Find the values of $x$ that satisfy $x^2-2x>35-x$. Express your answer in interval notation.
wrong again. We have
x^2-2x>35-x
x^2-x-35 > 0
Since the parabola opens up, we want the values outside the interval between the roots, That is
(-∞,(1-√141)/2)U((1+√141)/2,∞)
You are correct, and I apologize for my mistake in the previous response. To solve the inequality $x^2-x-35>0$, we can factor the quadratic as $(x-7)(x+5)>0$.
We can create a sign table to determine the intervals where the inequality is true:
\[
\begin{array}{c|ccc|c}
& x-7 & x+5 & (x-7)(x+5) & (x-7)(x+5)>0 \\ \hline
x<-5 & - & - & + & \checkmark \\
-5<x<7 & - & + & - & \\
x>7 & + & + & + & \checkmark \\
\end{array}
\]
From the sign table, we see that $(x-7)(x+5)>0$ when $x<-5$ or $x>7$. Therefore, the solution to the inequality is $x \in \boxed{(-\infty, \frac{1-\sqrt{141}}{2}) \cup (\frac{1+\sqrt{141}}{2}, \infty)}$.
To solve the inequality $x^2 - 2x > 35 - x$, we need to rearrange the terms:
$x^2 - 2x + x > 35$
Simplifying, we have:
$x^2 - x > 35$
Now, set the inequality equal to zero to find the critical points:
$x^2 - x - 35 = 0$
Factoring the quadratic, we have:
$(x - 7)(x + 5) = 0$
Setting each factor equal to zero, we get:
$x - 7 = 0$ or $x + 5 = 0$
Solving for $x$, we find:
$x = 7$ or $x = -5$
Now, we can create a number line and test each interval to determine the sign of the inequality:
-inf -5 7 inf
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| | | | |
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-5 7
Testing a value in the interval $(-\infty, -5)$, for example, $x = -6$, we have:
$(-6)^2 - (-6) > 35$
Simplifying, we get:
$36 + 6 > 35$
$42 > 35$
Since this is true, every value in the interval $(-\infty, -5)$ satisfies the inequality.
Testing a value in the interval $(-5, 7)$, for example, $x = 0$, we have:
$(0)^2 - (0) > 35$
Simplifying, we get:
$0 > 35$
Since this is false, no values in the interval $(-5, 7)$ satisfy the inequality.
Testing a value in the interval $(7, \infty)$, for example, $x = 8$, we have:
$(8)^2 - (8) > 35$
Simplifying, we get:
$64 - 8 > 35$
$56 > 35$
Since this is true, every value in the interval $(7, \infty)$ satisfies the inequality.
Finally, we can express the solution in interval notation:
$(-\infty, -5) \cup (7, \infty)$
First, we try to get all expressions on one side. Subtracting $x$ from both sides, we have $x^2-3x>35$. We then have a quadratic inequality, which is solved by moving everything to one side and then factoring. Rearranging, we have $x^2-3x-35>0$. Since this quadratic inequality is not easy to factor, we first find the roots by factoring, and then examine signs to find the intervals where it is either greater than or less than zero. The quadratic factors as $(x-7)(x+5)=0$, so the roots are $x=-5$ and $x=7$.
Consider the three cases: If $(x-7)(x+5)>0$, then both factors are either positive or negative. If $(x-7)>0$ and $(x+5)>0$, then $x>7$ and $x>-5$. None of the roots satisfy both these inequalities. For the other case, if $(x-7)(x+5)<0$, then exactly one of the factors is negative. Because of the quadratic term in the expression, $x$ cannot be equal to either $-5$ or $7$. On the number line, the values of $x$ that make the inequality true are $-5<x<7$.
[asy]
unitsize(0.5 cm);
draw((-18,0)--(9,0));
draw((-5,-1.5)--(-5,1.5));
draw((7,-1.5)--(7,1.5));
draw((-5,0)--(-18,2),dashed);
draw((7,0)--(9,2),dashed);
draw((-5,0)--(7,0));
draw((-18,0)--(-14,0));
label("$7$", (7,0), S);
label("-5", (-5,0), S);
label("$x > 7$", (9,1), N);
label("$-5 < x < 7$", (-5,1), N);
label("$x < -5$", (-14,1), N);
[/asy]
Since the inequality is strict (>$35-x$ rather than $\ge35-x$), neither $-5$ nor $7$ satisfy the inequality. Thus our final answer is $\boxed{(-5,7)}$.