Quadratic equation 13

1.It is given that the product of the roots of the quaratic equation -px^2 - (p + 3)x + 3(p - 1) = 0 is - 9/4
(a) Find the value of p.
Product of roots = 3(p - 1)/(-p) = - 9/4
12(p - 1) = 9p
12p - 12 = 9p
3p = 12
p = 4
(b)(i) Find the value of the discriminant of the equation.
The equation becomes - 4x^2 - 7x + 9 = 0
Discrimiinant = 7^2 - (4)(-4)(9) = 193
(b)(ii)Hence,determine the nature of the roots of the equation.
Since discriminant > 0 there are two distinct real roots
2. If α and β are the roots of the quadratic equation x^2 + 12x + 5 = 0, find the values
of α^2 + β^2.
Sum of roots α + β = - 12
Product of roots αβ = 5
α^2 + β^2
= α^2 + 2αβ + β^2 - 2αβ
= (α + β)^2 - 2αβ
= ( - 12)^2 - (2)(5)
= 144 - 10
= 134
3. It is given than 3α and 3β are the roots of the quadratic equation x^2 + px + q = 0. Form a quadratic equation in x whose roots are α + 2β and β + 2α, in terms of p and q.
Sum of roots 3α + 3β = - p = > α + β = - p/3
Product of roots = (3α)(3β) = 9αβ = q = > αβ = q/9
New sum of roots = α + 2β + β + 2α = 3α + 3β = - p
New product of roots = (α + 2β)(β + 2α)
= αβ + 2β^2 + 2α^2 + 4αβ
= αβ + 2(α^2 + 2αβ + β^2)
= αβ + 2(α + β)^2
= q/9 + 2( - p/3)^2
= q/9 + 2p^2/9
The new equation is x^2 + px + (q + 2p^2)/9 = 0, or
9x^2 + 9px + q + 2p^2 = 0
4. Prove that the quadratic equation (x - k)[x - (k + 1)] = 1 has two distinct real roots of any real values of k.
(x - k)[x - (k + 1)] = 1
x^2 - kx - (k + 1)x + k(k + 1) - 1 = 0
x^2 - (2k + 1)x + (k^2 + k - 1) = 0
Discriminant = (2k + 1)^2 - 4(k^2 + k - 1)
= 4k^2 + 4k + 1 - 4k^2 - 4k + 4
= 5 > 0
Hence the equation has two distinct real roots for any real values of k

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