Find the zeros of the function. Enter the solutions from least to greatest. f(x)=(x-10)^(2)-49 lesser x= greater x=

Set function equal to zero: To find the zeros of the function, we need to set the function equal to zero and solve for x. So, we set f(x) = 0: 0 = (x - 10)^2 - 49Isolate the squared term: Next, we add 49 to both sides of the equation to isolate the squared term: (x - 10)^2 = 49Take the square root: Now, we take the square root of both sides of the equation. Remember that taking the square root gives us two solutions, one positive and one negative: x - 10 = ±√49Solve for x: The square root of 49 is 7, so we have: x - 10 = ±7Positive solution: We will solve for x by adding 10 to both sides of each equation: For the positive solution: x = 7 + 10 x = 17Negative solution: For the negative solution: x = -7 + 10 x = 3Final solutions: We have found two solutions for x. The lesser value is 3 and the greater value is 17. lesser x = 3 greater x = 17

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Find the zeros of the function. Enter the solutions from least to greatest.

f(x)=(x-10)^(2)-49
lesser
x=
greater
x=

Find the zeros of the function. Enter the solutions from least to greatest. $\newline$ $f(x)=(x-10)^{2}-49$ $\newline$ lesser $x=$ $\newline$ greater $x=$

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Math Problems

Algebra 1

Transformations of absolute value functions: translations and reflections

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Q. Find the zeros of the function. Enter the solutions from least to greatest.

\newline

f(x)=(x-10)^{2}-49

\newline

lesser

x=

\newline

greater

x=

Set function equal to zero: To find the zeros of the function, we need to set the function equal to zero and solve for x. $\newline$ So, we set $f(x) = 0$ : $\newline$ $0 = (x - 10)^2 - 49$
Isolate the squared term: Next, we add $49$ to both sides of the equation to isolate the squared term: $\newline$ $(x - 10)^2 = 49$
Take the square root: Now, we take the square root of both sides of the equation. Remember that taking the square root gives us two solutions, one positive and one negative: $\newline$ $x - 10 = \pm\sqrt{49}$
Solve for x: The square root of $49$ is $7$ , so we have: $x - 10 = \pm7$
Positive solution: We will solve for $x$ by adding $10$ to both sides of each equation: $\newline$ For the positive solution: $\newline$ $x = 7 + 10$ $\newline$ $x = 17$
Negative solution: For the negative solution: $\newline$ x = $-7 + 10$ $\newline$ x = $3$
Final solutions: We have found two solutions for $x$ . The lesser value is $3$ and the greater value is $17$ . $\newline$ lesser $x = 3$ $\newline$ greater $x = 17$