Question

Type the correct answer in each box. Use numerals instead of words. Consider this quadratic equation. x2 + 2x + 7 = 21 The number of positive solutions to this equation is . The approximate value of the greatest solution to the equation, rounded to the nearest hundredth, is .

Answer 1

Answer:

(a) 1

(b) 1

(c) 3.03

Step-by-step explanation:

The given quadratic equation is

Subtract 27 from both sides.

Taking out common factor.

Divide both sides by 4.

If an expression is , then we need to add , to make it perfect square.

Here, b=2, so

Add 1 on both sides.

Taking square root on both sides.

Subtract 1 from both sides.

and

and

Only one solution is positive.

Greatest solution is 3.031, therefore the approximate value of this solution is 3.03.

Answer 2

Answer:

This quadratic equation has only 1 positive solution, and the greatest solution is 2.87, rounded to the nearest hundredth.

Step-by-step explanation:

The given expression is

x^{2}+2x+7=21

To solve this expression, we need to pass all terms to the left side

$x^{2}+2x+7=21\\x^{2}+2x+7-21=0\\x^{2}+2x-14=0$

Now, we solve the equation using the quadratic formula

$x_{1,2}=\frac{-b\±\sqrt{b^{2}-4ac} }{2a}$

Where

$a=1\\b=2\\c=-14$

Replacing these values, we have

$x_{1,2}=\frac{-2\±\sqrt{2^{2}-4(1)(-14)} }{2(1)}\\x_{1,2}=\frac{-2\±\sqrt{4+56} }{2}=\frac{-2\±\sqrt{60} }{2} \\x_{1}\approx 2.9\\x_{2}\approx -4.9$

Therefore, this quadratic equation has only 1 positive solution, and the greatest solution is 2.87, rounded to the nearest hundredth.