If $f(x)=-2x^{2}+7$ and $g(x)=|x-1|$ , find all values of $x$ , to the nearest tenth, for which $f(x)=g(x)$ ,

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Q. If

f(x)=-2x^{2}+7

and

g(x)=|x-1|

, find all values of

x

, to the nearest tenth, for which

f(x)=g(x)

Consider Cases: We need to solve the equation $f(x) = g(x)$ , which means we need to solve $-2x^2 + 7 = |x - 1|$ . We will consider two cases for the absolute value: one where $x - 1$ is non-negative ( $x \geq 1$ ) and one where $x - 1$ is negative ( $x < 1$ ).
Non-Negative Case: First, let's consider the case where $x - 1$ is non-negative, which means $|x - 1| = x - 1$ . So, we need to solve $-2x^2 + 7 = x - 1$ .
Quadratic Equation: Rearrange the equation to bring all terms to one side: $-2x^2 - x + 8 = 0$ .
Calculate Discriminant: Now, we need to solve the quadratic equation $-2x^2 - x + 8 = 0$ . We can use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = -2$ , $b = -1$ , and $c = 8$ .
Calculate Solutions: Calculate the discriminant: $b^2 - 4ac = (-1)^2 - 4(-2)(8) = 1 + 64 = 65$ .
Negative Case: Since the discriminant is positive, we have two real solutions. Calculate the solutions using the quadratic formula: $x = \frac{1 \pm \sqrt{65}}{-4}$ .
Quadratic Equation: The two solutions are $x = \frac{1 + \sqrt{65}}{-4}$ and $x = \frac{1 - \sqrt{65}}{-4}$ . To the nearest tenth, these are approximately $x = -2.5$ and $x = 1.5$ .
Calculate Discriminant: Now, let's consider the case where $x - 1$ is negative, which means $|x - 1| = -(x - 1)$ . So, we need to solve $-2x^2 + 7 = -(x - 1)$ .
Calculate Solutions: Rearrange the equation to bring all terms to one side: $-2x^2 + 7 = -x + 1$ , which simplifies to $-2x^2 + x + 6 = 0$ .
Check Valid Solutions: We need to solve the quadratic equation $-2x^2 + x + 6 = 0$ . Again, we can use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = -2$ , $b = 1$ , and $c = 6$ .
Final Valid Solutions: Calculate the discriminant: $b^2 - 4ac = (1)^2 - 4(-2)(6) = 1 + 48 = 49$ .
Final Valid Solutions: Calculate the discriminant: $b^2 - 4ac = (1)^2 - 4(-2)(6) = 1 + 48 = 49$ . Since the discriminant is a perfect square, we have two real solutions. Calculate the solutions using the quadratic formula: $x = [-1 \pm \sqrt{49}] / (-4)$ .
Final Valid Solutions: Calculate the discriminant: $b^2 - 4ac = (1)^2 - 4(-2)(6) = 1 + 48 = 49$ . Since the discriminant is a perfect square, we have two real solutions. Calculate the solutions using the quadratic formula: $x = [-1 \pm \sqrt{49}] / (-4)$ . The two solutions are $x = (-1 + 7) / (-4)$ and $x = (-1 - 7) / (-4)$ . Simplifying, we get $x = -1.5$ and $x = 2.0$ .
Final Valid Solutions: Calculate the discriminant: $b^2 - 4ac = (1)^2 - 4(-2)(6) = 1 + 48 = 49$ . Since the discriminant is a perfect square, we have two real solutions. Calculate the solutions using the quadratic formula: $x = [-1 \pm \sqrt{49}] / (-4)$ . The two solutions are $x = (-1 + 7) / (-4)$ and $x = (-1 - 7) / (-4)$ . Simplifying, we get $x = -1.5$ and $x = 2.0$ . We must check which of the solutions fall into the correct domain for each case. For the first case $(x \geq 1)$ , the solution $x = 1.5$ is valid. For the second case $(x < 1)$ , the solution $x = -1.5$ is valid. The solution $x = 2.0$ is not valid for the second case because it does not satisfy $x = [-1 \pm \sqrt{49}] / (-4)$ $1$ . The solution $x = [-1 \pm \sqrt{49}] / (-4)$ $2$ is not valid for the first case because it does not satisfy $x = [-1 \pm \sqrt{49}] / (-4)$ $3$ .
Final Valid Solutions: Calculate the discriminant: $b^2 - 4ac = (1)^2 - 4(-2)(6) = 1 + 48 = 49$ . Since the discriminant is a perfect square, we have two real solutions. Calculate the solutions using the quadratic formula: $x = [-1 \pm \sqrt{49}] / (-4)$ . The two solutions are $x = (-1 + 7) / (-4)$ and $x = (-1 - 7) / (-4)$ . Simplifying, we get $x = -1.5$ and $x = 2.0$ . We must check which of the solutions fall into the correct domain for each case. For the first case $(x \geq 1)$ , the solution $x = 1.5$ is valid. For the second case $(x < 1)$ , the solution $x = -1.5$ is valid. The solution $x = 2.0$ is not valid for the second case because it does not satisfy $x = [-1 \pm \sqrt{49}] / (-4)$ $1$ . The solution $x = [-1 \pm \sqrt{49}] / (-4)$ $2$ is not valid for the first case because it does not satisfy $x = [-1 \pm \sqrt{49}] / (-4)$ $3$ . Therefore, the valid solutions to the nearest tenth for the equation $x = [-1 \pm \sqrt{49}] / (-4)$ $4$ are $x = 1.5$ and $x = -1.5$ .