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Solve the following quadratic and linear equations: 5.1 2x2 = -x 5.2 x2 − 25 = 0 5.3 x2 + 5x − 6 = 0 5.4 racx+53 + racx+16 = racx3 - 2

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5 parts
5.1

x=0 or x=−12x = 0 \text{ or } x = -\frac{1}{2}

5.2

x=5 or x=−5x = 5 \text{ or } x = -5

5.3

x=−6 or x=1x = -6 \text{ or } x = 1

5.4

x=−23x = -23

5.5
36

Step-by-step solution

14 steps

Part 5.1· 3 steps

2x2=−x2x^2 = -x

  1. Move all terms to one side to set the equation to zero.

    2x2+x=02x^2 + x = 0
  2. Factor out the common term xx.

    x(2x+1)=0x(2x + 1) = 0
  3. Apply the Zero Factor Property.

    x=0or2x+1=0x = 0 \quad \text{or} \quad 2x + 1 = 0 2x=−12x = -1 x=−12x = -\frac{1}{2}

    The solutions are x=0 or x=−12\boxed{x = 0 \text{ or } x = -\frac{1}{2}}.

Part 5.2

x=5 or x=−5x = 5 \text{ or } x = -5

x2−25=0x^2 - 25 = 0

  1. Add 25 to both sides of the equation.

    x2=25x^2 = 25
  2. Take the square root of both sides.

    Remember to include both positive and negative roots.

    x2=±25\sqrt{x^2} = \pm\sqrt{25} x=±5x = \pm 5

    The solutions are x=5 or x=−5\boxed{x = 5 \text{ or } x = -5}.

Part 5.3

x=−6 or x=1x = -6 \text{ or } x = 1

x2+5x−6=0x^2 + 5x - 6 = 0

  1. Factor the quadratic expression.

    Find two numbers that multiply to −6-6 and add to 5. These numbers are 6 and −1-1.

    (x+6)(x−1)=0(x + 6)(x - 1) = 0
  2. Apply the Zero Factor Property.

    x+6=0orx−1=0x + 6 = 0 \quad \text{or} \quad x - 1 = 0 x=−6orx=1x = -6 \quad \text{or} \quad x = 1

    The solutions are x=−6 or x=1\boxed{x = -6 \text{ or } x = 1}.

Part 5.4

x=−23x = -23

x+53+x+16=x3−2\frac{x+5}{3} + \frac{x+1}{6} = \frac{x}{3} - 2

  1. Find the least common denominator (LCD) for all fractions, which is 6.

    Multiply every term by the LCD to clear the denominators.

    6(x+53)+6(x+16)=6(x3)−6(2)6 \left(\frac{x+5}{3}\right) + 6 \left(\frac{x+1}{6}\right) = 6 \left(\frac{x}{3}\right) - 6(2) 2(x+5)+(x+1)=2x−122(x+5) + (x+1) = 2x - 12
  2. Distribute and simplify both sides of the equation.

    2x+10+x+1=2x−122x + 10 + x + 1 = 2x - 12 3x+11=2x−123x + 11 = 2x - 12
  3. Gather xx terms on one side and constant terms on the other.

    3x−2x=−12−113x - 2x = -12 - 11 x=−23x = -23

    The solution is x=−23\boxed{x = -23}.

Part 5.536

An amount of R34,80 is made up of 50c and 20c coins. How many 50c coins are there out of a total of 120 coins?

  1. Define variables and set up the equations.

    Let ff be the number of 50c coins.
    Let tt be the number of 20c coins.
    Total number of coins:

    f+t=120(1)f + t = 120 \quad (1)

    Total value (in Rands):

    0.50f+0.20t=34.80(2)0.50f + 0.20t = 34.80 \quad (2)
  2. Solve equation (1) for tt.

    t=120−f(3)t = 120 - f \quad (3)
  3. Substitute equation (3) into equation (2).

    0.50f+0.20(120−f)=34.800.50f + 0.20(120 - f) = 34.80
  4. Distribute and solve for ff.

    0.50f+24−0.20f=34.800.50f + 24 - 0.20f = 34.80 0.30f+24=34.800.30f + 24 = 34.80 0.30f=34.80−240.30f = 34.80 - 24 0.30f=10.800.30f = 10.80 f=10.800.30f = \frac{10.80}{0.30} f=36f = 36

    There are 36\boxed{36} 50c coins.

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