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s2008m [1.1K]
3 years ago
5

Describe the nature of the roots for this equation

Mathematics
1 answer:
vitfil [10]3 years ago
5 0

First, fint the roots of the equation 2x^2+5x-7=0.

1.

D=5^2-4\cdot 2\cdot (-7)=25+56=81,\\\sqrt{D}=\sqrt{81}=9,\\ \\x_1=\dfrac{-5-9}{2\cdot 2}=-\dfrac{14}{4}=-3.5,\\ \\x_2=\dfrac{-5+9}{2\cdot 2}=\dfrac{4}{4}=1.

2. You can see that this equation has two different real roots. Note that you can make this statement without finding roots, only knowing the value of discriminant: since D=81>0, then the equation has two different real roots.

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A positive real number is 2 less than another. When 4 times the larger is added to the square of the smaller, the result is 49.
Kobotan [32]

Answer:

The numbers are

-2+3\sqrt{5}   and 3\sqrt{5}

Step-by-step explanation:

Let

x -----> the smaller positive real number

y -----> the larger positive real number

we know that

A positive real number is 2 less than another

so

x=y-2

y=x+2 ----> equation A

When 4 times the larger is added to the square of the smaller, the result is 49

so

4y+x^2=49 ----> equation B

substitute equation A in equation B

4(x+2)+x^2=49

solve for x

x^2+4x-41=0

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}

in this problem we have

x^2+4x-41=0

so

a=1\\b=4\\c=-41

substitute in the formula

x=\frac{-4\pm\sqrt{4^{2}-4(1)(-41)}} {2(1)}

x=\frac{-4\pm\sqrt{180}} {2}

x=\frac{-4\pm6\sqrt{5}} {2}

x=-2\pm3\sqrt{5}

so

The positive real number is

x=-2+3\sqrt{5}

Find the value of y

y=x+2

y=-2+3\sqrt{5}+2

y=3\sqrt{5}

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A website had 826,140 hits last month what is the value of the 8 in 826,140
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3 years ago
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pishuonlain [190]
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7 0
3 years ago
Write three fractions equivalent to -4/4
Aleonysh [2.5K]

-2/2, -3/3, -1/1

Work: -4 ×-2=8

6 0
3 years ago
Read 2 more answers
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