[22] If (A) × (B) = 0, then either A = 0 OR B = 0.
Solutions
x(x + 3) = 0 We have a product = 0. Therefore put each factor = 0 x = 0 or x + 3 = 0 x = 0 3 or x = –3 (2)
x(2x – 5) = 12 We need a product = 0. Therefore, multiply out brackets and write in standard form with all the terms on one side and equal to 0 (2x + 3)(x – 4) = 0 Factorise 2x + 3 = 0 or x – 4 = 0 Put each factor = 0 2x = –3 or x = 4 x = -3/2 x = 4 (2)
2x2 + x − 6 = 0 (2x − 3)(x + 2) = 0 33 ∴ 2x = 3 or x = − 2 Find the solutions by putting each factor equal to zero ∴ x = -3/2 or x = −2 (4)
2x2 = 32 2x2 – 32 = 0 Write in standard form with all the terms on one side and equal to 0 x2 – 16 = 0 Divide every term on both sides by 2 (x + 4)(x – 4) = 0 3 Factorise (the difference of two squares) ∴ x + 4 = 0 or x – 4 = 0 ∴ x = –4 or ∴ x = 4 (4)
3x + 1/x = 4, x ≠ 0 Multiply through by x to get rid of the denominator 3x2 + 1 = 4x Write in standard form with all the terms on one side and equal to 0 3x2 – 4x + 1 = 0 Factorise (the trinomial) (3x – 1)(x – 1) = 0 ∴ 3x – 1 = 0 or x – 1 = 0 ∴ 3x = 1 or x = 1 ∴ x = 1/3 or ∴ x = 1 (5)
2 √x − 3 = x − 3 Square both sides (2 √x − 3)2 = (x − 3)2 4(x − 3 ) = ( x − 3 ) ( x − 3 ) 4x − 12 = x2 − 6x + 9 0 = x2 − 10x + 21 0 = ( x − 7 ) ( x − 3 ) ∴ x − 7 = 0 or x − 3 = 0 ∴ x = 7 or x = 3 (5) [22]
Check your answers: x = 7 LHS = 2 √7 − 3 = 2 √4 = 2(2) = 4 RHS = 7 − 3 = 4 ∴ x = 7 is a solution x = 3 LHS = 2 √3 − 3 = 2 √0 = 0 RHS = 3 − 3 = 0 ∴ x = 3 is a solution If a quadratic equation cannot be factorised, there are other ways to find the roots or solutions. Sometimes the solutions do not exist!
Activity 3
What term can be added to the following equations to make a complete square?
0 = x2 – 8x + ?
y = x2 + 9x + ?
y = x2-b/a x + ?
Solve for x by using the method of completing the square.
–3x2 + 5x + 4 = 0
ax2 + bx + c = 0 [17]
Activity 4: Interpret a graph Solve for x (correct to two decimal places): 4x2 – 8x = 7 2x (3x + 5) – 11 = 0 [9]
Solutions 1. 4x2 – 8x = 7 Write the equation in standard form (ax2 + bx + c = 0) 4x2 – 8x = 7 = 0 List the values of a, b and c a = 4; b = – 8; c = –7 Write down the formula x = − b ± √b2 − 4ac Substitute the values for a, b and c into the formula. 2 x = – ( –8 ) ± √(–8 ) 2 – 4(4)(–7) Simplify the value under the square root sign 2(4) x = 8 ± √176 Separate the positive and negative value of the square root 8 x = 8 + √176 or x = 8 – √176 Answers in surd form 8 8 x = 2,66 3 or x = –0,66 3 Answers correct to two decimal places (5) 2. 2x(3x + 5) – 11 = 0 Write the equation in standard form 6x2 + 10x – 11 = 0 (ax2 + bx + c = 0) x = –10 ± √100 + 26 12 = –10 ± √364 12 = –5 ± √91 6 (4) [9] These roots are irrational. Unless the question asks for decimal values, leave them in surd form (the square root form).
