First slide

Theory of equations

Question

If $a + b = 2$ and $a^{4} + b^{4} = 272$ then a quadratic equation whose roots are a and b is

Moderate

A

$x^{2} - 2 x + 16 = 0$
B

$x^{2} - 2 x + 8 = 0$
C

$x^{2} - 2 x + 32 = 0$
D

$x^{2} - 2 x - 16 = 0$

Solution

We have,
$a + b = 2$ and $a^{4} + b^{4} = 272$
Now, $a^{4} + b^{4} = 272$
$\begin{array}{l} \Rightarrow {(a^{2} + {\dot{b}}^{2})}^{2} - 2 a^{2} b^{2} = 272 \\ \Rightarrow {\{(a + b)^{2} - 2 a b\}}^{2} - 2 a^{2} b^{2} = 272 \\ \Rightarrow (4 - 2 a b)^{2} - 2 a^{2} b^{2} = 272 \\ \Rightarrow 16 - 16 a b + 2 a^{2} b^{2} = 272 \\ \Rightarrow (a b)^{2} - 8 (a b) - 128 = 0 \\ \Rightarrow (a b - 16) (a b + 8) = 0 \Rightarrow a b = 16 or, - 8 \end{array}$
Thus, we have $a + b = 2$ and $a b = 16$ , or -8.
So, the quadratic equations whose roots are a and b are
$x^{2} - 2 x + 16 = 0 and x^{2} - 2 x - 8 = 0$

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