Question Video: Solving Quadratic Inequalities in One Variable | Nagwa Question Video: Solving Quadratic Inequalities in One Variable | Nagwa

Question Video: Solving Quadratic Inequalities in One Variable Mathematics • First Year of Secondary School

Solve the inequality −2𝑥² + 𝑥 ≥ −6.

04:08

Video Transcript

Solve the inequality negative two 𝑥 squared plus 𝑥 is greater than or equal to negative six.

If we take a look at this inequality, we can actually see that we’re gonna have a quadratic involved. So therefore, the first step I’m going to actually complete is to rearrange our inequality, so that actually it’s in the form of a quadratic equal to zero. So to achieve this, first of all I’m gonna add two 𝑥 squared to each side. And then, I’m gonna subtract 𝑥. So this is gonna give us the inequality zero is greater than or equal to two 𝑥 squared minus 𝑥 minus six. So great, we’ve got it in a form now that we recognize and that we can actually solve.

What we’re gonna see now though is we’re actually gonna make our inequality into an equation. So we’re actually gonna make it equal to zero because this is gonna help us find our critical values. And our critical values are gonna be the solutions to the equation two 𝑥 squared minus 𝑥 minus six is equal to zero. And we’d solve this equation by factoring, which will give me two 𝑥 plus three multiplied by 𝑥 minus two is equal to zero. And I’ve got these factors because I knew that we had to have a two 𝑥 and an 𝑥 at the start of the parentheses. And that’s because two 𝑥 multiplied by 𝑥 gives us two 𝑥 squared. I then knew that one parenthesis had to include a positive sign and one had to include a negative sign. And that’s because, and that’s because the second term in each parenthesis had to multiply together to give me negative six. So therefore, we know that it’s gotta be a positive and a negative to multiply together to give us a negative.

And then to find the second terms, I knew that the product had to be negative six. But then also, the sum of one of the numbers multiplied by two 𝑥 and one of the numbers multiplied by 𝑥 had to be negative one. So last, as said, that left us with two 𝑥 plus three multiplied by 𝑥 minus two. So now, to find the critical values and solve this, what we need to do is think about well, what would the value of 𝑥 need to be to make each of the brackets zero. So therefore, my critical values are 𝑥’s equal to negative three over two and two. And we got negative three over two because two multiplied by negative one and a half gives us negative three, while negative three plus three is zero. So that works for that parenthesis. And the second one is 𝑥 minus two. Well, if 𝑥 is equal to two, two minus two is also equal to zero. So that’s that parenthesis.

Okay, great. So we’ve got our critical values. But how does this help? Well, if I draw a rough sketch of a equation, what we can see is that it actually cross the 𝑥-axis at negative three over two and two. So it’s at this point we actually go back to our inequality because this is actually what we’re trying to solve. And we can see that our inequality says that zero is greater than or equal to two 𝑥 squared minus 𝑥 minus six. So therefore, we’re actually looking for the region of the graph of two 𝑥 squared minus 𝑥 minus six is equal to zero. We want the region that is actually below zero. So I’ve shaded in pink the area that we’re interested in.

So therefore, we can say that the set of values which actually solve our inequality are between negative three over two and two. And just to draw your attention to the fact, we’ve actually used brackets when we’ve represented the set, so the same between negative three over two and two. The reason to use brackets is because if we look back to the original inequality, it actually says “or equal to”. And that means it also includes negative three over two and includes two.

Okay, so we’ve got our final answer. But one last thing to remember is, always include a graph or a sketch to show that how you’ve arrived at the region that you’re using for your solution of the inequality cause this will gain you the full marks in any question of this type.

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