Activity 5 Solve for x if
(x+3)(x – 5) ≤–12
–x ≤ 2x2 – 3 [10]
Solutions
(x+3)(x – 5) ≤ –12 x 2 – 2x – 15 +12 ≤ 0 Get into the standard form (ax2 + bx + c ≤ 0) x 2 – 2x –3 ≤ 0 3 Factorise the trinomial: (x –3)(x + 1) ≤ 0 3 Critical values: x = 3 and x –1 We now indicate 3 and –1 on a number line. We know that the expression x2 – 2x –3 = 0 at x = 3 and x = –1. We can indicate this on the number line. Our next step is to choose values less than –1, values between –1 and 3 and values greater than 3 and substitute it into the expression x 2 – 2x –3. If the answer is positive, then we indicate + on the number line. If the answer is negative, we indicate – on the number line (If x = –10, then 2(–10)² +(–10) – 3 = 187 > 0 ∴ + If x = 0, then 2(0)² – 0 –3 = –3 < 0 ∴ – If x = 3, then 2(3)² + 3 – 3 = 18 > 0 ∴ +) We have to solve for x where x2 – 2x –3 ≤ 0. The solution on the number line is the interval where we see zero and a negative. This happen when the x values are less than or equal to 3 and are also more than or equal to –1. Therefore the solution is : − 1 ≤ × ≤ 3 (5)
–x ≤ 2x2 – 3 –2x2 – x + 3 ≤ 0 Get into the standard form (ax2 + bx + c ≤ 0) –2x – x + 3 ≥ 0 -1 -1 -1 -1 Divide both sides by –1 to change the coefficient of x2 to a positive 2x2 + x – 3 ≥ 0 3 Factorise the trinomial (2x + 3)(x – 1) ≥ 0 3 Critical values: x = –3/2 and x = 1 We now indicate –3/2 and 1 on a number line. We know that the expression 2x2 + x – 3 = 0 at x = –3/2 and x = 1. We can indicate this on the number line. Our next step is to choose values less than –3/2, values between–3/2 and 1 and values greater than 1 and substitute it into the expression 2x2 + x – 3. If the answer is positive, then we indicate + on the number line. If the answer is negative, we indicate – on the number line (Whenever we multiply or divide an inequality by a negative, the inequality sign changes i.e. the less or equal to sign changes to a greater or equal to sign. If x = –10, then (–10)² 2(–10) – 3 =117 > 0 ∴ + If x = 1, then (1)² – 2(1) – 3 =–4 < 0 ∴ – If x = 5, then (5)² –2(5) – 3 = 12 > 0 ∴ +) We have to solve for x where 2x2 + x – 3 ≥ 0. The solution on the number line is the interval where we see zero and a positive. This happens for the x values less than or equal to –3/2 and for the x values greater than or equal to 1. Therefore the solution is : x ≤ − –3/2 or x ≥ 1 33 (5)
OR ALTERNATIVE METHOD by using a rough sketch of the parabola: –x < 2x2 – 3 Get into the standard form ax2 + bx + c < 0 –2x2– x + 3 < 0 Divide both sides by –1. –2x – x + 3 > 0 This is necessary to draw the rough sketch of a “positive” parabola -1 -1 -1 2x2 + x – 3 > 0 3 With 0 on RHS (2x + 3)(x – 1) > 0 3 Factorise LHS Critical values of x: –3/2 and = 1 Make a rough sketch of a parabola If (***)(***) > 0 (it means where y is positive) Read off the x values of the graph above the x-axis (5) [10]
Activity 6 Solve the following equations simultaneously. 1. 2x + y = 3 and x2 + y + x = y2 2. y = − 6 − 2 and y = − 3x + 2 x + 1 [14]
Solutions
2x + y = 3 ..................... eqn (1) x2 + y + x = y2 .............. eqn (2) y = –2x + 3 .............. eqn (3) Use the linear equation (1) to write y alone on one side of the equation. Substitute eqn (3) into eqn (2), to eliminate the y variable. x2 + (–2x + 3) + x = (–2x + 3)2 3 Simplify both sides. x2 – x + 3 = 4x2 – 12x + 9 Factorise the trinomial. 0 = 3x2 – 11x + 6 0 = (3x – 2)(x – 3) ∴ 3x – 2 = 0 or x – 3 = 0 ∴ x = 2/3 or ∴ x = 3
y = − 6 − 2 and y = − 3x + 2 x + 1 y = − 6 − 2.....................(eqn 1) x + 1 y = –3x + 2.......................(eqn 2) y is alone on one side of both equations. ∴ y = − 6 − 2 = –3x + 2 ............LCD = x + 1 x + 1 ∴ –6 –2(x + 1) = –3x(x + 1) + 2(x + 1) ∴ –6 –2x –2 = –3x2 – 3x + 2x + 2 ∴ 3x2 – x – 10 = 0 ∴ (3x + 5)(x – 2) = 0 ∴ x = -5/3 or x = 2
[14]
Activity 7
Show that the roots of x2 – 2x – 7 = 0, are irrational, without solving the equation. (3)
Show that x2 + x + 1 = 0 has no real roots. (3)
If x = 2 is a root of the equation 3x2 – 5x – 2k = 0, determine the value of k. (2)
The solutions to a quadratic equation are: x = 5 ± √12 − 3a . For which value(s) of a will the equation have equal roots. (3)
Determine the value(s) of k for which the equation 3x2 + (k + 2) x + k = 0 has equal roots (4) [15]
Solutions
a = 1; b = –2 ; c = –7 Δ = b2 – 4ac = (–2)2 – 4(1)(–7) = 4 + 28 = 32 ∴ The roots will be irrational (Δ > 0 and not a perfect square) (3)
a = 1; b = 1; c = 1 Δ = b2 – 4ac = (1)2 – 4 (1)(1) = 1 – 4 = – 3 ∴ There are no real roots (Δ < 0 ) (3)
If 2 is a root of the equation, then x = 2. Therefore, we can substitute x = 2 into the equation. 3x2 – 5x – 2k = 0 ∴ 3(2)2 – 5(2) –2k = 0 ∴ 12 − 10 − 2k = 0 ∴ 2k = 2 ∴ k = 1 3 (2)
The equation will have equal roots if Δ = 0 Δ = 12 – 3a 0 3 = 12 − 3a − 12 = − 3a ∴ a = 4 (3)
3x2 + (k + 2)x + k = 0 ∴ a = 3 ; b = (k + 2) ; c = k ∴ Δ = b2 – 4ac = (k + 2)2 – 4(3)(k) = k2 + 4k + 4 – 12k = k2 – 8k + 4 For equal roots the Δ = 0 ∴ k2 – 8k + 4 = 0 ∴ k = 8± √(–8 ) 2 –4(1)(4) 2(1) ∴ k = 8± √48 2 ∴ k = 7,46 or k = –0,54 (4) [15]
