Question

what is the natural roots of the quadratic equation 4x^2-12x+3

Answer 1

$\mathfrak{\huge{Answer:}}$

To find the nature of root, what we need is the value of $\tt{bx^{2} - 4ac}$

Now, the values of a, b and c are as follows :

a = coeffient of $x^{2}$ = 4

b = coeffient of x = ( -12 )

c = constant = 3

Put the values :

$\tt{(-12)^{2} - 4 (4)(3)}$

=》 $\tt{144 - 48}$

=》 96

We know that when :

$\tt{bx^{2} - 4ac}$ > 0 , the real roots exist

$\tt{bx^{2} - 4ac}$ < 0 , there are no real roots

$\tt{bx^{2} - 4ac}$ = 0 , the roots are equal to one another.

Here, we have the case : $\tt{bx^{2} - 4ac}$ > 0

Thus, the nature of roots = $\bold{Real\:Roots}$

Answer 2

Question : What is the natural root of the quadratic equation 4x^2-12x+3?

Firstly, we know that :

The roots of the quadratic equation ax^2 + bx + c = 0 are determined by the value of b^2 - 4ac.

Hence, b^2 - 4ac is called the discriminant of the quadratic equation. It is denoted by delta ( ∆ ).

Coming to the question,

Comparing the given equation with ax^2 + bx + c = 0, we get :

a = 4

b = -12

c = 3

b^2 - 4ac = (-12)^2 - 4 × 4 × 3

= 144 - 48

= 96

Now, ∆ = 96

We know that,

• If b^2 - 4ac = 0, roots are real and equal.

• If b^2 - 4ac > 0, roots are real and unequal.

• If b^2 - 4ac < 0, roots are not real.

Hence, in the answer,  

b^2 - 4ac > 0

Thus, natural roots of the quadratic equation 4x^2-12x+3 is real and unequal.

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