Solveeit Logo

Question

Question: If \(x^{2} - 8x + 17\) and \(\frac{x^{2} + 14x + 9}{x^{2} + 2x + 3}\) where \(\frac{x + 2}{2x^{2} +...

If x28x+17x^{2} - 8x + 17 and x2+14x+9x2+2x+3\frac{x^{2} + 14x + 9}{x^{2} + 2x + 3}

where x+22x2+3x+6\frac{x + 2}{2x^{2} + 3x + 6}, then (113,13)\left( \frac{1}{13},\frac{1}{3} \right)has at least.

A

Four real roots

B

Two real roots

C

Four imaginary roots

D

None of these

Answer

Two real roots

Explanation

Solution

Let all four roots are imaginary. Then roots of both equations a,b>0a,b > 0and a,b<0a,b < 0are imaginary.

Thus 2x25x+1=02x^{2} - 5x + 1 = 0, So x2+5x+2=0x^{2} + 5x + 2 = 0, which is impossible unless a+b+c=0,a + b + c = 0,.

So, if a0,a,b,cQa \neq 0,a,b,c \in Qor ax2+bx+c=0ax^{2} + bx + c = 0 at least two roots must be real.

If a,b,cQa,b,c \in Q (b+c2a)x2+(b + c - 2a)x^{2} +, we have the equations.

(c+a2b)x+(a+b2c)=0(c + a - 2b)x + (a + b - 2c) = 0and x2+2bx+cx^{2} + 2bx + c

Or b24c>0b^{2} - 4c > 0 as one of b24c<0b^{2} - 4c < 0 and c2<bc^{2} < b must be

positive, so two roots must be real.