Let p(x) = x⁷-6x⁶+7x⁵+5x²+2x+2
It has only one change of sign. Now,
p(-x) = (-x)⁷ -6(-x)⁶ +7(-x)⁵ +5(-x)² +2(-x)+2
= -x⁷ -6x⁶ -7x⁵ + 5x² - 2x + 2
It has two, change of sign.
Hence, p(x) has one positive root and has at least two negative roots.

We have \(x=\sqrt { 2+x } \)
\(\Rightarrow { x }^{ 2 }=2+x \Rightarrow { x }^{ 2 }-x-2=0\)
\(\Rightarrow x=\frac { 1\pm \sqrt { 1+8 } }{ 2 } \Rightarrow x=\frac { 1\pm 3 }{ 2 } \)
\(\Rightarrow x=\frac { 1+3 }{ 2 } ,\frac { 1-3 }{ 2 } \Rightarrow x=2,-1\)
Also x>0, we get x = 2

The quadratic equation 3x² + 2(a² + 1)x + (a² - 3a + 2)
Will have two roots of opposite sign if it has real roots and the product of the roots is negative.
⇒ 4(a²+1)²-12(a²-3a+2)\(\ge\) 0 and \(\frac { { a }^{ 2 }-3a+2 }{ 3 } <0\)
Both of these conditions are true if 
⇒ a²-3a+2< 0
⇒ (a-1)(a-2)< 0
⇒ 1<a<2

Let ∝, β are the roots of the equation
Sum of the roots \(\alpha +\beta =\frac { -b }{ a } \)
 \(=\frac { [-(a-2)] }{ 1 } =a-2\)
Product of the roots \(=\alpha \beta =\frac { c }{ a } \)
\(=\frac { -(a+1) }{ 1 } =-(a+1)\)
we have \({ \alpha }^{ 2 }{ \beta }^{ 2 }=({ \alpha +\beta ) }^{ 2 }-2\alpha \beta \)
\(={ (a-2) }^{ 2 }+2(a+1)\)
\(={ a }^{ 2 }-4a+4+2a+2\)
\(=(a-1{ ) }^{ 2 }+5\)
Thus \(\\ { \alpha }^{ 2 }+{ \beta }^{ 2 }\) is least if a = 1

Sum of the roots = sin ∝ + cos ∝ = \(\frac{-b}{a}\)
Product of the roots = sin ∝ cos ∝ = \(\frac{c}{a}\)
Now 1 = cos²∝ + sin² ∝
= (sin ∝ +cos ∝)² - 2 sin ∝ cos ∝
\(1=\frac { { b }^{ 2 } }{ { a }^{ 2 } } -\frac { 2c }{ a } \Rightarrow 1=\frac { { b }^{ 2 }-2ac }{ { a }^{ 2 } } \)
⇒ a² = b² - 2ac ⇒ a² + 2ac = b²
Adding c² both sides, a² +2ac+c² = b²+c²
⇒ (a+c)² = b² + c²