Question

9th grade math please help its urgenttt

Answer 1

Answer:

B

Step-by-step explanation:

observe

$2x^2-5x+5=0

\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\\=\frac{5\pm \sqrt{(-5)^2-4(2)(5)}}{2(2)}

\\=\frac{5\pm \sqrt{-15}}{4}

\\=\frac{5\pm i\sqrt{15}}{4}$

Answer 2

Answer:

B

Step-by-step explanation:

(see attached for reference)

for a quadratic equation:

ax²+bx+c = 0

The solution given by the quadratic formula is :

x = [ -b ± √(b²-4ac) ] / 2a

comparing this to the given equation

2x² - 5x + 5 = 0

it is clear that a = 2, b = -5 and c = 5,

substituting the values into the quadratic formula:

x = [ -(-5) ± √( (-5)²-4(2)(5) ) ] / 2(2)

x = [ 5 ± √( 25 - 40 ) ] / 4

x = [ 5 ± √( -15 ) ] / 4    

x = [ 5 ± √( (-1)(15) ) ] / 4

x = [ 5 ± √( -1) √(15 ) ] / 4   (recall that √(-1) = i )

x = [ 5 ± i √(15 ) ] / 4   (answer B)

Hope this helps!