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Solve for x (x^2-2)^2-10(x^2-2)+21=0

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Final answer:

To solve the equation (x^2-2)^2-10(x^2-2)+21=0, we can rewrite it as a quadratic equation and solve for the possible values of x.

Step-by-step explanation:

To solve the equation (x^2-2)^2-10(x^2-2)+21=0, we can notice that it can be rewritten as a quadratic equation with respect to (x^2-2). Let's substitute (x^2-2) with a variable, let's say y. The equation becomes y^2 - 10y + 21 = 0. Now, we can solve this quadratic equation using factoring, completing the square, or the quadratic formula.

Factoring the quadratic equation (y^2 - 10y + 21), we get (y-3)(y-7) = 0. So y-3 = 0 or y-7 = 0. Solving for y, we get y = 3 or y = 7. Since y = x^2-2, we have two cases to consider: x^2-2 = 3 and x^2-2 = 7.

In the first case, x^2-2 = 3, we solve for x by adding 2 on both sides, resulting in x^2 = 5. Taking the square root on both sides, we get x = ±√5.

In the second case, x^2-2 = 7, we solve for x by adding 2 on both sides, resulting in x^2 = 9. Taking the square root on both sides, we get x = ±3.

Therefore, the solutions for the given equation are x = ±√5 and x = ±3.

User Alessandro Alinone
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