Factorise- x2 - 1-4x2 - 1 - 7x - 7-2x

The correct option is C (x + 1/2x) (x + 1/2x 7) #160; #160; x^2 + 14x^2 + 1 7x 72x =x^2+(12x)^2+2(x)(12x) 7(x+12x) =(x+12x)^2 7( ...

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Byju's Answer

Standard IX

Mathematics

Perfect Square Trinomials

Factorise: ...

Question

Factorise: $x^{2} + \frac{1}{4 x^{2}} + 1 - 7 x - \frac{7}{2 x}$

(x + \frac{9}{2 x}) (x + \frac{1}{2 x} - 3)

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(x - \frac{2}{x}) (x + \frac{1}{2 x} - 3)

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(x + \frac{1}{2 x}) (x + \frac{1}{2 x} - 7)

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(x - \frac{1}{x}) (x + \frac{1}{2 x} - 7)

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Solution

The correct option is C $(x + \frac{1}{2 x}) (x + \frac{1}{2 x} - 7)$
$x^{2} + \frac{1}{4 x^{2}} + 1 - 7 x - \frac{7}{2 x}$
$= x^{2} + (\frac{1}{2 x})^{2} + 2 (x) (\frac{1}{2 x}) - 7 (x + \frac{1}{2 x})$
$= (x + \frac{1}{2 x})^{2} - 7 (x + \frac{1}{2 x})$
$= (x + \frac{1}{2 x}) (x + \frac{1}{2 x} - 7)$

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Similar questions

Solve :

$(i) \frac{1}{3} x - 6 = \frac{5}{2} (i i) \frac{2 x}{3} - \frac{3 x}{8} = \frac{7}{12} (i i i) (x + 2) (x + 3) + (x - 3) (x - 2) - 2 x (x + 1) = 0 (i v) \frac{1}{10} - \frac{7}{x} = 35 (v) 13 (x - 4) - 3 (x - 9) - 4 (x + 4) = 0 (v i) x + 7 - \frac{8 x}{3} = \frac{17 x}{6} - \frac{5 x}{8} (v i i) \frac{3 x - 2}{4} - \frac{2 x + 3}{3} = \frac{2}{3} - x (v i i i) \frac{x + 2}{6} - (\frac{11 - x}{3} - \frac{1}{4}) = \frac{3 x - 4}{12} (i x) \frac{2}{5 x} - \frac{5}{3 x} = \frac{1}{15} (x) \frac{x + 2}{3} - \frac{x + 1}{5} = \frac{x - 3}{4} - 1 (x i) \frac{3 x - 2}{3} + \frac{2 x + 3}{2} = x + \frac{7}{6} (x i i) x - \frac{x - 1}{2} = 1 - \frac{x - 2}{3} (x i i i) \frac{9 x + 7}{2} - (x - \frac{x - 2}{7}) = 36 (x i v) \frac{6 x + 1}{2} + 1 = \frac{7 x - 3}{3}$

Find the value of $x : (x + 2) (x + 3) + (x - 3) (x - 2) - 2 x (x + 1) = 0$

Check whether the following are quadratic equations :
(i) ${(x + 1)}^{2} = 2 (x - 3)$
(ii) $x^{2} - 2 x = (- 2) (3 - x)$
(iii) $(x - 2) (x + 1) = (x - 1) (x + 3)$
(iv) $(x - 3) (2 x + 1) = x (x + 5)$
(v) $(2 x - 1) (x - 3) = (x + 5) (x - 1)$
(vi) $x^{2} + 3 x + 1 = {(x - 2)}^{2}$
(vii) ${(x + 2)}^{3} = 2 x (x^{2} - 1)$
(viii) $x^{3} - 4 x^{2} - x + 1 = {(x - 2)}^{3}$

Q. Solve

(x + 2) (x + 3) + (x - 3) (x - 2) - 2 x (x + 1) = 0

Q. Solve:

(x + 2) (x + 3) + (x - 3) (x - 2) - 2 x (x + 1) = 0

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Factorise: x2+14x2+1−7x−72x

The correct option is C (x+12x)(x+12x−7) x2+14x2+1−7x−72x=x2+(12x)2+2(x)(12x)−7(x+12x)=(x+12x)2−7(x+12x)=(x+12x)(x+12x−7)

Factorise: $x^{2} + \frac{1}{4 x^{2}} + 1 - 7 x - \frac{7}{2 x}$