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Mekhanik [1.2K]
2 years ago
13

What is the discriminant of the quadratic equation, 4x^2-8x-5=0​

Mathematics
2 answers:
uysha [10]2 years ago
7 0

Answer:

Step-by-step explanation:

4x^2-8x-5=0

The discriminant is the number under the √. In this case it's 144

Gennadij [26K]2 years ago
4 0

Answer:

\Delta=144

Step-by-step explanation:

For a quadratic in standard form, the discriminant is given by:

\Delta=b^2-4ac

We have the equation:

4x^2-8x-5=0

Hence, our a=4; b=-8; and c=-5.

Substituting into our formula, we acquire:

\Delta=(-8)^2-4(4)(-5)

Evaluate:

\Delta=64+80

Add:

\Delta=144

Therefore, our discriminant is 144.

Notes:

Since our discriminant is a positive value, this tells us that our quadratic has two real roots.

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8 0
2 years ago
Which expression has a value of 15 when n = 7? 43 minus 5 n 3 n minus 5 6 n minus 28 19 minus StartFraction 28 Over n EndFractio
andrey2020 [161]

Answer:

19-\dfrac{28}{n} is the expression with value of 15 when n = 7.

Step-by-step explanation:

To find the expression whose value is 15, substitute the value of n = 7 in each given expression.

Expression 1:

43-5n

Substituting the value of n,

43-5\left(7\right)

Simplifying,

43-35=8

Since the value of the expression is 8 which is not equal 15.

Hence expression 43-5n does not have value of 15 when n = 7.

Expression 2:

3n-5

Substituting the value of n,

3\left(7\right)-5

Simplifying,

21-5=16

Since the value of the expression is 16 which is not equal 15.

Hence expression 3n-5 does not have value of 15 when n = 7.

Expression 3:

6n-28

Substituting the value of n,

6\left(7\right)-28

Simplifying,

42-28=14

Since the value of the expression is 14 which is not equal 15.

Hence expression 6n-28 does not have value of 15 when n = 7.

Expression 4:

19-\dfrac{28}{n}

Substituting the value of n,

19-\dfrac{28}{7}

Simplifying,

19-4=15

Since the value of the expression is 15 which is equal 15.

Hence expression 19-\dfrac{28}{n} has the value of 15 when n = 7.

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3 years ago
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Express with radical signs instead of fractional exponents. Rationalize the denominator.
CaHeK987 [17]

x^{\frac{1}{2}} = \sqrt[2]{x} and 3^{-\frac{1}{2}} = \frac{1}{3^{\frac{1}{2}} } = \frac{1}{\sqrt[2]{3}} so

3^{-\frac{1}{2}} x^{\frac{1}{2}} = \frac{1}{\sqrt{3}} \cdot \sqrt{x}  = \frac{\sqrt{x}}{\sqrt{3}}

and you rationalize denominator by multiplying numerator and denominatr by \sqrt{3} so that gives--

\frac{\sqrt{x}}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3x}}{3}

Your answer is \frac{\sqrt{3x}}{3}

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