Question

The roots of the equation \((p+2 \sqrt{q})^{x^{3}-4 x+1}+(p-2 \sqrt{q})^{x^{3}-4 x+1}=2 p\) where, \(p^{2}-4 q=1\) are (a) \(\pm \sqrt{2}, 2 \pm \sqrt{2}\) (b) \(0,4,2 \pm \sqrt{2}\) (c) \(\pm \sqrt{20}, \pm 2\) (d) \(0, \pm 2\)

Short answer

The roots of the given equation are: \(x = 2, x = 1 + \sqrt{2},\) and \(x = 1 - \sqrt{2}\).

Step-by-step solution

Find the relationship between p and q from the given condition

Given that \(p^2 - 4q = 1\), we can write it as: \(p^2 = 4q + 1\) Now, let's express \(q\) in terms of \(p\): \(q = \frac{1}{4}(p^2 - 1)\)

Substitute the value of q in the main equation

Now, substitute the above expression of \(q\) in the main equation: \((p + 2\sqrt{\frac{1}{4}(p^2 - 1)})^{x^3 - 4x + 1} + (p - 2\sqrt{\frac{1}{4}(p^2 - 1)})^{x^3 - 4x + 1} = 2p\)

Simplify the equation

Notice that \(2\sqrt{\frac{1}{4}(p^2 - 1)} = \sqrt{p^2 - 1}\), and we can rewrite the above equation as: \((p + \sqrt{p^2 - 1})^{x^3 - 4x + 1} + (p - \sqrt{p^2 - 1})^{x^3 - 4x + 1} = 2p\) Now, let \(A = p + \sqrt{p^2 - 1}\), and \(B = p - \sqrt{p^2 - 1}\). So, the equation becomes: \(A^{x^3 -4x +1} + B^{x^3 - 4x +1} = 2p\) Since \(A\) and \(B\) are conjugates (related through the difference of squares), we know that: \(A \cdot B = (p + \sqrt{p^2 -1})(p - \sqrt{p^2 -1}) = p^2 - (p^2 -1) = 1\) Let \(y = x^3 - 4x + 1\), then our equation becomes: \(A^y + B^y = 2p\) As we have \(A \cdot B = 1\), we can rewrite the equation as: \(\frac{A^{y+1} - B^{y+1}}{A - B} = 2p\)

Recognize the form of the equation

Observe that we have a difference of powers in the numerator and the form of the equation is: \(\frac{A^{y+1} - B^{y+1}}{A - B} = 2p\) This form is recognizable as the derivative of \(A^y - B^y\). Taking the derivative of both sides with respect to \(y\) gives us: \(\frac{d}{dy} (A^y - B^y) = \frac{d}{dy}(2py)\) Now, calculate the derivatives: \((y+1)(A^y - B^y) = 2px\)

Solve the simplified equation for x

Now, we have a simplified equation: \((y+1)(A^y - B^y) = 2px\) Substitute \(y = x^3 - 4x + 1\): \((x^3 - 4x + 2)(A^{x^3 - 4x + 1} - B^{x^3 - 4x + 1}) = 2px\) To find the roots, we can set each factor equal to zero: 1. \(x^3 - 4x + 2 = 0\) 2. \(A^{x^3 - 4x + 1} - B^{x^3 - 4x + 1} = 0\) The second equation simplifies to \(A^y = B^y\), which is only true when \(y = 0\), as \(A\) and \(B\) are not equal. This implies that \(x^3 - 4x + 1 = 0\), which is the first equation. Thus, we only need to solve the first equation for the roots of the given equation. Now we have a cubic equation to solve: \(x^3 - 4x + 2 = 0\) By trial and error, we can find that \(x = 2\) is a root of the equation. And by synthetic division, we get a quadratic equation left: \(x^2 - 2x - 1 = 0\) Now, we can use the quadratic formula to find the other roots: \(x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-1)}}{2(1)}\) \(x = \frac{2 \pm \sqrt{8}}{2}\) \(x = 1 \pm \sqrt{2}\)

Write the answer

Now, we have found all the roots of the equation: 1. \(x = 2\) 2. \(x = 1 + \sqrt{2}\) 3. \(x = 1 - \sqrt{2}\) So the correct answer is: (a) \(\pm \sqrt{2}, 2 \pm \sqrt{2}\)

Key concepts

Cubic Equations

Cubic equations are polynomial equations of degree three. These equations take the general form \(ax^3 + bx^2 + cx + d = 0\), where \(a\), \(b\), \(c\), and \(d\) are constants, and \(a eq 0\). The main challenge of solving cubic equations lies in finding the roots or solutions for the variable \(x\).

• Cubic equations can have three real roots or one real root and two complex conjugate roots, depending on the discriminant value.
• The solution often begins with identifying any real roots through trial and error, which might then be used to factor the equation further.
• Once a real root is discovered, synthetic division or polynomial division can be employed to reduce the cubic equation to a quadratic equation.

Understanding the properties and possible configurations of roots is key to solving cubic equations effectively.

Conjugate Roots

Conjugate roots refer to a pair of numbers in the form \(a \pm bi\), where \(a\) and \(b\) are real numbers and \(i\) is the imaginary unit, such that \(i^2 = -1\). Conjugate pairs often appear in the solutions of polynomial equations with real coefficients.

• If a polynomial equation with real coefficients has a complex root \(a + bi\), its conjugate \(a - bi\) will also be a root.
• This property ensures that all non-real roots of such polynomials appear in pairs, maintaining the equation's real coefficients.
• The product of conjugate roots, \((a + bi)(a - bi)\), is always a real number, \(a^2 + b^2\).

Recognizing conjugate roots is crucial because it simplifies the process of polynomial factorization and helps ensure completeness of solutions.

Quadratic Formula

The quadratic formula is a solution method for quadratic equations, which are equations of the form \(ax^2 + bx + c = 0\). The formula is:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula enables us to find the roots of any quadratic equation, provided that the discriminant, \(b^2 - 4ac\), is non-negative.

• If the discriminant is positive, the equation will have two distinct real roots.
• If the discriminant equals zero, there is exactly one real root, known as a repeated or double root.
• If the discriminant is negative, the roots will be complex conjugates.

Using the quadratic formula simplifies root finding by providing a direct computation method rather than requiring factoring. This mathematical tool is vital for breaking down quadratic components within larger equations, such as cubic equations reduced to quadratic ones through synthetic division.