#### Topic Content:

- Finding a Quadratic Equation whose Roots are Given

If α and β are the roots of a quadratic equation, then either x = α or x = β

i.e. x – α = 0 or x – β = 0

∴ (x – α)(x – β) = 0

By expansion we have

\( \scriptsize x^2 \: – \: αx \: – \: xβ\: + \: αβ = 0\)= \( \scriptsize x^2 \: -\: \left(\: α \: + \: β \right)x \: + \: αβ = 0\)

**Note that: **

(i) The coefficient of x^{2} is 1

(ii) α + β = Sum of the roots.

Observe that it is the coefficient of x with the sign changed

(iii) αβ = Product of roots. Observe that it is the constant term

Thus we have:

\( \scriptsize x^2 \: – \: \left(\: α \: + \: β \right)x \: + \: αβ = \scriptsize x^2\: +\: \normalsize \frac{b}{a}\scriptsize x\: +\: \normalsize \frac{c}{a}\)

Comparing coefficients, we have

### Example 6.1.1:

Find:**(i) **a quadratic equation whose roots are 5 & -4, \( \frac{2}{5}\) and \( \scriptsize -1 \normalsize \frac{1}{2} \)**(ii)** the sum and product of roots of 15y^{2} – 17y + 5 = 0 and 6x^{2} = 2x + 20**(iii)** If α = \( – \frac{1}{2}\) is a root of the quadratic equation 8x^{2} – bx – 3 = 0, find the value of

(a) b

(b) the other root, β

(c) \( \left (\frac{1}{α} \: – \: \frac{1}{β} \right)^2 \)**(iv)** If one root of the quadratic equation 6x^{2} – kx + 48 = 0 is the square of the other, find the two roots and hence find the value of k.

**(i) **a quadratic equation whose roots are 5 & -4, \( \frac{2}{5}\) and \( \scriptsize -1 \normalsize \frac{1}{2} \)

**Solution **

**a.** Let x = 5

You are viewing an excerpt of this Topic. Subscribe Now to get **Full Access** to **ALL** this Subject's Topics and Quizzes for this Term!

Click on the button **"Subscribe Now"** below for Full Access!

### Subscribe Now

**Note:** If you have **Already Subscribed** and you are seeing this message, it means you are logged out. Please **Log In** using the **Login Button Below** to **Carry on Studying!
**

## Responses