Quadratic Expressions - I

Quadratic Expressions - I

10. Discriminant of a quadratic equation:
If ax²+bx+c=0 is a quadratic equation then we know that \( x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }} {{2a}} \)
Here, the value of b² -4ac is called "The discriminant" of the quadratic equation ax² +bx+c=0
The value of b²-4ac decides the nature of the roots of ax²+bx+c=0.
The value generally denoted by \( \Delta \)
\( i.e\boxed{\Delta = b^2 - 4ac} \)

11. Nature of roots of quadratic equation
Consider ax²+bx+c=0
I.    If a,b,c are real and 
    i) \(\Delta \) >0, then the roots are real and distinct
    ii) \(\Delta \)=0, then the roots are real and equal
    iii) \(\Delta \)<0, then the roots are non -real conjugate complex numbers i.e\( \alpha \pm i\beta \)
II.    If a,b,c are rational and     
    i)\(\Delta \) >0 and is a perfect square then the roots are rational and distinct
    ii)\(\Delta \)>0 and is not a perfect square then the roots are conjugate surds \( i.e.,\,\,\,\alpha \pm \sqrt \beta \,(where\,\,\beta \ne 0) \)
    iii)\(\Delta \)=0, then the roots are equal and rational
    iv)\(\Delta \) <0, then the roots are non real conjugate complex numbers i.e.,\( \alpha \pm i\beta \)

Problem (iv) If the roots of the quadratic equation \( x^2 - 4x - \log _3^a = 0 \) are real, then the least value of a is ?
Sol : Since the roots of \( x^2 - 4x - \log _3^a = 0 \) are real, we have
    Discriminant \( = \Delta \geqslant 0 \)
    i.e \( b^2 - 4ac \geqslant 0 \)
    \( \Rightarrow ( - 4)^2 - 4(1)\left( { - \log _3^a } \right) \geqslant 0 \)
    \( \Rightarrow 16 + 4\log _3^a \geqslant 0 \)
    \( \begin{gathered} \Rightarrow 4 + \log _3^a \geqslant 0 \hfill \\ \Rightarrow \log _3^a \geqslant - 4 \hfill \\ \Rightarrow a \geqslant 3^{ - 4} \hfill \\ \Rightarrow a \geqslant \frac{1} {{81}} \hfill \\ \end{gathered} \)
    \( \therefore \) The least value of a is \( \frac{1} {{81}} \)

Problem (V) If a+b+c=0, then the nature of the roots of 3ax²+2bx+c=0 is
Sol : Given 3ax²+2bx+c=0
discriminant \( = \Delta = b^2 - 4ac \) 
    = (2b)²-4(3a)(c)
    =4b² - 12ac
    =4(b²-3ac)
    =4((-a-c)² -3ac)    since a+b+c=0 b=-a-c
    =4(a²+c²-ac)
    =2(a²+c²+(a-c)²)>0
    The roots are real and distinct

Problem (vi) :If a,b,c are in A.P. the nature of the roots of the equation ax²+2bx+c=0 is ?
Solu: Given a,b,c are in A.P
     2b=a+c        .......(1)
     Now, ax²+2bx+c =0
    Discriminant =\( \Delta \) 
            = B² -4AC
            = (2b)² -4.a.c
            = (a+c)²-4ac (from (1))
            = (a-c)²>0
    \( \therefore \) The roots are real and distinct