Question

Find the roots of p(x) =√x2-4x+3 + √x2-9 -√4x2-14x+6

Answer 1

Solution:

$p(x)=\sqrt{x^2-4x+3}+\sqrt{x^2-9}-\sqrt{4x^2-14x+6}\\\\p(x)=\sqrt{(x-1)(x-3)}+\sqrt{x^2-3^2}-\sqrt{2(2x^2-7x+3)}\\\\p(x)=\sqrt{(x-1)(x-3)}+\sqrt{(x-3)(x+3)}-\sqrt{2(2x-1)(x-3)}\\\\p(x)=\sqrt{x-1}\sqrt{x-3}+\sqrt{x-3}\sqrt{x+3}-\sqrt{2}\sqrt{2x-1}\sqrt{x-3}\\\\p(x)=\sqrt{x-3}[\sqrt{x-1}+\sqrt{x+3}-\sqrt{2}\sqrt{2x-1}]$

put x=3 we get, p(3)=0

Therefore 3 is a root of p(x)

Answer 2

Answer:

3

Step-by-step explanation:

To find the roots of the equation p(x) =√x2-4x+3 + √x2-9 -√4x2-14x+6,

we need to equate p(x) = 0

=>√x2-4x+3 + √x2-9 -√4x2-14x+6 = 0

=>√x2-4x+3 + √x2-9 = √4x2-14x+6

=>√(x-1)(x-3) + √(x-3)(x+3) = √(x-3)(4x-2)

Squaring on both sides, we get

(x-1)(x-3) + (x-3)(x+3) + 2√(x-3)²(x-1)(x+3) =2(x-3)(2x-1)

=>(x-1)(x-3) + (x-3)(x+3) + 2(x-3)√(x-1)(x+3) = 2(x-3)(2x-1)

=>(x-1)(x-3) + (x-3)(x+3) + 2(x-3)√(x-1)(x+3) - 2(x-3)(2x-1) = 0

Taking (x-3) as common , we get

(x-3)[x-1 + x+3 + 2√(x-1)(x+3) - 2(2x-1)] = 0

=>(x-3)[2x+2+ 2√(x-1)(x+3) - 2(2x-1)] = 0

=> x - 3 = 0 or  2√(x-1)(x+3) - 2(2x-1) = 0

=> x = 3 or

2√(x-1)(x+3) = 2(2x-1)

=>√(x-1)(x+3) = (2x-1)

Now, Squaring on both sides, we get

(x-1)(x+3) = (2x-1)²

=> x²+2x-3 = 4x² + 1 -4x

=>3x² -6x + 4 = 0

For the above equation, Discriminant = 6² - 4*3*4 = -12<0

So, there are no real roots

Hence, roots are complex.

Complex roots are given by

(6±√-12)/6

=( 6±i2√3)/6

=1 ± i/√3.

Hence the  roots of the given equation are 3, 1 ± i/√3.

If only real roots were asked, then x = 3 is the only root.