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Re: Doubt on Solution to Inequalities [#permalink]

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24 Jul 2013, 21:55

Hi Bunuel

Thanks a lot for those links and your wonderful contribution.

I realized that I am struggling specifically with the below:

For example, let us take the equation (x-1)(x-4)=0

So, when I look at the expression (x-1)(x-4), I know that it is a parabola passing through x=1 and x=4.

But, I don't know whether this is V-shaped parabola or an inverted V-shaped parabola. Currently, the way I am doing this is to substitute a value between the roots (for eg. x=2 in the above case) and if it is negative, I know that the curve is a V-shaped parabola. Hence the curve is +ve outside the points x=1 and x=4 and is -ve between the roots.

So, for the inequality (x-1)(x-4)>0 the solution is 4<x<1 (where is the curve +ve) So, for the inequality (x-1)(x-4)<0 the solution is 1<x<4 (where is the curve -ve)

If you see the above, the crucial thing is to identify whether the curve is V-shaped or inverted V-shaped?

Currently, I am doing it using substitution method, and I feel it is prone to errors. So, I want to know is there any other way apart from substituting values to find out the shape of the curve?

I am extremely sorry for the repeated questions, but I really need some handholding on this topic. Kindly help me on this.

Thanks a lot for those links and your wonderful contribution.

I realized that I am struggling specifically with the below:

For example, let us take the equation (x-1)(x-4)=0

So, when I look at the expression (x-1)(x-4), I know that it is a parabola passing through x=1 and x=4.

But, I don't know whether this is V-shaped parabola or an inverted V-shaped parabola. Currently, the way I am doing this is to substitute a value between the roots (for eg. x=2 in the above case) and if it is negative, I know that the curve is a V-shaped parabola. Hence the curve is +ve outside the points x=1 and x=4 and is -ve between the roots.

So, for the inequality (x-1)(x-4)>0 the solution is 4<x<1 (where is the curve +ve) So, for the inequality (x-1)(x-4)<0 the solution is 1<x<4 (where is the curve -ve)

If you see the above, the crucial thing is to identify whether the curve is V-shaped or inverted V-shaped?

Currently, I am doing it using substitution method, and I feel it is prone to errors. So, I want to know is there any other way apart from substituting values to find out the shape of the curve?

I am extremely sorry for the repeated questions, but I really need some handholding on this topic. Kindly help me on this.

Re: Doubt on Solution to Inequalities [#permalink]

Show Tags

25 Jul 2013, 07:29

Quote:

Intersection points are the roots of the equation x^2-4x+3=0, which are x_1=1 and x_2=3. "<" sign means in which range of x the graph is below x-axis. Answer is 1<x<3 (between the roots).

If the sign were ">": x^2-4x+3>0. First find the roots (x_1=1 and x_2=3). ">" sign means in which range of x the graph is above x-axis. Answer is x<1 and x>3 (to the left of the smaller root and to the right of the bigger root).

This approach works for any quadratic inequality. For example: -x^2-x+12>0, first rewrite this as x^2+x-12<0 (so that the coefficient of x^2 to be positive. It's possible to solve without rewriting, but easier to master one specific pattern).

x^2+x-12<0. Roots are x_1=-4 and x_1=3 --> below ("<") the x-axis is the range for -4<x<3 (between the roots).

Again if it were x^2+x-12>0, then the answer would be x<-4 and x>3 (to the left of the smaller root and to the right of the bigger root).

Yes, this solves my doubt. You're suggesting that whatever the equation is convert it into the form ax^2+bx+c such that the "a" (coeff of x^2) is always +ve. So, for example, if we have to find where is the inverted-V curve +ve, we are actually finding where is the V-curve(mirror image) negative.

My only question is: Can we safely assume that a curve with a +ve coeff of x^2 is always a V-shaped curve?

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