Quadratic in Form Polynomials
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What are the solutions of the equation x4-5x2-14 = 0? Use factoring to solve.
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x=±√√7 and x=±√√√2
x-ti√7 and x-ti√√
x-ti√7 and x-±√√2
O x-t√7 and x-ti√√2



Answer :

To find the solutions to the equation \( x^4 - 5x^2 - 14 = 0 \) by factoring, we will follow these steps: **Step 1: Recognize the pattern.** The equation can be seen as a quadratic in form, where \( x^2 \) is taking the place of a usual \( x \) in a quadratic equation. Let's let \( u = x^2 \), which gives us a new equation \( u^2 - 5u - 14 = 0 \). **Step 2: Factor the new quadratic equation.** We need to factor the equation \( u^2 - 5u - 14 = 0 \). This can be factored into \( (u - 7)(u + 2) = 0 \) since \( -7 \) and \( +2 \) are the numbers that multiply to \( -14 \) and add to \( -5 \). **Step 3: Solve for \( u \).** We now have two simpler equations to solve: \( u - 7 = 0 \) which gives \( u = 7 \) and \( u + 2 = 0 \) which gives \( u = -2 \) **Step 4: Replace \( u \) with \( x^2 \) and solve for \( x \).** We go back to our substitution \( u = x^2 \) to find the values of \( x \). For \( u = 7 \), we have \( x^2 = 7 \) which means \( x = \sqrt{7} \) or \( x = -\sqrt{7} \). For \( u = -2 \), we have \( x^2 = -2 \) which generally gives us complex solutions since we cannot have the square root of a negative number in the set of real numbers. Therefore, if we are only looking for real solutions, we can disregard this case. **Step 5: State the final solutions.** The real solutions to the equation \( x^4 - 5x^2 - 14 = 0 \) are \( x = \sqrt{7} \), \( x = -\sqrt{7} \). Thus the correct option, if we have to choose from the ones provided, is x = ±√7 and there are no real solutions corresponding to x = ±√(-2) since that would involve imaginary numbers.