Polynomials 1

Algebra Level 5

A quadratic polynomial f ( x ) = x 2 + a x + b f\left( x \right) ={ x }^{ 2 }+ax+b is formed with one of its zeros being 4 + 3 3 2 + 3 \frac { 4+3\sqrt { 3 } }{ 2+\sqrt { 3 } } where a a and b b are integers. Also g ( x ) = x 4 + 2 x 3 − 10 x 2 + 4 x − 10 g\left( x \right) ={ x }^{ 4 }+{ 2x }^{ 3 }-{ 10x }^{ 2 }+4x-10 is a biquadratic polynomial such that g ( 4 + 3 3 2 + 3 ) = c 3 + d g\left( \frac { 4+3\sqrt { 3 } }{ 2+\sqrt { 3 } } \right) =\quad c\sqrt { 3 } +d where c c and d d are integers. Find the value of a + b + c + d a+b+c+d .

For set , click here .


The answer is -6.

View wiki
Utkarsh Bansal by Utkarsh Bansal , 23, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Chew-Seong Cheong
Apr 26, 2015

The zero = 4 + 3 3 2 + 3 = ( 4 + 3 3 ) ( 2 − 3 ) ( 2 + 3 ) ( 2 − 3 ) = 8 + 2 3 − 9 1 = 2 3 − 1 = \dfrac {4+3\sqrt{3}}{2+\sqrt{3}} = \dfrac {(4+3\sqrt{3})(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})} = \dfrac {8+2\sqrt{3}-9}{1} = 2\sqrt{3}-1

f ( x ) = x 2 + a x + b = 0 ⇒ x = − a ± a 2 − 4 b 2 f(x) = x^2+ax+b = 0 \quad \Rightarrow x = \dfrac {-a \pm \sqrt{a^2-4b}}{2}

Assuming that − a + a 2 − 4 b 2 = − 1 + 2 3 = − 2 + 48 2 \dfrac {-a + \sqrt{a^2-4b}}{2} = -1 + 2\sqrt{3} = \dfrac {-2+\sqrt{48}}{2}

⇒ a = 2 ⇒ a 2 − 4 b = 48 ⇒ b = − 11 \Rightarrow a = 2 \quad \Rightarrow a^2-4b = 48 \quad \Rightarrow b = -11

Therefore, f ( x ) = x 2 + 2 x − 11 f(x) = x^2+2x-11

g ( x ) = x 4 + 2 x 3 − 10 x 2 + 4 x − 10 = x 2 ( x 2 + 2 x − 10 ) + 4 x − 10 = x 2 ( x 2 + 2 x − 11 ) + x 2 + 4 x − 10 = x 2 ( x 2 + 2 x − 11 ) + x 2 + 2 x − 11 + 2 x + 1 = ( x 2 + 1 ) f ( x ) + 2 x + 1 ⇒ g ( 2 3 − 1 ) = ( 2 3 − 1 ) 2 f ( 2 3 − 1 ) + 2 ( 2 3 − 1 ) + 1 = 0 + 4 3 − 1 = 4 3 − 1 \begin{aligned} g(x) & = x^4+2x^3-10x^2+4x-10 \\ & = x^2(x^2+2x-10) + 4x - 10 \\ & = x^2(x^2+2x-11) + x^2 + 4x - 10 \\ & = x^2(x^2+2x-11) + x^2 + 2x - 11 + 2x + 1 \\ & = (x^2+1)f(x) + 2x +1 \\ \Rightarrow g (2\sqrt{3}-1) & = (2\sqrt{3}-1)^2f(2\sqrt{3}-1) + 2(2\sqrt{3}-1)+1 \\ & = 0 + 4 \sqrt{3} -1 = 4 \sqrt{3} -1 \end{aligned}

⇒ c = 4 ⇒ d = − 1 ⇒ a + b + c + d = 2 − 11 + 4 − 1 = − 6 \Rightarrow c = 4\quad \Rightarrow d = -1\quad \Rightarrow a+b+c+d = 2-11+4-1 = \boxed{-6}

5 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...