Question

The number of positive integral solutions of

${x}^{2} + 5x + 4 < 5 \sqrt{ {x}^{2} + 5x + 28 }$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given inequality is

$\rm \: {x}^{2} + 5x + 4 < 5 \sqrt{ {x}^{2} + 5x + 28 }$

$\rm \: {x}^{2} + 5x + 28 - 28 + 4 < 5 \sqrt{ {x}^{2} + 5x + 28 }$

$\rm \: ({x}^{2} + 5x + 28) - 24 < 5 \sqrt{ {x}^{2} + 5x + 28 }$

Let assume that

$\rm \: \sqrt{ {x}^{2} + 5x + 28 } = y$

So, above expression can be rewritten as

$\rm \: {y}^{2} - 24 < 5y$

$\rm \: {y}^{2} - 5y - 24 < 0$

$\rm \: {y}^{2} - 8y + 3y - 24 < 0$

$\rm \: y(y - 8) + 3(y - 8) < 0$

$\rm \: (y - 8)(y + 3) < 0$

$\rm\implies \: - 3 < y < 8$

$\rm\implies \: 0 < {y}^{2} < 64$

$\rm\implies \: 0 < {x}^{2} + 5x + 28 < 64$

Now, 2 cases arises

Consider Case :- 1

$\rm \: {x}^{2} + 5x + 28 > 0$

Since, its a quadratic equation such that a = 1 > 0 and Discriminant, D = 25 - 4 × 28 × 1 = 25 - 112 = - 87 < 0

$\rm \:\rm\implies \: {x}^{2} + 5x + 28 > 0 \: is \: true \: \forall \: x \in \: R - - - (1)$

Consider Case :- 2

$\rm \: {x}^{2} + 5x + 28 < 64$

$\rm \: {x}^{2} + 5x + 28 - 64 < 0$

$\rm \: {x}^{2} + 5x - 36 < 0$

$\rm \: {x}^{2} + 9x - 4x - 36 < 0$

$\rm \: x(x + 9) - 4(x + 9) < 0$

$\rm \: (x + 9)(x - 4) < 0$

$\rm\implies \: - 9 < x < 4 - - - (2)$

From equation (1) and (2), we concluded that

$\rm\implies \:x \: \in \: R \: \cap \: ( - 9,4)$

$\rm\implies \:x \: \in \: ( - 9,4)$

Now, we need to find positive integral solutions. So, x can assume the values as

$\rm \: x = \{1, \: 2, \: 3 \}$

$\red{\rm\implies\boxed{ \sf{ \:Number \: of \: + ve \: integral \: solutions \: = \: 3}} }$

$\rule{190pt}{2pt}$

Basic Concept Used :-

Let us consider a quadratic expression f(x) = ax² + bx + c, such that a > 0 and Discriminant, D = b² - 4ac < 0, then f(x) > 0

If a and b are two positive real numbers such that a < b, then (x - a)(x - b) < 0 implies a < x < b.

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{ |x| < y\rm\implies \: - y < x < y}\\ \\ \bigstar \: \bf{ |x| \leqslant y\rm\implies \: - y \leqslant x \leqslant y}\\ \\ \bigstar \: \bf{ |x| > y\rm\implies \: x < - y \: or \: x > y} \: \\ \\ \bigstar \: \bf{ |x| \geqslant y\rm\implies \:x \leqslant - y \: or \: x \geqslant y}\\ \\ \bigstar \: \bf{ |x - a| < y\rm\implies \:a - y < x < a + y}\\ \\ \bigstar \: \bf{ |x - a| \leqslant y\rm\implies \:a - y \leqslant x \leqslant a + y}\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

EXPLANATION.

Number of positive integrals solution.

⇒ x² + 5x + 4 < 5√(x² + 5x + 28).

As we know that,

We can write equation as,

⇒ x² + 5x + 4 = x² + 5x + 28 - 24.

We can write as,

⇒ x² + 5x + 28 - 24 < 5√(x² + 5x + 28).

Let we assume that,

⇒ √(x² + 5x + 28) = t.

Squaring on both sides of the equation, we get.

⇒ (x² + 5x + 28) = t².

Put the values in the equation, we get.

⇒ t² - 24 < 5t.

⇒ t² - 5t - 24 < 0.

As we know that,

Factorizes the equation into middle term splits, we get.

⇒ t² - 8t + 3t - 24 < 0.

⇒ t(t - 8) + 3(t - 8) < 0.

⇒ (t + 3)(t - 8) < 0.

Plot this point on wavy curve methods, we gt.

⇒ t + 3 = 0.

⇒ t = - 3.

⇒ t - 8 = 0.

⇒ t = 8.

⇒ t ∈ (- 3, 8).

⇒ - 3 < t < 8.

Put the values of √(x² + 5x + 28) = t in the equation, we get.

⇒ - 3 < √(x² + 5x + 28) < 8.

Squaring on both sides of the equation, we get.

⇒ (- 3)² < (x² + 5x + 28) < (8)².

⇒ 0 < (x² + 5x + 28) < 64.

Now we considered,

Two cases in this situations.

Case = 1.

⇒ x² + 5x + 28 > 0.

This equation is true for ∀ x ∈ R. - - - - - (1).

Case = 2.

⇒ x² + 5x + 28 < 64.

⇒ x² + 5x + 28 - 64 < 0.

⇒ x² + 5x - 36 < 0.

Factorizes the equation into middle term splits, we get.

⇒ x² + 9x - 4x - 36 < 0.

⇒ x(x + 9) - 4(x + 9) < 0.

⇒ (x - 4)(x + 9) < 0.

Put the points on wavy curve method, we get.

⇒ x - 4 = 0.

⇒ x = 4.

⇒ x + 9 = 0.

⇒ x = - 9.

We get,

⇒ x ∈ (- 9, 4).

⇒ - 9 < x < 4. - - - - - (2).

From equation (1) and (2), we get.

⇒ x ∈ (- 9, 4).

All points are : {- 8, - 7 , - 6, - 5, - 4, - 3, - 2, - 1, 0, 1, 2, 3}.

But we want only positive integrals,

∴ Number of positive integrals = 3.