Question

Show that the quadratic 2 X square + 5 root 3 X + 6 has real root and always solve it​ 2

Answer 1

Answer:

answer is this question by a u abhi

Answer 2

Correct Question :-

Show that the equation 2x² + 5√3x + 6 = 0 has real roots and solve it.

Given :-

• 2x² + 5√3x + 6 = 0

To show :-

• 2x² + 5√3x + 6 = 0 has real roots

Solution :-

we know that,this equation is of the form ax² + bx + c = 0

• a = 2
• b = 5√3
• c = 6

we know that,

D = b² - 4ac

⤇ D = (5√3)² - 4 × 2 × 6

⤇ D = 75 - 48

⤇ D = 27 > 0

So, the given equation has real roots.

Now,

$\rm\alpha\:=\dfrac{-b+\sqrt{D}}{2a}$

$\rm\longrightarrow\alpha\:=\dfrac{-5\sqrt{3}+\sqrt{27}}{2\times{2}}$

$\rm\longrightarrow\alpha\:=\dfrac{-5\sqrt{3}+3\sqrt{3}}{4}$

$\rm\longrightarrow\alpha\:=\dfrac{-2\sqrt{3}}{4}$

$\rm\longrightarrow\alpha\:=\dfrac{-\sqrt{3}}{2}$

Then,

$\rm\beta\:=\dfrac{-b-\sqrt{D}}{2a}$

$\rm\longrightarrow\beta\:=\dfrac{-5\sqrt{3}-\sqrt{27}}{2\times{2}}$

$\rm\longrightarrow\beta\:=\dfrac{-5\sqrt{3}-3\sqrt{3}}{4}$

$\rm\longrightarrow\beta\:=\dfrac{-8\sqrt{3}}{4}$

$\rm\longrightarrow\beta\:=-2\sqrt{3}$

Hence, $\rm\:\dfrac{-\sqrt{3}}{2}\:and\:-2\sqrt{3}$ are the roots of the given equation.