Complex numbers and quadratic equations is a segment of maths that deals with crucial theorems and concepts along with various formulae. It comprises of linear and quadratic equations along with roots related to the complex number's set (known as complex roots)..

**Q1.**In the given figure, vertices of ∆ABC lie on y=f(x)=ax^2+bx+c. The ∆ABC is right angled isosceles triangle whose hypotenuse AC=4√2 unitsy=f(x) is given by

**Q2.**Consider the inequality 9

^{x}- a3

^{x}- a + 3 ≤ 0, where ‘a’ is a real parameter

The given inequality has at least one negative solution for a∈

**Q3.**Consider the in equation x

^{2}+x+a-9< 0

The value of the real parameter ‘a’ so that the given in equation has at least one positive solution:

**Q4.**‘af(Î¼)< 0’ is the necessary and sufficient condition for a particular real number Î¼ to lie between the roots of a quadratic equation f(x) = 0, where f(x)=ax

^{2}+bx+c. Again if f(Î¼

_{1})f(Î¼

_{2})< 0, then exactly one of the roots will lie between Î¼

_{1}and Î¼

_{2}

If |b|>|a+c|, then

Solution

(b)

b

⇒ (a+c-b)(a+c+b)< 0

⇒ f(-1)f(1)< 0

So, there is exactly one root in (-1,1)

(b)

b

^{2}>(a+c)^{2}⇒ (a+c-b)(a+c+b)< 0

⇒ f(-1)f(1)< 0

So, there is exactly one root in (-1,1)

**Q5.**The real numbers x

_{1},x

_{2},x

_{3}satisfying the equation x

^{3}-x

^{2}+Î²x+Î³=0 are in A.P.

All possible values of Î² are

Solution

(a)

From the question, the real roots of x

x

⇒a-d+a+a+d=1

⇒a=1/3 (1)

x

⇒(a-d)a+a(a+d)+(a+d)(a-d)=Î² (2)

x

⇒(a-d)a(a+d)=-Î³ (3)

From (1) and (2), we get

3a

⇒ 3 1/9-d

From (1) and (3), we get

1/3 (1/9-d

⇒ Î³=1/3 (d

Î³∈(-1/27,+∞)

(a)

From the question, the real roots of x

^{3}-x^{2}+Î²x+Î³=0 are x_{1},x_{2},x_{3}and they are in A.P. As x_{1},x_{2},x_{3}are in A.P., let x_{1}=a-d,x_{2}=a,x_{3}=a+d. Now,x

_{1}+x_{2}+x_{3}=-(-1)/1=1⇒a-d+a+a+d=1

⇒a=1/3 (1)

x

_{1}x_{2}+x_{2}x_{3}+x_{3}x_{1}=Î²/1=Î²⇒(a-d)a+a(a+d)+(a+d)(a-d)=Î² (2)

x

_{1}x_{2}x_{3}=-Î³/1=-Î³⇒(a-d)a(a+d)=-Î³ (3)

From (1) and (2), we get

3a

^{2}-d^{2}=Î²⇒ 3 1/9-d

^{2}=Î², so Î²=1/3-d^{2}< 1/3From (1) and (3), we get

1/3 (1/9-d

^{2})=-Î³⇒ Î³=1/3 (d

^{2}-1/9)>1/3 (-1/9)=-1/27Î³∈(-1/27,+∞)