Solve the following- x2 - 13x - 40-19x - x2 -78 - 0-

x^2+13x+40√(19x x^2+78)≤ 0 (x^2+13x+40)√(19x x^2+78)19x x^2+78≤ 0 ⇒(x^2+13x+40)√(19x x^2+78)≤ 0 √(19x x^2+78) 19x x^2+78≥ 0 (x^2+1 ...

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Byju's Answer

Standard VIII

Mathematics

Factors of Algebraic Expressions and Factorisation

Solve the fol...

Question

Solve the following:
$\frac{x^{2} + 13 x + 40}{\sqrt{19 x - x^{2} + 78}} = 0.$

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Solution

$\frac{x^{2} + 13 x + 40}{\sqrt{19 x - x^{2} + 78}} \leq 0$

$\frac{(x^{2} + 13 x + 40) \sqrt{19 x - x^{2} + 78}}{19 x - x^{2} + 78} \leq 0$

$\Rightarrow (x^{2} + 13 x + 40) \sqrt{19 x - x^{2} + 78} \leq 0$

$\sqrt{19 x - x^{2} + 78}$

$19 x - x^{2} + 78 \geq 0$

$(x^{2} + 13 x + 40) \sqrt{19 x - x^{2} + 78} \leq 0$

$(- x^{2} + 19 x + 78) {(x^{2} + 13 x + 40)}^{2} \leq 0$

$- x^{6} - 7 x^{5} + 323 x^{4} + 5719 x^{3} + 37582 x^{2} + 111520 x + 124800 \leq 0$

$- {(x + 5)}^{2} {(x + 8)}^{2} (x^{2} - 19 x - 78) \leq 0$

${(x + 5)}^{2} {(x + 8)}^{2} (x^{2} - 19 x - 78) \geq 0$

$- 19 x + x^{2} - 78 \leq 0$

$x^{2} - 19 x - 78 = 0$

$x = \frac{- (- 19) \pm \sqrt{{(- 19)}^{2} - 4 \cdot 1 (- 78)}}{2 \cdot 1} : \frac{19 \pm \sqrt{673}}{2}$

$- 19 x + x^{2} - 78 \leq 0$

$x^{2} - 19 x - 78 = 0$

$x = \frac{- (- 19) \pm \sqrt{{(- 19)}^{2} - 4 \cdot 1 (- 78)}}{2 \cdot 1} : \frac{19 \pm \sqrt{673}}{2}$ $

$x = \frac{19 - \sqrt{673}}{2}, x = \frac{19 + \sqrt{673}}{2}$

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Similar questions

Using the remainder Theorem, factorise each of the following completely:
(i) $3 x^{3} + 2 x^{2} - 19 x + 6$
(ii) $2 x^{3} + x^{2} - 13 x + 6$
(iii) $3 x^{3} + 2 x^{2} - 23 x - 30$
(iv) $4 x^{3} + 7 x^{2} - 36 x - 63$
(v) $x^{3} + x^{2} - 4 x - 4.$

Q. Factorise:

x^{2} + 13 x + 40

Q. Solve:

\sqrt{40 - x^{2}} - 6 = 0

If $({5 x}^{2} + 14 x + 2)^{2}$ – $({4 x}^{2} - 5 x + 7)^{2}$ is divided by $(x^{2} + x + 1)$ , then quotient q and remainder r are given by:

Q. x² – 13x – 30 = 0

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Solve the following:x2+13x+40√19x−x2+78=0.

Solve the following:
$\frac{x^{2} + 13 x + 40}{\sqrt{19 x - x^{2} + 78}} = 0.$