Example Question - real number solutions
Here are examples of questions we've helped users solve.
Solving Absolute Value Inequality
To solve the inequality \( 9 - 8|r + 5| > -11 \), we need to isolate the absolute value expression on one side. First, let's move the constant term on the left to the other side by subtracting 9 from both sides: \( 9 - 8|r + 5| - 9 > -11 - 9 \) \( -8|r + 5| > -20 \) Next, to get \( |r + 5| \) by itself, divide both sides of the inequality by -8. Remember that dividing by a negative number flips the direction of the inequality: \( \frac{-8|r + 5|}{-8} < \frac{-20}{-8} \) \( |r + 5| < \frac{20}{8} \) \( |r + 5| < \frac{5}{2} \) or \( |r + 5| < 2.5 \) The absolute value inequality \( |r + 5| < 2.5 \) means that \( r + 5 \) must be less than 2.5 and greater than -2.5 (because the distance from zero is less than 2.5). So we can break this into two separate inequalities: \( r + 5 < 2.5 \) \( r + 5 > -2.5 \) Let's solve each of these: For \( r + 5 < 2.5 \): Subtract 5 from both sides: \( r < 2.5 - 5 \) \( r < -2.5 \) For \( r + 5 > -2.5 \): Subtract 5 from both sides: \( r > -2.5 - 5 \) \( r > -7.5 \) Combining both inequalities, we get the solution for \( r \): \( -7.5 < r < -2.5 \) This is the set of all real numbers greater than -7.5 and less than -2.5.
Determining Range of Slope for Intersection of Line and Curve
为了找到直线 y = mx - m 与曲线 y = 2x^2 + 3x - 3 在不同的两个点相交时的斜率 m 的取值范围，我们可以将两个方程等同起来来找出 x 的值。如果直线与曲线在两个不同点相交，那么由它们形成的方程会产生两个不同的实数解。下面用中文解答。 令直线与曲线的 y 值相等，得到以下等式： mx - m = 2x^2 + 3x - 3 将所有项移到等式一边来形成一个关于 x 的二次方程： 2x^2 + 3x - 3 - mx + m = 0 整理一下可得到： 2x^2 + (3-m)x + (m-3) = 0 为了使这个二次方程有两个不同的实数解，判别式（b^2 - 4ac）必须大于 0。带入 a = 2，b = 3-m 和 c = m-3 可得： (3-m)^2 - 4(2)(m-3) > 0 展开平方项和乘法得到： 9 - 6m + m^2 - 8m + 24 > 0 合并同类项得： m^2 - 14m + 33 > 0 这个不等式是一个开口向上的抛物线，我们需要找到使这个不等式成立的 m 值，即该抛物线在 x 轴上方的部分。为了确定这个范围，我们需要计算该二次方程的根。 抛物线与 x 轴交点的 m 值就是二次方程的根，我们通过求解二次方程来计算： m^2 - 14m + 33 = 0 求解这个二次方程，我们用配方法或公式法，得到： m = (14 ± √(14^2 - 4*1*33)) / (2*1) m = (14 ± √(196 - 132)) / 2 m = (14 ± √64) / 2 m = (14 ± 8) / 2 所以解为： m = (14 + 8) / 2 = 22 / 2 = 11 m = (14 - 8) / 2 = 6 / 2 = 3 因此，使 m^2 - 14m + 33 > 0 成立的 m 的取值范围是 m < 3 或 m > 11。这是直线 y = mx - m 与曲线 y = 2x^2 + 3x - 3 在不同的两点相交时 m 的取值范围。
Solving Equations with Absolute Values and Radicals
The image shows two equations that we are asked to solve over the real number set \(\mathbb{R}\): 1) \(x^2 - 2x - 2|x - 1| - 2 = 0\) 2) \((x + 2022)(\sqrt{x^2 + 2021} - 1) = 2021\) Let's solve them one by one. For the first equation: To solve this equation, we must consider the absolute value function which splits the domain into two parts, depending on the sign of the expression inside the absolute value function. Absolute value function rules: - If \(x - 1 \geq 0\) (which means if \(x \geq 1\)), then \(|x - 1| = x - 1\). - If \(x - 1 < 0\) (which means if \(x < 1\)), then \(|x - 1| = -(x - 1) = -x + 1\). So, we need to consider two cases: Case 1: \(x \geq 1\) If \(x \geq 1\), then \(|x - 1| = x - 1\), so our equation becomes \(x^2 - 2x - 2(x - 1) - 2 = 0\), which simplifies to: \(x^2 - 2x - 2x + 2 - 2 = 0\) \(x^2 - 4x = 0\) \(x(x - 4) = 0\) This gives us two solutions: \(x = 0\) and \(x = 4\). However, since \(x = 0\) is not in the domain where \(x \geq 1\), we discard it. So the valid solution in this case is \(x = 4\). Case 2: \(x < 1\) If \(x < 1\), then \(|x - 1| = -x + 1\), so our equation becomes \(x^2 - 2x - 2(-x + 1) - 2 = 0\), which simplifies to: \(x^2 - 2x + 2x - 2 - 2 = 0\) \(x^2 - 4 = 0\) \(x^2 = 4\) Taking the square root of both sides, we get \(x = \pm 2\). Since \(x\) must be less than \(1\), we discard \(x = 2\) and only \(x = -2\) is within the valid domain. So, the solution to equation 1) based on the split due to the absolute value is \(x = 4\) and \(x = -2\). Now for the second equation: The second equation is \((x + 2022)(\sqrt{x^2 + 2021} - 1) = 2021\). To solve this, we can set each factor equal to zero. If \(x + 2022 = 0\), then \(x = -2022\). For the other factor \(\sqrt{x^2 + 2021} - 1 = 0\), we add \(1\) to both sides and then square both sides: \(\sqrt{x^2 + 2021} = 1\) \(x^2 + 2021 = 1\) \(x^2 = 1 - 2021\) \(x^2 = -2020\) Since \(x^2\) cannot be negative in the set of real numbers, there are no real solutions from the second factor. Therefore, the only solution from the second equation is \(x = -2022\). To summarize: 1) \(x = 4\) and \(x = -2\) 2) \(x = -2022\)
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