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Morgarella [4.7K]
3 years ago
10

Use the discriminant to determine the number of solutions to the quadratic equation 3x^2+5x=-1

Mathematics
1 answer:
kari74 [83]3 years ago
4 0

Answer:

Two real distinct solutions

Step-by-step explanation:

Hi there!

<u>Background of the Discriminant</u>

The discriminant b^2-4ac applies to quadratic equations when they are organised in standard form: ax^2+bx+c=0.

All quadratic equations can be solved with the quadratic formula: x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}}.

When b^2-4ac is positive, it is possible to take its square root and end up with two real, distinct values of x.

When it is zero, we won't be taking the square root at all and we will end up with two real solutions that are equal, or just one solution.

When it is negative, it is impossible to take the square root and we will end up with two non-real solutions.

<u>Solving the Problem</u>

<u />3x^2+5x=-1<u />

We're given the above equation. It hasn't been organised completely in ax^2+bx+c=0, but we can change that by adding 1 to both sides to make the right side equal to 0:

3x^2+5x+1=0<u />

Now that we can identify the values of a, b and c, we can plug them into the discriminant:

D=b^2-4ac\\D=(5)^2-4(3)(1)\\D=25-4(3)(1)\\D=25-12\\D=13

Therefore, because the discriminant is positive, the equation has two real, distinct solutions.

I hope this helps!

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1 cm of cable weighs = \frac{52}{8.5}grams
                                      
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Then, 

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<h2><em>Answer:</em></h2><h2><em>x^</em><em>2</em><em>+</em><em>4</em><em> </em><em>and </em><em>X+</em><em>2</em></h2>

<em>Solution,</em>

<em>x^</em><em>4</em><em>-</em><em>1</em><em>6</em>

<em>=</em><em> </em><em>(</em><em>X^</em><em>2</em><em>)</em><em>^</em><em>2</em><em>-</em><em>(</em><em>4</em><em>)</em><em>^</em><em>2</em>

<em>Use </em><em>the </em><em>formula(</em><em>a^</em><em>2</em><em>-</em><em>b</em><em>^</em><em>2</em><em>)</em>

<em>=</em><em> </em><em>(</em><em>x^</em><em>2</em><em>+</em><em>4</em><em>)</em><em>(</em><em>x^</em><em>2</em><em>-</em><em>4</em><em>)</em>

<em>=</em><em>(</em><em>x^</em><em>2</em><em>+</em><em>4</em><em>)</em><em>(</em><em>x^</em><em>2</em><em>-</em><em>2</em><em>^</em><em>2</em><em>)</em>

<em>=</em><em>(</em><em>x^</em><em>2</em><em>+</em><em>4</em><em>)</em><em>(</em><em>X+</em><em>2</em><em>)</em><em>(</em><em>x-2)</em>

<em>Hope </em><em>it</em><em> helps</em>

<em>Good </em><em>luck</em><em> on</em><em> your</em><em> assignment</em>

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The mistake could be in operation however here is the all operation

Step-by-step explanation:

Getting each term and sum each one

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Check replacing in the original equation:

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