Which of the following is true about the real solutions

Question

Which of the following is true about the real solutions for √4-x² - x +1 = 0

a. It has no real solutions

b. It has exactly one real solution c, and c>0

c. It has exactly one real solution c, and c<0

d. It has two real solutions a,b, and ab>0

e. It has two real solutions a,b, and ab<0

Answer 1

√4-x² - x +1 = 0

√4-x² = x - 1

4 - x² = ( x - 1 )² , raising both sides to power 2

4 - x² = x² - 2 x + 1

4 - x² - 4 = x² - 2 x + 1 -4 , adding (-4) to both sides

- x² = x² - 2 x - 3

0 = 2 x² - 2 x -3 adding x² to both sides

Use the discriminate to state the number and type of solutions for the equation ax² +b x+ c = 0

where a =2 , b = -2 , c = -3 , x = ( -b ± √ b²- 4 a c ) / 2 a

x = [ -(-2) ± √ 4 - 4 * 2* -3 ] / 2 * 2

x =( 2 ± √ 28 ) / 4

x = ( 2 ± 2√7) /4 , where 28 = √ 4*7 = 2√ 7

x = 1/2 ± √7/2

either x= 1/2 + √7/2 or x = 1/2 - √7/2

x ≈ 1.82 , or x ≈ - 0.82

Let us check the two roots in the original equation

√4-x² - x +1 = 0

If x ≈ 1.82, then

√4-x² - x +1 =√4-1.82² -1.82 +1 ≈ 0

If x ≈ - 0.82

then √4-x² - x +1= √4-0.82² +0.82 +1 ≈ 3.6 ≠ 0

So the only root is x= 1/2 + √7/2

we have one real solution

Answer 2

The solutions to any equation of the form ax²+bx+c=0 are

x=(-b±√(b²-4ac))/2a. One solution is using the plus sign, and the other the minus sign, unless you get the same value with either plus or minus, in which case you would just have that one solution, which would only happen if b²-4ac=0.

We can simplify the equation given in the problem using √4+1=3 to get

-x²-x+3=0. So a=-1, b=-1, and c=3. Putting those values in the quadratic formula you can find the solutions. Determine whether the solutions are real or imaginary. They are real if the value in the square root is greater than equal to zero. They are imaginary if the value in the square root is negative. Then of the real solutions, decide whether they are positive or negative to find the answer.

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