Question Use inspection to solve the equation x2=81 , then select the complete solution set below. If there are no real solutions, select “There are no real solutions.”(1 point) Responses {9} left brace 9 right brace {−9,9} left brace negative 9 comma 9 right brace There are no real solutions. There are no real solutions. {−9}
To solve the equation x^2 = 81, we can take the square root of both sides to isolate x.
√(x^2) = √81
|x| = 9
Since the absolute value of x equals 9, x could be either positive 9 or negative 9.
So, the complete solution set is {−9, 9}.
To solve the equation (x−18)^2 = 196, we can take the square root of both sides to isolate x−18.
√[(x−18)^2] = √196
|x−18| = 14
Now, we can split this equation into two cases:
1) x−18 = 14
Solving for x, we get x = 14 + 18 = 32
2) -(x−18) = 14
Simplifying, we get -x + 18 = 14
Subtracting 18 from both sides, we get -x = -4
Multiplying both sides by -1, we get x = 4
So, the solution set is {32, 4}.
To solve the equation (x+6)^2/2 = 8, we can first multiply both sides of the equation by 2 to remove the fraction:
(x+6)^2 = 2 * 8
(x+6)^2 = 16
Next, let's take the square root of both sides to isolate x + 6:
√((x+6)^2) = √16
|x + 6| = 4
Now, let's split this equation into two cases:
1) x + 6 = 4
Solving for x, we get x = 4 - 6 = -2
2) -(x + 6) = 4
Simplifying, we get -x - 6 = 4
Subtracting 6 from both sides, we get -x = 10
Multiplying both sides by -1, we get x = -10
So, the solution set is {-2, -10}.
To solve the equation (x+36)^2 - 38 = 251, we need to isolate the square term.
(x+36)^2 = 251 + 38
(x+36)^2 = 289
Next, let's take the square root of both sides to isolate x + 36:
√((x+36)^2) = ±√289
|x + 36| = ±17
Now, let's split this equation into two cases:
1) x + 36 = 17
Solving for x, we get x = 17 - 36 = -19
2) -(x + 36) = 17
Simplifying, we get -x - 36 = 17
Subtracting 36 from both sides, we get -x = 53
Multiplying both sides by -1, we get x = -53
So, the solution set is {-19, -53}.
Use inspection to solve the equation x2=36/169 , then select the correct solution set below. If there are no real solutions, select “There are no real solutions.”(1 point) Responses {−613} left brace negative Start Fraction 6 over 13 End Fraction right brace There are no real solutions. There are no real solutions. {−613,613} left brace negative Start Fraction 6 over 13 End Fraction comma Start Fraction 6 over 13 End Fraction right brace {613}
Solve the following quadratic equation using square roots: (x−18)^2=196 .(1 point) Responses {−4,−32} left brace negative 4 comma negative 32 right brace {214,−178} left brace 214 comma negative 178 right brace {21,5} left brace 21 comma 5 right brace {32,4} left brace 32 comma 4 right brace
Solve the following quadratic equation using square roots: (x+6)^2/2=8 .(1 point) Responses {10,−22} left brace 10 comma negative 22 right brace {−1.17,−6.83} left brace negative 1.17 comma negative 6.83 right brace {−2,−10} left brace negative 2 comma negative 10 right brace {−0.34,−11.66}
Solve the following quadratic equation using square roots: (x+36)^2−38=251 .(1 point) Responses {17.84,−13.84} left brace 17.84 comma negative 13.84 right brace {−19,−53} left brace negative 19 comma negative 53 right brace {53,19} left brace 53 comma 19 right brace {253,−325}
Use the table to answer the question. (x+3)(x+4)=0 x−3=0 x+4=0 x=3 x=−4 While using the Zero Product Property to find the values of x that make the quadratic equation (x+3)(x+4) equals 0, Oliver completed the work provided in the table. Is Oliver’s work accurate?(1 point) Responses No, he should have a third column where he should set x=0. No, he should have a third column where he should set x equals 0 . No, x−3=0 in the second row should be x+3=0. No, x minus 3 equals 0 in the second row should be x plus 3 equals 0 . No, x+4=0 in the second row should be x−4=0. No, x plus 4 equals 0 in the second row should be x minus 4 equals 0 . Yes, the work is accurate.
No, Oliver's work is not accurate.
In the second row, where x-3=0 and x+4=0, the signs should be switched. It should be x+3=0 and x-4=0.
So, the correct table should be:
x+3=0 x+4=0
x=-3 x=-4
Therefore, the answer is "No, x−3=0 in the second row should be x+3=0. No, x+4=0 in the second row should be x−4=0."