SOLUTION: Find an integer c such that the equation 4x^2+cx=9=0 has a double real root. please help me with this problem thank you.

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Question 939556: Find an integer c such that the equation 4x^2+cx=9=0 has a double real root.
please help me with this problem thank you.
Found 4 solutions by stanbon, josgarithmetic, MathLover1, MathTherapy:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Find an integer c such that the equation 4x^2+cx+9=0 has a double real root.
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Condition required to have double real root:: b^2-4ac = 0
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Solve:: c^2 -4*4*9 = 0
c^2 -144 = 0
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(c-12)(c+12) = 0
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c = 12 or c = -12
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Cheers,
Stan H.
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Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Want two solutions.
You wrote two equality symbols, so I am using the first of them as a PLUS sign.

Try discriminant. You do not want a square, but you want two different solutions for x.

c%5E2-4%2A4%2A9

The discriminant needs to be positive.

c%5E2-16%2A9
c%5E2=16%2A9
c=12------------the value if exactly one real root. But you want TWO real roots, and want c as integer.

c can be any of c=13 or c=14, or ANY integer greater than 12.

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

A double root occurs when the quadratic is a perfect square trinomial.
4x^2+cx=9=0 I guess you have 4x%5E2%2Bcx%2B9=0+
use a rule a perfect square %28a%2Bb%29%5E2=a%5E2%2B2ab%2Bb%5E2
note that a=2x, 2ab=c, and b=3
%282x%29%5E2%2Bcx%2B3%5E2=0+ => c=2%2A2%2A3=12 ; so, the integer is 12
so, a perfect square trinomial is
4x%5E2%2B12x%2B9=0+
or %282x%2B3%29%5E2=0+ and solution is a double root
%282x%2B3%29%5E2=0+ =>%282x%2B3%29%282x%2B3%29=0+ => each factor is %282x%2B3%29=0+ and it is equal to zero if 2x=-3=>x=-3%2F2

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

Find an integer c such that the equation 4x^2+cx=9=0 has a double real root.
please help me with this problem thank you.


4x%5E2+%2B+cx+%2B+9+=+0 ----- Assuming that the extra “=” sign is actually “+”

Let c be p to prevent confusion, so the equation now becomes: 4x%5E2+%2B+px+%2B+9+=+0

For this equation to have a double real root, its discriminant: b%5E2+-+4ac will equal 0 (zero)

Therefore, with a being 4; b being p; and c being 9, we get:

p%5E2+-+4%284%29%289%29+=+0

p%5E2+-+144+=+0

p%5E2+=+144

p+=+%22+%22+%2B-sqrt%28144%29, or p, or highlight_green%28c+=+%22+%22+%2B-+12%29



Therefore, in 4x%5E2+%2B+cx+%2B+9+=+0, c will have a value of %22+%22+%2B-+12 in order for the equation to have a double root solution.