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How do you solve #2x^2 - 10x + 7 = 0#?

Answer 1

Avery Alexander

The solutions are:
#color(blue)(x=(5+sqrt(11))/2#

#color(blue)(x=(5-sqrt(11))/2#

#2x^2−10x+7=0#

The equation is of the form #color(blue)(ax^2+bx+c=0# where: #a=2, b=-10, c=7#

The Discriminant is given by:

#color(blue)(Delta=b^2-4*a*c#

# = (-10)^2-(4* 2 * 7)#

# = 100 -56=44#

The solutions are found using the formula #color(blue)(x=(-b+-sqrtDelta)/(2*a)#

#x = (-(-10)+-sqrt(44))/(2*2) = (10+-2sqrt(11))/4#

#x=(cancel2( 5+-sqrt(11)))/cancel4#

#color(blue)(x=(5+sqrt(11))/2#

#color(blue)(x=(5-sqrt(11))/2#

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Answer 2

Abigail Allen

To solve the quadratic equation 2x^2 - 10x + 7 = 0, you can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

Where: a = 2 b = -10 c = 7

Plugging in the values:

x = (-(-10) ± √((-10)^2 - 4(2)(7))) / (2(2))

x = (10 ± √(100 - 56)) / 4

x = (10 ± √44) / 4

x = (10 ± 2√11) / 4

So, the solutions are:

x = (10 + 2√11) / 4 x = (10 - 2√11) / 4

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.