Respuesta :
we know that
The formula to solve a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
Step 1
Solve the quadratic equation by the formula
in this problem we have
[tex]x^{2} -10x+25=0[/tex]
so
[tex]a=1\\b=-10\\c=25[/tex]
substitute in the formula
[tex]x=\frac{-(-10)(+/-)\sqrt{(-10)^{2}-4(1)(25)}} {2(1)}[/tex]
[tex]x=\frac{10(+/-)\sqrt{(100)-100}}{2}[/tex]
[tex]x=\frac{10}{2}=5[/tex]
Step 2
Statements about Marcus work
case A) Marcus should have substituted [tex]-10[/tex] for b, not [tex]10[/tex]
The statement is True
The value of b is [tex]-10[/tex]
See the procedure in Step [tex]1[/tex]
case B) The denominator should be 1, not 2
The statement is False
The denominator is [tex]2[/tex]
See the procedure in Step [tex]1[/tex]
case C) Marcus should have subtracted [tex]4(1)(25)[/tex] in the square root
The statement is True
See the procedure in Step [tex]1[/tex]
case D) This equation, when solved correctly, only has 1 real number solution
The statement is True
See the procedure in Step [tex]1[/tex]
The solution is [tex]x=5[/tex]
case E) This equation has [tex]2[/tex] real number solutions
The statement is False
The equation has only one real number solution
See the procedure in Step [tex]1[/tex]
The root of the quadratic equation [tex]x^2-10x+25=0[/tex] are two real number values both being 5.
Given to us
[tex]x^2-10x+25=0[/tex]
What are the roots of a quadratic equation?
The roots of a general quadratic equation [tex]ax^2+bx+c[/tex] can be found using the formula,
[tex]x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
Using the same formula for the quadratic equation,
[tex]x^2-10x+25=0[/tex]
a= 1, b=-10, and c =25,
substitute the values,
[tex]x = \dfrac{-(-10)\pm\sqrt{(-10)^2-4(1)(25)}}{2(1)}[/tex]
[tex]x = \dfrac{10\pm\sqrt{100-100}}{2}[/tex]
[tex]x = \dfrac{10}{2}[/tex]
[tex]x = 5[/tex]
thus, the quadratic equation will have two real number values both being 5.
Learn more about Quadratic equations:
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