2+{5}, 1 are roots of the cubic equation given by

Correct Answer is : x3−5x2+3x+1=0x³-5 x²+3 x+1=0x3−5x2+3x+1=0

Correct Answer is : x3−5x2+3x+1=0x³-5 x²+3 x+1=0x3−5x2+3x+1=0

Test Index

AP EAPCET 20-Aug-2021 Shift 2 Paper

Section: Mathematics

Question : 17 of 160

Marks: +1, -0

2+\sqrt{5}, 1

are roots of the cubic equation given by

Solution:

Roots of cubic equations are 1 and

2+\sqrt{5}

Since,

2+\sqrt{5}

is root, then its conjugate also become roots, i.e.

2-\sqrt{5}

is root of cubic equation.

\therefore

Roots

=1,2+\sqrt{5}, 2-\sqrt{5}

i.e.

\alpha, \beta, \gamma

Let cubic equation be

x^{3}+b x^{2}+c x+d=0

Then,

Sum of Roots

=-b=1+2+\sqrt{5}+2-\sqrt{5}=5

\Rightarrow b=-5

Product of Roots

=-d=(1)(2+\sqrt{5})(2-\sqrt{5})

=4-5=-1

\Rightarrow d=1

\therefore \alpha\beta+\beta\gamma+\gamma\alpha=c

(1)(2+\sqrt{5})+(2+\sqrt{5})(2-\sqrt{5})+(2-\sqrt{5})(1)=c

\Rightarrow 2+\sqrt{5}+2-\sqrt{5}+4-5=6

\Rightarrow 3=c

∴ Cubic equation is

x^{3}-5 x^{2}+3 x+1=0

Go to Question:

Prev Question

Next Question