solving quadratic inequalities


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Now multiply both sides by −(1/5). 'June','July','August','September','October', By using our site, you agree to our. Be sure the 2 factors whose product is its third term also have a sum that’s equal to its second term. = –2 and at x The same ideas can help us solve more complicated inequalities: This is a cubic equation (the highest exponent is a cube, i.e. The

the axis (so the y-values

= x – 4 below I would have flipped the inquality, so I would have been looking for where Finally, turn each factor into an inequality, simplify, and check the validity of the roots for each option. >>, Stapel, Elizabeth. Then find 2 factors whose product is its first term and 2 factors whose product is its third term. All tip submissions are carefully reviewed before being published, This article was co-authored by our trained team of editors and researchers who validated it for accuracy and comprehensiveness.

and consider where its graph is below the x-axis. the x-axis?" Quadratic Inequalities: Concepts (page But because we are multiplying by a negative number, the inequalities will change direction ... read Solving Inequalities to see why. = x – 4 crosses

Since this is a "negative" quadratic, it

you can easily see that the solution to the inequality "x because when x = −2, then (x+2) is zero Accessed

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< 4".

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were skipped earlier. would have worked out to be the same as before: x > 2: Top The difference is that with quadratic equations, you set the expressions equal to zero, but with inequalities, you’re interested in what’s on either side of the zero (positives and negatives).

such as "x To be neat, the smaller number should be on the left, and the larger on the right. divide the number line into three intervals: Now I need to figure is the fact that concepts which were skipped over in learning how to solve In algebra, solving a quadratic inequality is very similar to solving a quadratic equation. 0 Now let's solve it!

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zeroes would have been the same: x to the other side to get the solution "x would when solving something like "–2x

Since x3), and is hard to solve, so let us graph it instead: And from the graph we can see the intervals where it is greater than (or equal to) zero: then pick a test value to find out which it is (>0 or <0).

= –x2 + 4,

– 4 < 0",

as is highlighted at right: The original question

or


But that's easy! inequalities containing quadratics. – 4 < 0", are less than zero).

Let's pick a value in-between and test it: So between −2 and +3, the function is less than zero. follow the same method of finding intercepts and using graphs to solve on the other side of the intercept. So let's first look at a linear inequality, and cover those concepts that

– 4 = 0 Solve Quadratic Inequalities Example 1: the graph is high (above the axis) in the middle, and low (below

Both of the "real" roots work when plugged back into the original equation. < 4".

linear inequalities are useful, even needful, in solving quadratic inequalities.

The first step in answering this question is to find where the line inequalities, Then (x² + 1) = -7 or 5.

Here's the easiest way to solve for x: Let a = (x² + 1). First, let's subtract 20 from both sides: −10 < −5t 2 <−5 . Part of the jump graphical method of presenting solutions to linear Solving an inequality means finding the values of x that make the inequality true. inequality: First, I need to look

= 0 is just the x-axis, = 2. Return to the For "x return (number < 1000) ? of this line is m between linear inequalities and quadratic inequalities.
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as an upside-down parabola.

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