inequalities

[neelam]

Question 7
(a) Factorise x² + 3x (1 mark)
(b) Sketch y = x² + 3x
Label the x values of the points of intersection with the x-axis. (2 marks)
(c) Hence, or otherwise, solve x² + 3x = 0 (2 marks)Mark Scheme
(a) x(x + 3) B1
(b) U-shaped parabola M1
0 and 3 labelled on x-axis (c) x <3 and x > 0
For part c of the question, I don't get how the answer is x<-3 x>0. I thought the answer was -3<x<0.

 

[JeffM]

Question 7
(a) Factorise x2 + 3x (1 mark)
(b) Sketch y = x2 + 3x
Label the x values of the points of intersection with the x-axis. (2 marks)
(c) Hence, or otherwise, solve x2 + 3x  0 (2 marks)Mark Scheme What is that box supposed to mean? Please use Preview Post function before posting to ensure that your questions are comprehensible.
(a) x(x + 3) B1 Yes.
(b) U-shaped parabola M1 Yes, but perhaps better to say concave upwards.
0 and 3 labelled on x-axis (c) x <3 and x > 0
For part c of the question, I don't get how the answer is x<-3 x>0. I thought the answer was -3<x<0.

If you do not show your work, how are we to determine where you are going astray?

When you sketched the parabola in part b, what was the lowest value of y and at what value of x did it occur?

 

[stapel]

For part c of the question, I don't get how the answer is x<-3 x>0. I thought the answer was -3<x<0.

Does the above mean that you've completed (but not shown your work or answers for) parts (a) and (b)? Also, on what basis did you conclude that the answer should be as you propose? Please show your work.

When you reply, please state whether you also need help on the other parts of this exercise. Thank you!

 

[HallsofIvy]

Question 7
(a) Factorise x² + 3x (1 mark)
(b) Sketch y = x² + 3x
Label the x values of the points of intersection with the x-axis. (2 marks)
(c) Hence, or otherwise, solve x² + 3x = 0 (2 marks)Mark Scheme
(a) x(x + 3) B1
(b) U-shaped parabola M1
0 and 3 labelled on x-axis (c) x <3 and x > 0
For part c of the question, I don't get how the answer is x<-3 x>0. I thought the answer was -3<x<0.

Clearly (c) has been miscopied. The solutions to \(\displaystyle x^2+ 3x= 0\) are x= -3 and x= 0. "x< -3, x> 0" is the solution set of the inequality \(\displaystyle x^2+ 3x> 0\) and "-3< x< 0" of \(\displaystyle x^2+ 3x< 0\). What is the actual question?