Question

Let p and q be two positive number such that p + q = 2 and $p^4 + q^4 = 272$. Then p and q are roots of the equation:

• A. $x^2 - 2x + 2 = 0$
• B. $x^2 - 2x + 8 = 0$
• C. $x^2 - 2x + 136 = 0$
• D. $x^2 - 2x + 16 = 0$

Answer 1

Answer: (D)

x^2 - 2x + 16 = 0

Answer 2

Let the two positive numbers be p and q. We are given the following conditions:

1. $p + q = 2$
2. $p^4 + q^4 = 272$

We need to find a quadratic equation whose roots are p and q. A quadratic equation with roots p and q can be written as: $x^2 - (p+q)x + pq = 0$

From the first condition, we already have the sum of the roots: $p+q = 2$

Now we need to find the product of the roots, $pq$. Let $pq = k$. We can express $p^4 + q^4$ in terms of $p+q$ and $pq$. First, let's find $p^2 + q^2$: $p^2 + q^2 = (p+q)^2 - 2pq$ Substitute the given value $p+q=2$ and $pq=k$: $p^2 + q^2 = (2)^2 - 2k = 4 - 2k$

Next, let's find $p^4 + q^4$: $p^4 + q^4 = (p^2)^2 + (q^2)^2$ Using the identity $a^2 + b^2 = (a+b)^2 - 2ab$, where $a=p^2$ and $b=q^2$: $p^4 + q^4 = (p^2 + q^2)^2 - 2(p^2)(q^2)$ $p^4 + q^4 = (p^2 + q^2)^2 - 2(pq)^2$

Now substitute the expressions for $p^2+q^2$ and $pq$: $p^4 + q^4 = (4 - 2k)^2 - 2(k)^2$

We are given $p^4 + q^4 = 272$. So, $272 = (4 - 2k)^2 - 2k^2$ Expand the term $(4 - 2k)^2$: $(4 - 2k)^2 = 4^2 - 2(4)(2k) + (2k)^2 = 16 - 16k + 4k^2$

Substitute this back into the equation: $272 = (16 - 16k + 4k^2) - 2k^2$ $272 = 16 - 16k + 2k^2$

Rearrange the terms to form a quadratic equation in k: $2k^2 - 16k + 16 - 272 = 0$ $2k^2 - 16k - 256 = 0$

Divide the entire equation by 2 to simplify: $k^2 - 8k - 128 = 0$

Now, solve this quadratic equation for k using the quadratic formula $k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: Here, $a=1$, $b=-8$, $c=-128$. $k = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(-128)}}{2(1)}$ $k = \frac{8 \pm \sqrt{64 + 512}}{2}$ $k = \frac{8 \pm \sqrt{576}}{2}$

Calculate the square root of 576: $\sqrt{576} = 24$. $k = \frac{8 \pm 24}{2}$

This gives two possible values for k: $k_1 = \frac{8 + 24}{2} = \frac{32}{2} = 16$ $k_2 = \frac{8 - 24}{2} = \frac{-16}{2} = -8$

The problem states that p and q are two positive numbers. If p and q are positive, their product $pq$ must also be positive. Therefore, $k = pq$ must be positive. So, we choose $k = 16$.

Now we have the sum of the roots $p+q=2$ and the product of the roots $pq=16$. Substitute these values into the quadratic equation form: $x^2 - (p+q)x + pq = 0$ $x^2 - 2x + 16 = 0$

Let's check the options provided: (1) $x^2 - 2x + 2 = 0$ (2) $x^2 - 2x + 8 = 0$ (3) $x^2 - 2x + 136 = 0$ (4) $x^2 - 2x + 16 = 0$

Our derived equation matches option (4).

Note: If we check the discriminant of the resulting quadratic equation $x^2 - 2x + 16 = 0$, it is $D = (-2)^2 - 4(1)(16) = 4 - 64 = -60$. Since the discriminant is negative, the roots are complex numbers ($1 \pm i\sqrt{15}$). This means that there are no real numbers p and q (let alone positive real numbers) that satisfy the given conditions. However, in multiple-choice questions of this type, the expectation is to follow the algebraic derivation to find the equation.