What is the value of the discriminant for the quadratic equation zero equals 2x^2+x-3
1 answer:
Discriminant for the quadratic equation is
How to find the d Discriminant of the quadratic equation ?
We know that the standard quadratic equation
Discriminant of the equation is
Equation given in the question is
So we can find the values of
Substitute the values
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The complete question is
"Substitute the values for a, b, and c into b2 – 4ac to determine the discriminant. Which quadratic equations will have two real number solutions? (The related quadratic function will have two x-intercepts.) Check all that apply.
0 = 2x^2 – 7x – 9
0 = 4x^ 2 – 3x – 1
The quadratic equations that have real number solutions are; 4x^2 – 3x – 1, and 2x^2 – 7x – 9.
<h3>What is the formula for Discriminant?</h3>The formula for finding the discriminant is
The solution contains the term which will be:
Real and distinct if the discriminant is positive
Real and equal if the discriminant is 0
Non-real and distinct roots if the discriminant is negative
For the quadratic equation
This equation has two real number solutions.
For the quadratic equation
This equation will have two real number solutions.
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Megan:
x to the one third power =
<span>x to the one twelfth power = </span>
<span>The quantity of x to the one third power, over x to the one twelfth power is:
</span>
<span>
Since </span>
then
Now, just subtract exponents:
1/3 - 1/12 = 4/12 - 1/12 = 3/12 = 1/4
Julie:
x times x to the second times x to the fifth = x * x² * x⁵
<span>The thirty second root of the quantity of x times x to the second times x to the fifth is
</span>![\sqrt[32]{x* x^{2} * x^{5} }](https://tex.z-dn.net/?f=%20%5Csqrt%5B32%5D%7Bx%2A%20x%5E%7B2%7D%20%2A%20x%5E%7B5%7D%20%7D%20)
<span>
Since </span>
Then![\sqrt[32]{x* x^{2} * x^{5} }= \sqrt[32]{ x^{1+2+5} } =\sqrt[32]{ x^{8} }](https://tex.z-dn.net/?f=%5Csqrt%5B32%5D%7Bx%2A%20x%5E%7B2%7D%20%2A%20x%5E%7B5%7D%20%7D%3D%20%5Csqrt%5B32%5D%7B%20x%5E%7B1%2B2%2B5%7D%20%7D%20%3D%5Csqrt%5B32%5D%7B%20x%5E%7B8%7D%20%7D)
Since![\sqrt[n]{x^{m}} = x^{m/n} }](https://tex.z-dn.net/?f=%20%5Csqrt%5Bn%5D%7Bx%5E%7Bm%7D%7D%20%3D%20x%5E%7Bm%2Fn%7D%20%7D%20)
Then![\sqrt[32]{ x^{8} }= x^{8/32} = x^{1/4}](https://tex.z-dn.net/?f=%5Csqrt%5B32%5D%7B%20x%5E%7B8%7D%20%7D%3D%20x%5E%7B8%2F32%7D%20%3D%20x%5E%7B1%2F4%7D%20)
Since both Megan and Julie got the same result, it can be concluded that their expressions are equivalent.
x to the one third power =
<span>x to the one twelfth power = </span>
<span>The quantity of x to the one third power, over x to the one twelfth power is:
</span>
<span>
Since </span>
then
Now, just subtract exponents:
1/3 - 1/12 = 4/12 - 1/12 = 3/12 = 1/4
Julie:
x times x to the second times x to the fifth = x * x² * x⁵
<span>The thirty second root of the quantity of x times x to the second times x to the fifth is
</span>
<span>
Since </span>
Then
Since
Then
Since both Megan and Julie got the same result, it can be concluded that their expressions are equivalent.