Quadratic Equation: Roots – Nature and Solution

Quadratic equation is one of the most important topic for CAT. Let us dive deep into this polynomial equation of degree 2.

Basics:

It is generally represented as: $\mathsf{a}{{\mathsf{x}}^{\mathsf{2}}}\mathsf{+bx+c}$

a belongs to R – {0} and b and c belong to R. (R is the set of real number).

Sum of roots =(-b)/a= – (coefficient of $\mathsf{x}$) / coefficient of ${{\mathsf{x}}^{\mathsf{2}}}$

Product of roots= c/a = (constant term)/ coefficient of ${{\mathsf{x}}^{\mathsf{2}}}$

Roots can be determined by the simple formula:

$\frac{\mathsf{-b-}\sqrt{\mathsf{D}}}{\mathsf{2a}}$  and  $\frac{\mathsf{-b+}\sqrt{\mathsf{D}}}{\mathsf{2a}}$

Here, D= discriminant of the given equation

$\mathsf{D=}{{\mathsf{b}}^{\mathsf{2}}}\mathsf{-4ac}$

Nature of roots:

1) Real and distinct :  if  D>0  or  ${{\mathsf{b}}^{\mathsf{2}}}\mathsf{-4ac}$ >0

a) Rational roots : if  ${{\mathsf{b}}^{\mathsf{2}}}\mathsf{-4ac}$ is a perfect square.

b)Irrational roots : if  ${{\mathsf{b}}^{\mathsf{2}}}\mathsf{-4ac}$ is not a perfect square.

Irrational roots always occur in conjugate pair of each other.

2) Real and equal :      if  D=0  or  ${{\mathsf{b}}^{\mathsf{2}}}\mathsf{-4ac}$=0.

The quadratic equation becomes a perfect square.

3) Imaginary:        if  D<0 or ${{\mathsf{b}}^{\mathsf{2}}}\mathsf{-4ac}$<0, then the equation has Complex roots and are conjugate pair .

Quadratic equation is one of the easiest and shortest topics in terms of conceptual understanding.

Similarly, we can find sum and product of the roots using the coefficients of $x^2, x$ and the constant term.

Let’s solve a few examples based on the concepts we have covered above.

EXAMPLE 1:  The values of a and b for which a and b are roots of the equation

${{\mathsf{x}}^{\mathsf{2}}}\mathsf{+ax+b}$

where (a$\ne$0 & b$\ne$0)

Solution 1:  Clearly, since a and b are roots of quadratic equation:

i) Here Sum of roots = $\mathsf{a+b}$

= ($\mathsf{-}$coefficient of x) / (coefficient of ${{\mathsf{x}}^{\mathsf{2}}}$)

$\mathsf{= (-a) / 1}$

This implies ,  $\mathsf{a+b=-a}$

$\mathsf{2a+b=0}$

ii) Product of roots$\mathsf{=ab}$

=(constant term)/coefficient of ${{\mathsf{x}}^{\mathsf{2}}}$

$\mathsf{=(b/1)}$

$\Rightarrow$ $\mathsf{ab=b}$

$\Rightarrow$ $\mathsf{ab-b=0}$

$\Rightarrow$  $\mathsf{b(a-1)=0}$

a=1 (since b$\ne$0)

Also,  $\mathsf{2a+b=0}$  or $2(1)+b=0$

$\mathsf{b= -2}$

So, final answer is $a=1$   & $\mathsf{b= -2}$.

EXAMPLE 2 : If the equation  ${{\mathsf{x}}^{\mathsf{2}}}\mathsf{+3x+4=0}$ and  $\mathsf{a}{{\mathsf{x}}^{\mathsf{2}}}\mathsf{+bx+c=0}$ ,  where a,b,c belongs to R, have a common root , then the ratio of a:b:c is

SOLUTION 2: Firstly, check the discriminant of the given quadratic equation

D= 32– 4(1)(4)=  9-16 = -7 (<0)

Since the discriminant is coming out to be negative then the given quadratic equation has imaginary roots and they are always a conjugate pair of each other.

Now  ${{\mathsf{x}}^{\mathsf{2}}}\mathsf{+3x+4=0}$ and $\mathsf{a}{{\mathsf{x}}^{\mathsf{2}}}\mathsf{+bx+c=0}$ have a common root but one of the given equation has imaginary root hence the equation  $\mathsf{a}{{\mathsf{x}}^{\mathsf{2}}}\mathsf{+bx+c=0}$ also has complex roots.

Now, complex roots always appear in conjugate pairs as stated above:

So, both the roots of given quadratic equation are same.

Hence ratio of a:b:c = 1:3:4.

EXAMPLE 3:  If a and b are roots of quadratic equation ${{\mathsf{x}}^{\mathsf{2}}}\mathsf{-3x+5}$ = 0 , then the equation whose roots are  ${{\mathsf{a}}^{\mathsf{2}}}\mathsf{-3a+9}$  and  $\mathsf{2}{{\mathsf{b}}^{\mathsf{2}}}\mathsf{-6b+14}$  is:

SOLUTION 3  :  Here a and b are roots of quadratic equation  ${{\mathsf{x}}^{\mathsf{2}}}\mathsf{-3x+5}$=0

$\Rightarrow {{\mathsf{a}}^{\mathsf{2}}}\mathsf{-3a+5}$ and  ${{\mathsf{b}}^{\mathsf{2}}}\mathsf{-3b+5=0}$ .

