Factorise - 2-2x2 - 9x - 5 -2-0

2√(2)x^2+9x+5√(2)=0 2√(2)x^2+4x+5x+5√(2)=0 2√(2)x(x+√(2))+5(x+√(2))=0 (2√(2)x+5)(x+√(2))=0 (2√(2)x+5)=0 #160; or #160; #160; ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Mathematics

Quadratic Formula

Factorise : ...

Question

Factorise : $2 \sqrt{2} x^{2} + 9 x + 5 \sqrt{2} = 0$

Open in App

Solution

$2 \sqrt{2} x^{2} + 9 x + 5 \sqrt{2} = 0$

$2 \sqrt{2} x^{2} + 4 x + 5 x + 5 \sqrt{2} = 0$

$2 \sqrt{2} x (x + \sqrt{2}) + 5 (x + \sqrt{2}) = 0$

$(2 \sqrt{2} x + 5) (x + \sqrt{2}) = 0$

$(2 \sqrt{2} x + 5) = 0$ or $(x + \sqrt{2}) = 0$

$x = - \frac{5}{2 \sqrt{2}} = - \frac{5}{4} \sqrt{2}$ or $x = - \sqrt{2}$

Suggest Corrections

Similar questions

Q. Factorise :

\sqrt{2} x^{2} + 9 x + 4 \sqrt{2}

Q. Factorise:

2 x^{2} - 9 x + 9

Q. Factorise

2 x^{2} - 9 x + 9

Q. Factorise

4 y^{2} + 9 x^{2} - 40 y - 54 x + 145 = 0

Using the FActor theroem, show that:
(i) (x-2) is a factor fo $x^{3} - 2 x^{2} - 9 x + 18$ Hence factories the expression
(ii) (x+5) is a factor of $2 x^{3} + 5 x^{2} - 28 x - 15$ Hence factorise fthe expression $2 x^{3} + 5 x^{2} - 28 x - 18$ completely.
(iii) (3x+2) is a factor fo $3 x^{3} + 2 x^{2} - 3 x - 2.$ Hence factorise the expression
$3 x^{3} + 2 x^{2} - 3 x - 2$ completely

Join BYJU'S Learning Program

Factorise : 2√2x2+9x+5√2=0

2√2x2+9x+5√2=02√2x2+4x+5x+5√2=02√2x(x+√2)+5(x+√2)=0(2√2x+5)(x+√2)=0(2√2x+5)=0 or (x+√2)=0x=−52√2=−54√2 or x=−√2

Factorise : $2 \sqrt{2} x^{2} + 9 x + 5 \sqrt{2} = 0$