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Lapatulllka [165]
3 years ago
10

Marcus solved for x in the quadratic equation x2 – 10x + 25 = 0. x = x = x = What is true about Marcus’s work? Check all that ap

ply. Marcus should have substituted –10 for b, not 10. The denominator should be 1, not 2. Marcus should have subtracted 4(1)(25) in the square root. This equation, when solved correctly, only has 1 real number solution. This equation has 2 real number solutions.
Mathematics
2 answers:
aniked [119]3 years ago
7 0
Marcus should have substituted -10for b, not 10
Marcus should have subtracted 4(1)(25) in the square root
This equation, when solved correctly, only has one real number solution


So A, C, D
agasfer [191]3 years ago
6 0

we know that

The formula to solve a quadratic equation of the form ax^{2} +bx+c=0 is equal to

x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}

Step 1

Solve the quadratic equation by the formula

in this problem we have

x^{2} -10x+25=0  

so

a=1\\b=-10\\c=25

substitute in the formula

x=\frac{-(-10)(+/-)\sqrt{(-10)^{2}-4(1)(25)}} {2(1)}

x=\frac{10(+/-)\sqrt{(100)-100}}{2}

x=\frac{10}{2}=5

Step 2

<u>Statements about Marcus work</u>

<u>case A)</u> Marcus should have substituted -10 for b, not 10

The statement is True

The value of b is -10

See the procedure in Step 1

<u>case B)</u> The denominator should be 1, not 2

The statement is False

The denominator is 2

See the procedure in Step 1

<u>case C)</u> Marcus should have subtracted 4(1)(25) in the square root

The statement is True

See the procedure in Step 1

<u>case D) </u>This equation, when solved correctly, only has 1 real number solution

The statement is True

See the procedure in Step 1

The solution is x=5

<u>case E)</u> This equation has 2 real number solutions

The statement is False

The equation has only one real number solution

See the procedure in Step 1


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