Question

Let p and q be two positive numbers 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

Hint: We know that the roots of the equation are the values at which the function's value becomes zero or the graph of the functions intersects the x-axis and the roots of an equation are those values on which an equation is equal to zero and an equation with two real roots is represented as:
\[{x^2} - \left( {sum\,\,of\,the\,\,roots} \right)\,x + \left( {product\,of\,the\,\,roots} \right) = 0\]

Formula used:
1. \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
2. \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{2}\]

Complete step-by-step solution:
We are given that \[p + q = 2...\left( 1 \right)\] and
\[{p^4} + {q^4} = 272...\left( 2 \right)\]
Now we add and subtract \[2{p^2}{q^2}\] in equation (2), and we get
 \[{p^4} + {q^4} + 2{p^2}{q^2} - 2{p^2}{q^2} = 272\]
Now we know that \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
So, \[{\left( {{p^2} + {q^2}} \right)^2} - 2{p^2}{q^2} = 272\]
Now we add and subtract \[2pq\] in the above equation, and we get
\[{\left( {{p^2} + {q^2} + 2pq - 2pq} \right)^2} - 2{p^2}{q^2} = 272\]
Now again we apply \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
\[{\left( {{{\left( {p + q} \right)}^2} - 2pq} \right)^2} - 2{p^2}{q^2} = 272\]
Now it is given that \[p + q = 2\]
Now we assume that \[pq\, = \,A\]
So,
 \[
  {\left( {4 - 2A} \right)^2} - 2{A^2}\, = 272 \\
  16 + 4{A^2} - 16A - 2{A^2} = 272 \\
  {A^2} - 8A - 128 = 0 \\
 \]
Now we find the real roots of the equation, we get
\[
  A = \dfrac{{8 \pm \sqrt {64 + 4 \times 1 \times 128} }}{2} \\
   = \dfrac{{8 \pm \sqrt {576} }}{2} \\
   = \dfrac{{8 \pm \,24}}{2} \\
   = 4 \pm 12 \\
 \]
\[A = 16\,,\, - 8\]
So, \[pq = \,16\,,\, - 8\]
Now we know that the standard equation of line for two real roots is:
\[{x^2} - \left( {sum\,\,of\,the\,\,roots} \right)\,x + \left( {product\,of\,the\,\,roots} \right) = 0\]
By substituting the above value in the standard equation, we get
When \[pq = 16\]
\[
  {x^2} - \left( {p + q} \right)x + pq = 0 \\
  {x^2} - 2x - 8 = 0 \\
   \\
 \]
and when \[pq = - 8\] that means both p and q can’t be positive which contradicts the given condition.
Hence, option (D) is correct

Note: Real roots of the equation do not exist if the quantity inside the square root, which is the discriminant, is negative. Students frequently overlook the negative sign and write the correct answer. We may also use the factorization approach to discover the roots of quadratic equations.