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Sagot :
To solve the equation [tex]\((x - 3)^2 + 2(x - 3) - 8 = 0\)[/tex] using the substitution method, follow these steps:
1. Perform the substitution:
Let [tex]\( u = (x - 3) \)[/tex]. This changes the equation to [tex]\( u^2 + 2u - 8 = 0 \)[/tex].
2. Recognize the standard quadratic form:
We now have a quadratic equation in terms of [tex]\(u\)[/tex]:
[tex]\[ u^2 + 2u - 8 = 0 \][/tex]
3. Solve the quadratic equation:
Use the quadratic formula [tex]\( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex].
Here, [tex]\(a = 1\)[/tex], [tex]\(b = 2\)[/tex], and [tex]\(c = -8\)[/tex].
4. Calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot (-8) = 4 + 32 = 36 \][/tex]
5. Find the roots of the quadratic equation:
[tex]\[ u_1 = \frac{-b + \sqrt{\Delta}}{2a} = \frac{-2 + \sqrt{36}}{2 \cdot 1} = \frac{-2 + 6}{2} = 2 \][/tex]
[tex]\[ u_2 = \frac{-b - \sqrt{\Delta}}{2a} = \frac{-2 - \sqrt{36}}{2 \cdot 1} = \frac{-2 - 6}{2} = -4 \][/tex]
6. Substitute back to find [tex]\(x\)[/tex]:
Recall that [tex]\( u = x - 3 \)[/tex].
So for [tex]\( u_1 = 2 \)[/tex]:
[tex]\[ 2 = x - 3 \implies x = 2 + 3 \implies x = 5 \][/tex]
And for [tex]\( u_2 = -4 \)[/tex]:
[tex]\[ -4 = x - 3 \implies x = -4 + 3 \implies x = -1 \][/tex]
Therefore, the solutions to the equation [tex]\((x - 3)^2 + 2(x - 3) - 8 = 0\)[/tex] are [tex]\( x = -1 \)[/tex] and [tex]\( x = 5 \)[/tex].
So, the correct answer is:
[tex]\[ x = -1 \text{ and } x = 5 \][/tex]
1. Perform the substitution:
Let [tex]\( u = (x - 3) \)[/tex]. This changes the equation to [tex]\( u^2 + 2u - 8 = 0 \)[/tex].
2. Recognize the standard quadratic form:
We now have a quadratic equation in terms of [tex]\(u\)[/tex]:
[tex]\[ u^2 + 2u - 8 = 0 \][/tex]
3. Solve the quadratic equation:
Use the quadratic formula [tex]\( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex].
Here, [tex]\(a = 1\)[/tex], [tex]\(b = 2\)[/tex], and [tex]\(c = -8\)[/tex].
4. Calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot (-8) = 4 + 32 = 36 \][/tex]
5. Find the roots of the quadratic equation:
[tex]\[ u_1 = \frac{-b + \sqrt{\Delta}}{2a} = \frac{-2 + \sqrt{36}}{2 \cdot 1} = \frac{-2 + 6}{2} = 2 \][/tex]
[tex]\[ u_2 = \frac{-b - \sqrt{\Delta}}{2a} = \frac{-2 - \sqrt{36}}{2 \cdot 1} = \frac{-2 - 6}{2} = -4 \][/tex]
6. Substitute back to find [tex]\(x\)[/tex]:
Recall that [tex]\( u = x - 3 \)[/tex].
So for [tex]\( u_1 = 2 \)[/tex]:
[tex]\[ 2 = x - 3 \implies x = 2 + 3 \implies x = 5 \][/tex]
And for [tex]\( u_2 = -4 \)[/tex]:
[tex]\[ -4 = x - 3 \implies x = -4 + 3 \implies x = -1 \][/tex]
Therefore, the solutions to the equation [tex]\((x - 3)^2 + 2(x - 3) - 8 = 0\)[/tex] are [tex]\( x = -1 \)[/tex] and [tex]\( x = 5 \)[/tex].
So, the correct answer is:
[tex]\[ x = -1 \text{ and } x = 5 \][/tex]
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