what method would you choose to solve the equation $2x^2 - 7 = 9$? explain why you chose this method.

what method would you choose to solve the equation $2x^2 - 7 = 9$? explain why you chose this method.

what method would you choose to solve the equation $2x^2 - 7 = 9$? explain why you chose this method.

Answer

Explanation:

Step1: Isolate the quadratic term

First, we want to get the (x^2) term by itself. Start with the equation (2x^2 - 7 = 9). Add 7 to both sides of the equation to isolate the term with (x^2). [ 2x^2 - 7 + 7 = 9 + 7 ] Simplifying both sides, we get: [ 2x^2 = 16 ]

Step2: Solve for (x^2)

Next, divide both sides by 2 to solve for (x^2). [ \frac{2x^2}{2} = \frac{16}{2} ] Simplifying, we have: [ x^2 = 8 ]

Step3: Solve for (x)

To solve for (x), take the square root of both sides. Remember that when we take the square root of a number, we get both a positive and a negative solution. [ x = \pm\sqrt{8} ] Simplify (\sqrt{8}) as (2\sqrt{2}), so: [ x = \pm 2\sqrt{2} ]

Why this method (Isolation and Square Root):

The equation (2x^2 - 7 = 9) is a quadratic equation, but it is in a form where we can easily isolate the (x^2) term (there is no (x) term, only (x^2), constant terms). This makes the square root method straightforward. If we had a quadratic with an (x) term (like (ax^2 + bx + c = 0) with (b\neq0)), we might use factoring, completing the square, or the quadratic formula, but since this is a pure quadratic (no linear term, (b = 0) in the standard (ax^2+bx + c = 0) form, here rewritten as (2x^2+0x - 16 = 0)), isolating (x^2) and taking the square root is the simplest approach.

Answer:

The method chosen is Isolation and Square Root Method. The solution steps are:

  1. Isolate (2x^2): (2x^2 - 7 + 7 = 9 + 7 \implies 2x^2 = 16)
  2. Solve for (x^2): (\frac{2x^2}{2}=\frac{16}{2}\implies x^2 = 8)
  3. Solve for (x): (x=\pm\sqrt{8}=\pm 2\sqrt{2})

This method is chosen because the quadratic equation has no linear term ((bx) term), so isolating (x^2) and taking the square root is the most straightforward approach. The solutions are (x = 2\sqrt{2}) and (x=- 2\sqrt{2}) (or approximately (x\approx\pm2.828)).