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3 years ago

Which of the following best describes the zeros of the quadratic function above?

Mathematics

1 answer:

bixtya [17]3 years ago

4 0

The zeros of a quadratic function are found where the graph intersects the x-axis. If the graph interects the x-axis in 2 places, we have 2 real solutions; if the graph intersects--or just touches--the x-axis in one place we have one real solution multiplicity 2; if the graph doesn't go through the x-axis at all we have 2 imaginary solutions. Ours goes through the x-axis in 2 places so we have 2 real solutions. Choice A.

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Which unit would you use to measure the weight of a car <br> A. mg <br> B. kg <br> C. g<br> D. km

kobusy [5.1K]

Answer:

A. Mg

Step-by-step explanation:

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4 0

3 years ago

Read 2 more answers

What is the solution of

kobusy [5.1K]

Answer:

Third option: x=0 and x=16

Step-by-step explanation:

$\sqrt{2x+4}-\sqrt{x}=2$

Isolating √(2x+4): Addind √x both sides of the equation:

$\sqrt{2x+4}-\sqrt{x}+\sqrt{x}=2+\sqrt{x}\\ \sqrt{2x+4}=2+\sqrt{x}$

Squaring both sides of the equation:

$(\sqrt{2x+4})^{2}=(2+\sqrt{x})^{2}$

Simplifying on the left side, and applying on the right side the formula:

$(a+b)^{2}=a^{2}+2ab+b^{2}; a=2, b=\sqrt{x}$

$2x+4=(2)^{2}+2(2)(\sqrt{x})+(\sqrt{x})^{2}\\ 2x+4=4+4\sqrt{x}+x$

Isolating the term with √x on the right side of the equation: Subtracting 4 and x from both sides of the equation:

$2x+4-4-x=4+4\sqrt{x}+x-4-x\\ x=4\sqrt{x}$

Squaring both sides of the equation:

$(x)^{2}=(4\sqrt{x})^{2}\\ x^{2}=(4)^{2}(\sqrt{x})^{2}\\ x^{2}=16 x$

This is a quadratic equation. Equaling to zero: Subtract 16x from both sides of the equation:

$x^{2}-16x=16x-16x\\ x^{2}-16x=0$

Factoring: Common factor x:

x (x-16)=0

Two solutions:

1) x=0

2) x-16=0

Solving for x: Adding 16 both sides of the equation:

x-16+16=0+16

x=16

Let's prove the solutions in the orignal equation:

1) x=0:

$\sqrt{2x+4}-\sqrt{x}=2\\ \sqrt{2(0)+4}-\sqrt{0}=2\\ \sqrt{0+4}-0=2\\ \sqrt{4}=2\\ 2=2$

x=0 is a solution

2) x=16

$\sqrt{2x+4}-\sqrt{x}=2\\ \sqrt{2(16)+4}-\sqrt{16}=2\\ \sqrt{32+4}-4=2\\ \sqrt{36}-4=2\\ 6-4=2\\ 2=2$

x=16 is a solution

Then the solutions are x=0 and x=16

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3 years ago

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Tresset [83]

Answer:

The answer is 0.46947

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3 years ago

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Ksivusya [100]

Answer:

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Step-by-step explanation:

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2 years ago

What values of a and b make the equation true? sqrt(648)=sqrt(2^a•3^b)

Anuta_ua [19.1K]

ANSWER

The value of a and b are:

$a=3,b=4$

EXPLANATION

The expression given to us is

$\sqrt{648} = \sqrt{ {2}^{a} \times {3}^{b} }$

We square both sides of the equation to obtain

$648= {2}^{a} \times {3}^{b}$

We now find the prime factorization of 648 to obtain,

${2}^{3} \times {3}^{4} = {2}^{a} \times {3}^{b}$

We now compare the exponents on both sides of the equation to obtain,

$a=3,b=4$

Therefore the correct option is C

3 0

2 years ago