'Rooty polynomials!(Part 3)

Algebra Level 3

Let $x_1 , x_2$ be the roots of equation:

$\sqrt{3x^2 + 5x + 8} - \sqrt{3x^2 + 5x + 1} = 1$

If , $x_1 > x_2$ ,

Find $x_1 - 9x_2$

1 solution

Nihar Mahajan
Feb 20, 2015

Let $3x^2 + 5x + 1 = a$ .Thus , the equation becomes:

$\sqrt{a + 7} - \sqrt {a} = 1$

Squaring both the sides gives us:

$a + 7 - 2\sqrt{a(a+7)} + x = 1$

$2a + 6 = 2\sqrt{a(a+7)}$

$a + 3 = \sqrt{a(a+7)}$

Squaring both the sides:

$a^2 + 6a + 9 = a^2 + 7a$

$\Rightarrow \boxed{a = 9}$

Hence , $3x^2 + 5x + 1 = 9$

$3x^2 + 5x - 8 = 0$

$(x -1)(3x + 8) = 0$

Thus , $x = (1 , \dfrac{-8}{3})$ where , $x_1 = 1 , x_2 = \dfrac{-8}{3}$

Therefore , $x_1 - 9x_2 = 1 + 24 = \huge\boxed{25}$

$3x^2 + 5x + 8 = a , 3x^2 + 5x + 1 = b$

$a - b = 7$

$(\sqrt{a} - \sqrt{b})(\sqrt{a} + \sqrt{b}) = 7$

$\sqrt{a} + \sqrt{b} = 7$

$\sqrt{a} - \sqrt{b} = 1$

$2\sqrt{a} = 8$

$\sqrt{3x^2 + 5x + 8} = 4$

$3x^2 + 5x - 8=0$

$(x-1)(3x + 8) = 0$

U Z - 6 years, 3 months ago

Can you explain how $a - b = 7$ ? It is given that $a - b = 1$ . I think radical sign must be removed in the assumption.

Nihar Mahajan - 6 years, 3 months ago

Yes sorry just copied your latex

U Z - 6 years, 3 months ago

Wow excellent solution.

Jesus Ulises Avelar - 6 years, 3 months ago

Did in the same way. Upvoted.

Sai Ram - 5 years, 6 months ago