Roots of Quadratic Equations

Posted on April 18, 2020 by user

Roots of Quadratic Equations

Roots of a quadratic equation are the values of variables/unknowns that satisfy the equation.

Example:
Determine whether 1, 2, and 3 are the roots of the quadratic equation

$x^{2} - 5 x + 6 = 0$ .

Answer:
When x = 1,

$\begin{array}{l} x^{2} - 5 x + 6 = 0 \\ (1)^{2} - 5 (1) + 6 = 0 \\ 2 = 0 \end{array}$
x = 1 does not satisfy the equation

When x = 2,

$\begin{array}{l} x^{2} - 5 x + 6 = 0 \\ (2)^{2} - 5 (2) + 6 = 0 \\ 0 = 0 \end{array}$
x = 2 satisfies the equation.

When x = 3

$\begin{array}{l} x^{2} - 5 x + 6 = 0 \\ (3)^{2} - 5 (3) + 6 = 0 \\ 0 = 0 \end{array}$
x = 3 satisfies the equation.

Conclusion:

2 and 3 satisfy the equation $x^{2} - 5 x + 6 = 0$ , hence there are the roots of the equation.
1 does not satisfy the equation $x^{2} - 5 x + 6 = 0$ , hence it is NOT the root of the equation.

SPM Practice (Paper 1)

Posted on April 18, 2020 by user

Question 11:
The quadratic equation

$x^{2} - 4 x - 1 = 2 p (x - 5)$ , where p is a constant, has two equal roots. Calculate the possible values of p.

Solution:

Question 12:
Find the range of values of k for which the equation

$x^{2} - 2 k x + k^{2} + 5 k - 6 = 0$ has no real roots.

Solution:

Question 13:
Find the range of values of p for which the equation

$5 x^{2} + 7 x - 3 p = 6$ has no real roots.

Solution:

SPM Practice (Paper 1)

Posted on April 18, 2020 by user

Question 6:
Write and simplify the equation whose roots are the reciprocals of the roots of

$3 x^{2} + 2 x - 1 = 0$ , without solving the given equation.

Solution:

Question 7:
Find the value of p if one root of

$x^{2} + p x + 8 = 0$ is the square of the other.

Solution:

Question 8:
If one root of

$2 x^{2} + p x + 9 = 0$ is twice the other, find the values of p.

Solution:

SPM Practice (Paper 1)

Posted on April 18, 2020 by user

Question 1:
Solve the following quadratic equations by factorisation.

$\begin{array}{l} (a) x^{2} - 5 x - 10 = - 4 \\ (b) 3 - x - 2 x^{2} = 0 \\ (c) 11 a = 2 a^{2} + 12 \\ (d) \frac{2 x + 7}{3 x - 2} = x \end{array}$

Solution:

Question 2:
Solve the following quadratic equations by completing the square.

$\begin{array}{l} (a) 5 x^{2} + 10 x - 3 = 0 \\ (b) 2 x^{2} - 5 x - 6 = 0 \end{array}$

Solution:

2.4 Discriminant of a Quadratic Equation

Posted on April 18, 2020 by user

The Discriminant

The expression

$b^{2} - 4 a c$ in the general formula is called the discriminant of the equation, as it determines the type of roots that the equation has.

Example
Determine the nature of the roots of the following equations.
a.

$5 x^{2} - 7 x + 3 = 0$
b.

$x^{2} - 4 x + 4 = 0$
c.

$- 2 x^{2} + 5 x - 9 = 0$

Answer:

Posted in Quadratic Equations, SPM Add Maths

2.3d Forming New Quadratic Equation given a Quadratic Equation (Example)

Posted on April 18, 2020 by user

Example

If the roots of $x^{2} - 3 x - 7 = 0$ are α and β , find the equation whose roots are $α^{2} β$ and $α β^{2}$ .

Solution

Part 1 : Find SoR and PoR for the quadratic equation in the question

Part 2 : Form a new quadratic equation by finding SoR and PoR

Posted in Quadratic Equations, SPM Add Maths

Finding the Sum of Roots (SOR) and Product of Roots (POR) of a Quadratic Equation (Example)

Posted on April 18, 2020 by user

Example
The roots of $2 x^{2} + 3 x - 1 = 0$ are α and β. Find the values of
(a) $(α + 1) (β + 1)$
(b) $\frac{1}{α} + \frac{1}{β}$
(c) $α^{2} β + α β^{2}$
(d) $\frac{α}{β} + \frac{β}{α}$

[Clue: $α^{2} + β^{2} = {(α + β)}^{2} - 2 α β$ ]

Posted in Quadratic Equations, SPM Add Maths

Finding the Sum of Roots (SoR) and Product of Roots (PoR)

Posted on April 18, 2020 by user

2.6 Finding the Sum of Roots (SoR) and Product of Roots (PoR)

Example
Find the sums and products of the roots of the following equations.
a. $x^{2} + 7 x - 3 = 0$
b. $x (x - 1) = 5 (1 - x)$

Answer:
(a)
$\begin{array}{l} x^{2} + 7 x - 3 = 0 \\ a = 1, b = 7, c = - 3 \\ Sum of Roots \\ α + β = - \frac{b}{a} = - \frac{7}{1} = - 7 \\ Product of Roots \\ α β = \frac{c}{a} = \frac{- 3}{1} = - 3 \end{array}$

(b)
$\begin{array}{l} x (x - 1) = 5 (1 - x) \\ x^{2} - x = 5 - 5 x \\ x^{2} - x + 5 x - 5 = 0 \\ x^{2} + 4 x - 5 = 0 \\ a = 1, b = 4, c = - 5 \\ Sum of Roots \\ α + β = - \frac{b}{a} = - \frac{4}{1} = - 4 \\ Product of Roots \\ α β = \frac{c}{a} = \frac{- 5}{1} - 5 \end{array}$

Posted in Quadratic Equations, SPM Add Maths

2.3a Forming Quadratic Equations from Given Roots

Posted on April 18, 2020 by user

2.5 Forming Quadratic Equations from Given Roots

Let α and β be the roots of the equation ax²+ bx + c = 0, this means

Example:

Find the quadratic equation from the respective roots below:
(a) 3, −1

(b) −2, ¼

(c) ⅔, ¼
(d) 3m, −2m

Solution:

Posted in Quadratic Equations, SPM Add Maths

2.2.2c Solving Quadratic Equations – Quadratic Formula

Posted on April 18, 2020 by user

The quadratic equation $a x^{2} + b x + c = 0$ can be solved by using the quadratic formula

$\frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

Example
Use the quadratic formula to find the solutions of the following equations.
a. $x^{2} + 5 x - 24 = 0$
b. $x (x + 4) = 10$

Answer
(a)
For the equation $x^{2} + 5 x - 24 = 0$

a = 1, b = 5, c = -24

$\begin{array}{l} x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ x = \frac{- (5) \pm \sqrt{{(5)}^{2} - 4 (1) (- 24)}}{2 (1)} \\ x = \frac{- 5 \pm \sqrt{121}}{2} \\ x = - 8 or x = 3 \end{array}$

(b)
$\begin{array}{l} x (x + 4) = 10 \\ x^{2} + 4 x - 10 = 0 \\ a = 1, b = 4, c = - 10 \\ x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ x = \frac{- (4) \pm \sqrt{{(4)}^{2} - 4 (1) (- 10)}}{2 (1)} \\ x = \frac{- 4 \pm \sqrt{56}}{2} \\ x = 1.742 or x = - 5.742 \end{array}$

Posted in Quadratic Equations, SPM Add Maths

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