying them can often unveil essential clues for solving equations.
In this exercise, after factoring the quadratics, we need to evaluate the square roots of these factored expressions.

• For \(\sqrt{x^2 + 3x - 4}\), once factored into \((x + 4)(x - 1)\), we see it becomes \(\sqrt{(x + 4)(x - 1)}\).
• Similarly, \(\sqrt{x^2 - 5x + 4}\) simplifies to \(\sqrt{(x - 4)(x - 1)}\).

By understanding the relationships between the terms inside the square roots, we gain insight into potential cancellation and simplification opportunities. In this problem, recognizing \(x-1\) as a common factor both in square roots and on the other side of the equation helps to further transform and simplify the equation.

Algebraic manipulation is the art of rearranging and simplifying algebraic expressions to solve equations.
In our exercise, after simplifying the square roots and factoring, we use several techniques to manipulate the equation.
We introduce a useful substitution by letting \(y = \sqrt{x-1}\), which transforms the equation into a more manageable form: \(y(x+4) - y(x-4) = x-1\). This allows us to recognize that \(y\) is a common factor, which can be factored out:

• After factoring out \(y\), the expression simplifies to \(y \cdot 8 = x-1\), which directly relates \(y\) to \(x\).
• Reversing the substitution for \(y\) involves solving \(\sqrt{x-1} = \frac{x-1}{8}\). By squaring both sides, we eliminate the square root and simplify further to get \((x-1)^2 = 64(x-1)\).

These transformations reduce a complex equation into a simpler quadratic form, which can then be solved by conventional methods like factoring or the quadratic formula.