Verified Answer :

samtoumi

you can solve all quadratic equations using this method:
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x^2 + 4x + 4 = 0
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the above quadratic equation is in standard form, with a=1, b=4, and c=4
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to solve the quadratic equation, by using the quadratic formula, copy and paste this:
1 4 4
into this solver: https://sooeet.com/math/quadratic-equation-solver.php
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this quadratic has ONE real root (the x-intercept), which is:
x = -2
x = -2
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x^2 + x = 0
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the above quadratic equation is in standard form, with a=1, b=1, and c=0

Answer: The correct equation is \((A)~x^2+4x+4=0.\)

Step-by-step explanation: We are given to select the quadratic equation that has only one solution.

We know that - for the quadratic equation \(ax^2+bx+c=0,a eq 0\) the type of solution can be determined by the discriminant \(D=b^2-4ac\) as follows :

(i) If D > 0, then the equation will have two distinct real solutions.

(ii) If D < 0, then the equation will have two distinct complex solutions.

(ii) If D = 0, then there is only one real solution.

(A) The first equation is \(x^2+4x+4=0.\)

Here, a = 1, b = 4 and c = 4.

So, the discriminant is given by

\(D=b^2-4ac=4^2-4\times 1\times 4=16-16=0.\)

Therefore, the equation will have only one real solution.

(B) The second equation is \(x^2+x=0.\)

Here, a = 1, b = 1 and c = 0.

So, the discriminant is given by

\(D=b^2-4ac=1^2-4\times 1\times 0=1-0=1>0.\)

Therefore, the equation will have two distinct real solutions.

(C) The third equation is \(x^2-1=0.\)

Here, a = 1, b = -1 and c = 0.

So, the discriminant is given by

\(D=b^2-4ac=(-1)^2-4\times 1\times 0=1-0=1>0.\)

Therefore, the equation will have two distinct real solutions.

Thus, \((A)~x^2+4x+4=0\) is the correct equation.

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