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3 years ago

Which of the following equations has one real solution?

Mathematics

1 answer:

erastovalidia [21]3 years ago

4 0

Answer:

4x2−12x+9=0

Step-by-step explanation:

The equation, 2x2+x+3=0, does not have a solution

The equation, x2+5x−3=0, equals -5/2+1/2√37 and -5/2+-1/2√37

The equation, x2−5x+4=0, equal 1 and 4

The equation, 4x2−12x+9=0, equals only 3/2

Hope this Helps

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Sedbober [7]

Answer:

$x=\frac{275}{62}$ , $y=\frac{127}{31}$

Step-by-step explanation:

<h3>I will use the elimination method.</h3>

We want to make the X's the same:

$24x-15y=45$

$24x+16y=172$

Because the signs of the X's are the same we subtract the 2 equations to make:

(I put the second one on top of the 1st)

$31y=127$

So:

$y=\frac{127}{31}$

Substitute y into either equation 1 or 2:

(I chose equation 1)

$8x-5(\frac{127}{31})=15$

$8x=5(\frac{127}{31}) + 15$

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So:

$x=\frac{275}{62}$

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Step-by-step explanation:

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Andreyy89

Answer:

B and D

Step-by-step explanation:

The quadratic formula is $x=\frac{-b+/-\sqrt{b^2-4ac} }{2a}$ . Here a=20, b=-26, and c=8.

Substitute and you'll have:

$x=\frac{-b+/-\sqrt{b^2-4ac} }{2a} =\frac{26+/-\sqrt{26^2-4(20)(8)} }{2(20)}=\frac{26+/-\sqrt{676-640} }{40)}$

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Answer:

Step-by-step explanation:

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———————

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