Here we have to form another equation whose roots are

${{\mathsf{a}}^{\mathsf{2}}}\mathsf{-3a+9}$ and  $\mathsf{2}{{\mathsf{b}}^{\mathsf{2}}}\mathsf{-6b+14}$

$\mathsf{=(}{{\mathsf{a}}^{\mathsf{2}}}\mathsf{-3a+5)+4}$  and $\mathsf{(2}{{\mathsf{b}}^{\mathsf{2}}}\mathsf{-6b+10)+4}$

$\mathsf{=(0)+4}$  and   $\mathsf{2(}{{\mathsf{b}}^{\mathsf{2}}}\mathsf{-3b+5)+4}$

$=4$    and    $\mathsf{2(0)+4}$

$=4$    and       $\mathsf{4}$

Hence the simplified roots of another quadratic equation are 4 and 4.

So, the equation is

${{\mathsf{x}}^{\mathsf{2}}}\mathsf{-(4+4)x+(4)(4) = 0}$

${{\mathsf{x}}^{\mathsf{2}}}\mathsf{-8x+16 = 0}$

EXAMPLE 4:  The coefficient of $\mathsf{x}$  in the given equation ${{\mathsf{x}}^{\mathsf{2}}}\mathsf{-px+q=0}$ was taken as 17 instead of 13 and its roots to be found were 2 and 15. Find the original roots of the quadratic equation.

SOLUTION 4:  The quadratic equation with roots 2 and 15 can be represented as:

${{\mathsf{x}}^{\mathsf{2}}}\mathsf{-}$ (Sum of roots) $\mathsf{+}$ (Product of roots) $\mathsf{=}$ 0

${{\mathsf{x}}^{\mathsf{2}}}\mathsf{-(2+15)x+(2)(15)=0}$

${{\mathsf{x}}^{\mathsf{2}}}\mathsf{-17x+30 = 0}$

Comparing it with ${{\mathsf{x}}^{\mathsf{2}}}\mathsf{-px+q=0}$ we get:

p = 17   and   q=30.

According to the question only the coefficient of x was taken to be incorrect(17 instead of 13).

Hence, the value of q (product of roots) must remain same .

q=30.

Hence, the original equation (correct coefficient) is

${{\mathsf{x}}^{\mathsf{2}}}\mathsf{-13x+30=0}$

${{\mathsf{x}}^{\mathsf{2}}}\mathsf{-10x-3x+30=0}$

$\mathsf{x(x-10)-3(x-10)=0}$

$\mathsf{(x-3)(x-10)=0}$

$\mathsf{x=3 }$

Hence original roots are 3 and 10.

EXAMPLE 5: Let $a$ be the roots of quadratic equation

${{\mathsf{x}}^{\mathsf{2}}}\mathsf{-1154x+1}$ =0

Then the value of  ${{a}^{{}^{1}\!\!\diagup\!\!{}_{4}\;}}$ +  ${{b}^{{}^{1}\!\!\diagup\!\!{}_{4}\;}}$ is equal to:

SOLUTION 5:   This is one of the most difficult and tricky question you will ever face.

Sum of roots =$a+b$  = -(-1154)/1 = 1154.

&    Product of roots= $ab$  = 1/1 =1.

Now, assume briefly  ${{a}^{{}^{1}\!\!\diagup\!\!{}_{4}\;}}$ + ${{b}^{{}^{1}\!\!\diagup\!\!{}_{4}\;}}$ = m

Our first target is to convert m into the form of a and b which is possible by removing the powers

So, on squaring both the sides we get,

${{({{a}^{{}^{1}\!\!\diagup\!\!{}_{4}\;}}+{{b}^{{}^{1}\!\!\diagup\!\!{}_{4}\;}})}^{2}}$ = m2

${{a}^{{}^{1}\!\!\diagup\!\!{}_{2}\;}}+b{}^{{}^{1}\!\!\diagup\!\!{}_{2}\;}+2{{(ab)}^{{}^{1}\!\!\diagup\!\!{}_{4}\;}}$  = m2

Here, since $ab$ =1

${{a}^{{}^{1}\!\!\diagup\!\!{}_{2}\;}}+{{b}^{{}^{1}\!\!\diagup\!\!{}_{2}\;}}$  + 2= m2

Again, assume  ${{a}^{{}^{1}\!\!\diagup\!\!{}_{2}\;}}+{{b}^{{}^{1}\!\!\diagup\!\!{}_{2}\;}}$= n

On squaring both the sides we get,

( ${{a}^{{}^{1}\!\!\diagup\!\!{}_{2}\;}}+{{b}^{{}^{1}\!\!\diagup\!\!{}_{2}\;}}$)2   =  n2

$a+b$+2$\sqrt{ab}$    = n2

Here $a+b$=1154  (sum of given equation)  and $ab$=1(product of given equation)

1154 + 2 = n2

1156 = n2

n=34   or        ${{a}^{{}^{1}\!\!\diagup\!\!{}_{2}\;}}+{{b}^{{}^{1}\!\!\diagup\!\!{}_{2}\;}}$ = 34

Coming back to our original equation

${{a}^{{}^{1}\!\!\diagup\!\!{}_{2}\;}}+{{b}^{{}^{1}\!\!\diagup\!\!{}_{2}\;}}$ + 2 = m2

n+2 = m2

34 + 2= m2

m=6       or        ${{a}^{{}^{1}\!\!\diagup\!\!{}_{4}\;}}$ + ${{b}^{{}^{1}\!\!\diagup\!\!{}_{4}\;}}$ = 6

In this way, you can solve questions from Quadratic Equations. Do check out our other blogs to learn more CAT concepts. You can also visit our website and start your CAT preparation for free.

Join our WhatsApp broadcast list and CAT Facebook group so that you never miss such useful articles. Subscribe to our YouTube channel for interesting videos shared every week.