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3 years ago

How to tell if a quadratic equation has a real solution?

Mathematics

2 answers:

Dennis_Churaev [7]3 years ago

7 0

The easiest way to determine if a quadratic equation has real solutions is by using the discriminant.

Start by finding a,b and c. You can find a by looking at the coefficient of x^2. You can find b by looking at the coefficient of x. You can find c by looking at the constant at the end.

Now use the discriminant formula.

b^2 - 4ac

After plugging in and solving, check the number. If it is negative, there are no real solutions. If it is positive there are 2 solutions. If it is 0, there is one real answer and it is a double root.

attashe74 [19]3 years ago

4 0

If the discriminant(b^2-4ac) is rational, then the quadratic is rational.

~ThePirc

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-17+ 1/4 x + 1/4 x + 31 = 24

wlad13 [49]

<h2><u>Answer:</u></h2><h2><u></u> $\boxed{x=20}$ <u></u></h2><h2><u></u></h2><h2><u>Solution Steps:</u></h2>

______________________________

<h3>1.) Combine like terms:</h3>

$\frac{1}{4}x+\frac{1}{4}x=\frac{1}{2}x$

<em> - We do this because we need 1 Number/Variable left over. </em>

<u>Equation at the end of Step 1:</u>

<u /> $-17+\frac{1}{2}x+31=24$ <u />

<em />

$-17+31=14$

<em> - We do this because we can only have 1 number on each side of the equals. (This doesn't count towards the number connected to the variable.)</em>

<u>Equation at the end of Step 2:</u>

<u /> $14+\frac{1}{2}x=24$ <u />

<h3>3.) Subtract 14 from both sides:</h3>

$14-14=$ Cancels Out

$24-14=10$

<em> - We do this because now we just need the 1 variable and the 1 number on opposite sides of the equal.</em>

<u>Equation at the end of Step 3:</u>

<u /> $\frac{1}{2}x=10$ <u />

<h3>4.) Multiply both sides by 2:</h3>

<em /> $\frac{1}{2}x$ <em> </em>× $2=x$ <em />

<em /> $10$ <em> </em>× $2=20$

<em> - We multiply by 2 since we have a fraction. In this case the reciprocal is 2 meaning we multiply by 2. </em>

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3 years ago

What is the sum of 2x^2 and 2x^3

jeyben [28]

For this case we have the following expression:

$2x ^ 2 + 2x ^ 3$

The terms are not similar, so they cannot be added.

<em>Examples of similar terms:</em>

$x ^ 3 + x ^ 3 = 2x ^ 3\\3x ^ 4 + 2x ^ 4 = 5x ^ 4$

Then, the expression given can only be rewritten as:

$2 (x ^ 2 + x ^ 3)$

This is taking common factor 2 to both terms.

Answer:

$2 (x ^ 2 + x ^ 3)$

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