# Quadratic Equation: Roots – Nature and Solution

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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  $latex {{a}^{{}^{1}\!\!\diagup\!\!{}_{4}\;}}$ +  $latex {{b}^{{}^{1}\!\!\diagup\!\!{}_{4}\;}}$ is equal to:

SOLUTION 5:   This one is difficult and tricky

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 $latex ab$=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

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