Question

Q.2 What is the nature of the roots of the quadratic equation 4² + 4√3 + 3 = 0.​

Answer 1

We know that D = b^2 - 4 * a * c

                          = (4root3)^2 - 4 * 4 * 3

                         = 16 * 3 - 48

                         = 48 - 48

                         = 0

So, given quadratic equation has equal real roots.

Hope this helps!

Answer 2

Answer:

CORRECT QUESTION:

What is the nature of roots of quadratic equation 2x²+3x-4=0?

SOLUTION:

Nature of the roots of any quadratic equation can be find by applying formula of discriminant. Nature of roots can be expressed as follows:-

\begin{gathered}\qquad\qquad\quad\large{\sf{Nature\: of\: roots}}\\\\\boxed{\begin{array}{|c||c|}\cline{1-2}\bf{Discriminant}&\bf{Nature}\\\cline{1-2}\sf{+ve}&\sf{2\: distinct\:roots}\\\cline{1-2}\sf{-ve}&\sf{No\:real\:roots}\\\cline{1-2}\sf{0}&\sf{2\: equal\:roots}\cline{1-2}\cline{1-2}\end{array}}\end{gathered}

Natureofroots

\cline1−2Discriminant

\cline1−2+ve

\cline1−2−ve

\cline1−20

Nature

2distinctroots

Norealroots

2equalroots\cline1−2\cline1−2

\rule{250}{2}

In the given quadratic equation:-

a=coefficient of x²=2

b=coefficient of x=3

c=constant term=-4

\rule{250}{2}

We have formula of discriminant:-

\underline{\boxed{\sf{D=b^{2}-4ac}}}\star

D=b

2

−4ac

⋆

\sf{D=3^{2}-4(2)(-4)}D=3

2

−4(2)(−4)

\sf{D=9+32}D=9+32

\sf{D=41}D=41

Here the value of D is +ve, so:-

\Large{\underline{\boxed{\bf{NATURE}\longrightarrow\sf{2\: distinct\:roots}}}}

NATURE⟶2distinctroots