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Find the set of values of $x$ for which (a) $4x - 5 > 15 - x$ (b) $x(x - 4) > 12$ - Edexcel - A-Level Maths Pure - Question 5 - 2012 - Paper 1

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Find-the-set-of-values-of-$x$-for-which--(a)-$4x---5->-15---x$--(b)-$x(x---4)->-12$-Edexcel-A-Level Maths Pure-Question 5-2012-Paper 1.png

Find the set of values of $x$ for which (a) $4x - 5 > 15 - x$ (b) $x(x - 4) > 12$

Worked Solution & Example Answer:Find the set of values of $x$ for which (a) $4x - 5 > 15 - x$ (b) $x(x - 4) > 12$ - Edexcel - A-Level Maths Pure - Question 5 - 2012 - Paper 1

Step 1

(a) $4x - 5 > 15 - x$

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Answer

To solve the inequality 4x5>15x4x - 5 > 15 - x, we first rearrange it to isolate xx:

  1. Add xx to both sides: 4x+x5>154x + x - 5 > 15 which simplifies to 5x5>155x - 5 > 15.

  2. Next, add 55 to both sides: 5x>205x > 20.

  3. Finally, divide both sides by 55: x>4x > 4.

Thus, the set of values for part (a) is: x>4x > 4.

Step 2

(b) $x(x - 4) > 12$

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Answer

To solve the inequality x(x4)>12x(x - 4) > 12, we rearrange it first:

  1. Rewrite the inequality as: x24x12>0x^2 - 4x - 12 > 0.

  2. Now, factor the quadratic: x24x12=(x6)(x+2)>0x^2 - 4x - 12 = (x - 6)(x + 2) > 0.

  3. Identify the critical points, which are x=6x = 6 and x=2x = -2.

  4. Create a number line and test intervals determined by these critical points, which are (o2)(- o -2), (2,6)(-2, 6), and (6,+oextinfinity)(6, + o ext{infinity}):

    • For x<2x < -2, choose x=3x = -3: (3)(34)>12(-3)(-3 - 4) > 12 (true).
    • For 2<x<6-2 < x < 6, choose x=0x = 0: (0)(04)>12(0)(0 - 4) > 12 (false).
    • For x>6x > 6, choose x=7x = 7: (7)(74)>12(7)(7 - 4) > 12 (true).
  5. Therefore, the solution set for part (b) is: x<2extorx>6x < -2 ext{ or } x > 6.

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