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2 years ago

Which value of b will cause the quadratic equation x2 + bx + 5 = 0 to have two real number solutions?

Mathematics

2 answers:

DedPeter [7]2 years ago

6 0

Answer:

The correct option is 1.

Step-by-step explanation:

The given quadratic equation is

$x^2+bx+5=0$

A quadratic equation $ax^2+bx+c=0$ have to real solution if the value of discriminant is garter than 0.

$b^2-4ac>0$

$b^2-4(1)(5)>0$

$b^2-20>0$

Add 20 on both the sides.

$b^2>20$

It is possible if

$b\sqrt{20}$

$b4.472$

Since -5<-.4.472, therefore option 1 is correct.

REY [17]2 years ago

5 0

Which value of b will cause the quadratic equation x2 + bx + 5 = 0 to have two real number solutions?

–5

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Step-by-step explanation:

Here we have , A customer at a store paid $64 for 3 large candles and 4 small candles. At the same store, a second customer paid $4 more than the first customer for 1 large candle and 8 small candles. The price of each large candle is the same, and the price of each small candle is the same. We need to find Which system of equations can be used to find the price in dollars of each large candle, x, and each small candle, y . Let's find out:

Let the price in dollars of each large candle, x, and each small candle, y .So

A customer at a store paid $64 for 3 large candles and 4 small candles

Equation is :

⇒ $3x+4y=64$ .....(1)

At the same store, a second customer paid $4 more than the first customer for 1 large candle and 8 small candles.

Equation is :

⇒ $x+8y=68$ .......(2)

3(2)-(1) i.e.

⇒ $3(x+8y)-(3x+4y)=68(3)-64$

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2 years ago

The first-, second-, and third-year enrollment values for a technical school are shown in the table below.

VARVARA [1.3K]

Answer:

The solution to f(x) = t(x) is x = 2010

Option 3 is true.

Step-by-step explanation:

The first-year , second-year , and third-year enrollment values for a technical school are shown in the table below.

Year (x) First Year f(x) Second Year s(x) Third Year t(x)

2009 785 756 756

2010 740 785 740

2011 690 710 781

2012 732 732 710

2013 781 755 800

Now we will check each option.

Option 1: The solution to f(x) = s(x) is x = 2,009

In year 2009, f(x)=s(x)

But 785≠756

Thus, False

Option 2: The solution to f(x) = s(x) is x = 785

x represents year, but 785 it no year

Thus, False

Option 3: The solution to f(x) = t(x) is x = 2010

In year 2010, f(x)=t(x)=740

But 740=740

Thus, True

Option 4: The solution to f(x) = t(x) is x =740

x represents year, but 740 it no year

Thus, False

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