Question

Solve the inequality using WAVY CURVE METHOD ONLY
(x² + 3x + 1) (x² + 3x + 3) ≥ 5

Answer 1

Given

(x²+3x+1)(x²+3x–3) ≥ 5

★Solution★

For making the given expression suitable for application of the wavy curve method, we must simplify it first.

On opening the brackets:

⇒(x⁴)+(3x³+3x³)+(9x²+x²–3x²)–6x–3 ≥ 5

⇒x⁴+6x³+7x²–6x–3 ≥ 5

Now, on subtracting 5 from both sides:

⇒x⁴+6x³+7x²–6x–8 ≥ 0

# Writing 6x³, 7x², & –6x as:

⇒x⁴+6x³+7x²–6x–8 ≥ 0

⇒x⁴+2x³+4x³+8x²–x²–2x–4x–8 ≥ 0

⇒x³(x+2)+4x²(x+2)–x(x+2)–4(x+2) ≥ 0

⇒(x+2)(x³+4x²–x–4) ≥ 0

# Writing 4x², & –x as:

⇒(x+2)(x³+4x²–x–4) ≥ 0

⇒(x+2)(x²(x+4)–1(x+4)) ≥ 0

⇒(x+2)(x²–1)(x+4) ≥ 0

[ ∵ a²–b²=(a–b)(a+b) ]

⇒(x+2)(x+4)(x–1)(x+1) ≥ 0

Now,

The roots are : x= –4, –2, –1, & 1

Plotting the roots on number line, and checking sign at one of the interval, except at the roots;

E.g. The value of (x+2)(x+4)(x–1)(x+1) at x=2, is 72>0(+ve). Now, changing the signs alternatively, we get:

+ve –ve +ve –ve +ve

–∞‹___,______,______,______,___›∞

–4 –2 –1 1

‡‡Answer‡‡

Therefore,

______________________________

x ∈ (–∞, –4 ] ∪ [ –2, –1 ] ∪ [ 1, ∞ )

______________________________

Answer 2

$\sf ( {x}^{2} + 3x + 1)( {x}^{2} + 3x + 3) \geq 5$

$\sf {x}^{2} + 3x : = t$

$\sf {t}^{2} + 4t + 3 \geq5$

$\sf {t}^{2} + 4t - 2 \geq 0$

$\sf \: (t - ( - 2 + \sqrt{ 6} ))(t - ( - 2 - \sqrt{6} )) \geq0$

$\sf( {x }^{2} + 3x + 2 - \sqrt{6} )( {x}^{2} + 3x + 2 + \sqrt{6} ) \geq0$

$\sf \because {x }^{2} + 3x + 2 + \sqrt{6} > 0 \: \: \forall \: x \in \real$

$\sf {x}^{2} + 3x + 2 - \sqrt{6} \geq0$

$\sf(x - ( \frac{ - 3 + \sqrt{1 + 4 \sqrt{6} } }{2} ))(x - ( \frac{ - 3 - \sqrt{1 + 4 \sqrt{6} } }{2} )) \geq0$

By using the Wavy Curve method to solve inequalities:

$\sf \boxed{ \small{x \in ( - \infty , \frac{ - 3 - \sqrt{1 + 4 \sqrt{6} } }{2} )∪ ( \frac{ \sqrt{1 + 4 \sqrt{6} } - 3}{2} ,+ \infty )}}$