If x² - px + 4 > 0 for all real values of x, then which one of the following is correct?

Concept:

Consider a quadratic equation: $\mathrm{a}{\mathrm{x}}^2+\mathrm{b}\mathrm{x}+\mathrm{c}=0$

Roots of quadratic equation are given by,

⇔ $\mathrm{x}=\frac{-\mathrm{b}\pm \sqrt{{\mathrm{b}}^2-4\mathrm{a}\mathrm{c}}}{2\mathrm{a}}$ 

The term  b²- 4ac is called discriminant, it is important as it tells about the nature of roots.

A parabola graphically represents a quadratic equation with a vertex located at the origin, below the x-axis, or above the x-axis. Therefore, a quadratic function may have one, two, or zero roots.

Calculation:

Given: ${\mathrm{x}}^2-\mathrm{p}\mathrm{x}+4>0\mathrm{\forall }\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{l}\mathrm{v}\mathrm{a}\mathrm{l}\mathrm{u}\mathrm{e}\mathrm{s}\mathrm{o}\mathrm{f}\mathrm{x}$ 

On comparing the given equation with the standard quadratic equation

$\mathrm{a}{\mathrm{x}}^2+\mathrm{b}\mathrm{x}+\mathrm{c}=0$, we get

a = 1, b = -p and c = 4

Given quadratic equation is more than zero that means it has no real roots If roots are not real (Then only the quadratic equation will not intersect the x-axis)

∴ ${\mathrm{b}}^2-4\mathrm{a}\mathrm{c}<0$ 

$\Rightarrow {\left(-\mathrm{p}\right)}^2-4\left(1\right)\left(4\right)<0$

$\Rightarrow {\left(\mathrm{p}\right)}^2-16<0$

$\Rightarrow {\left(\mathrm{p}\right)}^2-4^2<0$

$\Rightarrow \left(\mathrm{p}+4\right)\left(\mathrm{p}-4\right)<0$                     (Using a²- b² = (a + b). (a - b))

This inequality holds if p ranges from -4 to 4 i.e., p = (-4, 4)

Only |p| < 4 satisfy the above condition

Hence, option 1 is correct.

Important Points

If the quadratic function is always more than zero, that means, when graphed on the coordinate system, the function never intersects the X-axis. That means that the quadratic equation has no real roots.