The graph of the equation $3 x^{2} + 4 x + 5$ = 0 intersects the x-axis at point(s).

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A 0
The point at which the graph of a polynomial intersects the $X$ -axis is called a root of the corresponding equation.
A quadratic polynomial will intersect the $X$ -axis, if it has real roots.
The nature of the roots of a polynomial $a x^{2} + b x + c$ depends on its discriminant, $D$ = $b^{2} - 4 a c .$
Consider the equation $3 x^{2} + 4 x + 5 = 0.$
$⟹ D = b^{2} - 4 a c = 4^{2} - 4 (3) (5)$
$= - 44$
Since $D = - 44 < 0,$ the given equation has no real roots.
Hence, there will be no point of intersection of the polynomial with the $X$ -axis.

Suggest Corrections

Similar questions

y = x^{3} - 3 x^{2} + 5 x - 3

f (x) = 2 x^{3} + 3 x^{2} + c x + 8

c

f

(- 4, 0) (\frac{1}{2}, 0)

(p, 0)

c

y = 4 x - x^{2}

x^{2} + y^{2} + 4 x - 4 y - 5 = 0

The equation of the circle which intersects circles $x^{2} + y^{2} + x + 2 y + 3 = 0, x^{2} + y^{2} + 2 x + 4 y + 5 = 0$ and

$x^{2} + y^{2} - 7 x - 8 y - 9 = 0$ at right angle, will be

Join BYJU'S Learning Program