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 x2 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 15y2 – 17y + 5 = 0 and 6x2 = 2x + 20
(iii) If α = \( – \frac{1}{2}\) is a root of the quadratic equation 8x2 – 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 6x2 – 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
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