Question

$\sqrt{2} {x}^{2} + 3x \times \sqrt{2}$
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Answer 1

Correct Question :

$\tt \sqrt{2} {x}^{2} + 3x + \sqrt{2}$ then, Find its Zeros.

Solution :

We have,

Let try to find there zeros ( By Splitting the Middle term method )

=> √2x² + 3x + √2

=> √2x² + 2x + x + √2

=> √2x( x + √2 ) + 1 ( x + √2 )

=> (√2x + 1 )( √2 + 1 )

$\rule{90}1$

Either,

=> √2x + 1 = 0.

=> x = - 1 /√2

$\rule{90}1$

Or,

=> √2x + 1 = 0.

=> x = - 1 /√2.

Therefore, the value of zeros be - 1 /√2 and - 1 /√2.

$\rule{200}3$

Verification :

Let,

• A = - 1/ √2
• B = - 1/ √2

$\rule{90}1$

Sum of Zeros.

=> A + B = -1/√2 - 1/√2

=> A + B = -2 √2.

$\rule{90}1$

Product of zeros.

=> A * B = -1/√2 * - 1/√2

=> A * B = 2/2

=> A * B = 1.

$\rule{90}1$

So,

K ( x² - Sx + P )

• S sum of zeros.
• P product of Zeros.
• K Constant term.

=> K [ x² - ( -2 /√2)x + 1 ]

=> K ( √2x² + 3x + √2 )

Hance Proved.