Question

find the zeroes
$4 {x}^{2} + 5 \sqrt{2} - 3$
$2 {x}^{2} \frac{7}{2} + \frac{3}{4}$
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Answer 1

Answer:

Step-by-step explanation:

4x²+5­­√2x-3

=4x² + 6­­√2x - √2x -3

=2√2x (√2x+3)-(√2x+3)

=(√2x+3) ( 2√2x-1)

√2x+3=0, x= -3/√2

2√2x-1=0

or x=1/2√2

zeros are: -3/√2 and 1/2√2

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2x²+7/2 x+3/4

=1/4(8x²+14x+3)

=1/4 ( 8x²+12x+2x+3)

=1/4 { 4x(2x+3)+(2x+3) }

=1/4( 4x+1)(2x+3)

now 4x+1=0,x=-1/4

2x+3=0, x=-3/2