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# explain why......... watch

1. explain why the folllowing has no real roots

3x^2 - 4x + 60 = 0

anyone??

thanks!!
2. (Original post by danr2)
explain why the folllowing has no real roots

3x^2 - 4x + 60 = 0

anyone??

thanks!!
You could draw the graph of 3x^2 -4x +60 and see why

Depending on the context and wording of the question
You could complete the square to discover the minimum point on the curve and also show there are no real solutions
You could use calculus to find the minimum point on the curve
You could use the quadratic formula and show no real solutions
You could use the discriminant to show there would be no real solutions

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3. (Original post by gdunne42)
You could draw the graph of 3x^2 -4x +60 and see why

Depending on the context of the question
You could complete the square to discover the minimum point on the curve and also show there are no real solutions
You could use calculus to find the minimum point on the curve
You could use the quadratic formula and show no real solutions
You could use the discriminant to show there would be no real solutions

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how would use the quadratic formula and the discriminant?
4. If B^2 is less than 4*a*c, then the equation has no real roots.

ax^2 + bx + c = 0
3x^2 - 4x + 60 = 0

Your equation is already in the form ax^2 + bx + c, so...

a = 3
b = 4
c = 60

b^2 = 16
4*a*c = 3*(-4)*60 = -720

-720 < 16, so the equation has no real roots.

As well as this, if b^2 > 4ac, then the equation is said to have real roots, and if b^2 = 4ac, then it is said to have equal roots.
5. You need to use the quadratic formula...google it. The discriminant is basically everything that is in the square root part. So normally the square root part consists of (b^2-4ac)^1/2 (the 1/2 is another way of writing square root). In your example, the a would be 3, the b would be-4 and the c would be 60. Sub these values in and you should get (16-4(3*60))^1/2. This simplifies to (-704)^1/2. You should know that the square root of a negative number doesn't give a true number (it is deemed imaginary) and so you don't get a valid solution for x (when you use the quadratic formula). This bit is the discriminant and it means that there are no x values and so there are no roots, therefore, its roots aren't 'real' unless you consider imaginary numbers.
6. (Original post by danr2)
how would use the quadratic formula and the discriminant?
You know the discriminant √(b2-4ac) right? When:

b2 > 4ac there are two real solutions
b2 = 4ac there is one solution
b2 < 4ac there are no real solutions (as is the case for this question)
7. (Original post by danr2)
how would use the quadratic formula and the discriminant?
What level of maths are you studying?
If you attempt to use the quadratic formula you will reach a point where you are required to find the square root of a negative number. There are no real solutions to the square root of a negative number so no real solutions to the quadratic.

The discriminant is the b^2 - 4ac part of the quadratic formula and can be used to show there will be no real roots. If b^2 - 4ac is negative (hence b^2 is less than 4ac) there will be no solutions.
http://www.examsolutions.net/maths-r...tutorial-1.php

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