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Softa [21]
2 years ago
12

Solve (2x + 5)= (2x + 3)(2x - 1)

Mathematics
2 answers:
____ [38]2 years ago
6 0

Answer:

x = -1± sqrt( 33)

    ----------------------------

             4

Step-by-step explanation:

(2x + 5)= (2x + 3)(2x - 1)

Distribute

2x+5 = 4x^2 +6x-2x-3

Combine like terms

2x+5 = 4x^2 +4x -3

Subtract 2x +5 from each side

0 = 4x^2 +4x-3-2x-5

Combine like terms

0 =4x^2 +2x -8

Using the quadratic formula

a=4  b=2  c=-8

x = -b± sqrt( b^2-4ac)

    ----------------------------

             2a

x = -2± sqrt( 2^2-4(4)(-8))

    ----------------------------

             2(4)

x = -2± sqrt( 4+128)

    ----------------------------

             8

x = -2± sqrt( 132)

    ----------------------------

             8

x = -2± 2sqrt( 33)

    ----------------------------

             8

x = -1± sqrt( 33)

    ----------------------------

             4

AlexFokin [52]2 years ago
4 0

(2x + 5)= (2x + 3)(2x - 1)\\2x+5=4x^2-2x+6x-3\\4x^2-2x+6x-3-2x-5=0\\4x^2+2x-8=0\\(4x+1-\sqrt{33})( 4x+1+\sqrt{33})=0 \\ x=\frac{-1+\sqrt{33} }{4} /or/x=\frac{-1-\sqrt{33} }{4}

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kumpel [21]

If the equation x² + 16x + c = 0 has only one real solution, the value of c is equal to 64.

<h3>Discriminant: </h3>

The quadratic formula includes the discriminant, which comes after the square root. The discriminant of a quadratic equation is important because it indicates the quantity and kind of solutions.  

The formula for the Discriminant of a Quadratic Equation ax²+bx +c = 0 is given by

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Here we have

x² + 16x + c = 0 equation has only one real solution

Compare given equation with ax²+ bx + c = 0

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As we know when an equation has only one real solution

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By the above values,

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=> 256 - 4c = 0

=> 4c = 256

=> c = 256/4

=> c = 64

Therefore,

If the equation x² + 16x + c = 0 has only one real solution, the value of c is equal to 64.

 

Learn more about Discriminant at

brainly.com/question/15884086

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