Question

Good Evening. Please I really need help for this Further Math Question. Thanks ​

Answer 1

You should be aware of the fundamental theorem of algebra (FTCoA). If a quadratic has roots at
`r_1` and
`r_2`, then we can factorize it as

`(x - r_1) (x - r_2)`

Expanding, this is equivalent to

`x^2 - (r_1 + r_2) x + r_1 r_2`

Now, if we can also write this

`x^2 + bx + c`

then we must have

`x^2 + bx + c = x^2 - (r_1 + r_2) x + r_1r_2 \\\\ ~~~~ \implies \begin{cases} b = -(r_1 + r_2) \\ c = r_1r_2 \end{cases}`

(By the way, these are known as Vieta's formulas.)

We're told that the quadratic,

`p(x) = 3x^2 - 6x + 8,`

has roots
`x=\alpha` and
`x=\beta`, so by the FTCoA, we can write

`p(x) = 3 (x - \alpha) (x - \beta)`

Expanding this last form, we have the identity and exact values

`3x^2 - 6x + 8 = 3 (x^2 - (\alpha + \beta) x + \alpha\beta) \\\\ ~~~~ \implies x^2 - 2x + \frac83 = x^2 - (\alpha + \beta) x + \alpha\beta \\\\ ~~~~ \implies \begin{cases} \boxed{\alpha + \beta = 2} \\\\ \boxed{\alpha\beta = \frac83} \end{cases}`

(i) We want to construct a new quadratic
`p_(\rm i)(x)` such that its roots are
`x=\alpha^2` and
`x=\beta^2`. By the FTCoA, we would have

`p_(\rm i)(x) = (x - \alpha^2) (x - \beta^2)`

Expanding,

`p_(\rm i)(x) = x^2 - (\alpha^2 + \beta^2) x + \alpha^2\beta^2`

Now, notice that

`\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 2^2 - 2\cdot\frac83 = -\frac43`

`\alpha^2\beta^2 = (\alpha\beta)^2 = \left(\frac83\right)^2 = \frac{64}9`

It follows that

`p_(\rm i)(x) = x^2 + \frac43 x + \frac{64}9`

which we can rewrite with integer coefficients by scaling each term by a factor of 9 to get

`p_(\rm i)(x) = \boxed{9x^2 + 12x + 64}`

(ii) Now we want a quadratic
`p_(\rm ii)(x)` with roots at
`2\alpha` and
`2\beta`. This means

`p_(\rm ii)(x) = (x - 2\alpha) (x - 2\beta) \\\\ ~~~~~~~~ = x^2 - 2 (\alpha + \beta)x + 4\alpha\beta`

`\implies p_(\rm ii)(x) = x^2 - 4x + \frac{32}3`

or more cleanly, scaling by 3,

`p_(\rm ii)(x) = \boxed{3x^2 - 12x + 32}`

(iii) Now the roots are
`x=\frac\alpha2` and
`x=\frac\beta2`, which gives

`p_(\rm iii)(x) = \left(x - \frac\alpha2\right) \left(x - \frac\beta2\right) \\\\ ~~~~~~~~ = x^2 - \frac{\alpha + \beta}2 x + \frac{\alpha\beta}4`

`\implies p_(\rm iii)(x) = x^2 - x + \frac8{12}`

`\implies p_(\rm iii)(x) = \boxed{12x^2 - 12x + 8}`

(iv) If the roots are
`x=\frac\alpha{\alpha+1}` and
`x=\frac\beta{\beta+1}`, we have

`p_(\rm iv)(x) = \left(x - \frac\alpha{\alpha+1}\right) \left(x - \frac\beta{\beta+1}\right) \\\\ ~~~~~~~~~= x^2 - \left(\frac\alpha{\alpha+1} + \frac\beta{\beta+1}\right) x + (\alpha\beta)/((\alpha+1)(\beta+1))`

Note that

`\frac\alpha{\alpha+1} + \frac\beta{\beta+1} = (\alpha(\beta+1) + \beta(\alpha+1))/((\alpha+1)(\beta+1)) \\\\ ~~~~~~~~~~~~~~~~~~~~ = (2\alpha\beta + \alpha + \beta)/(\alpha\beta + \alpha + \beta + 1) \\\\ ~~~~~~~~~~~~~~~~~~~~ = \frac{\frac{16}3 + 2}{\frac83 + 2 + 1} \\\\ ~~~~~~~~~~~~~~~~~~~~ = (22)/(17)`

`(\alpha\beta)/((\alpha+1)(\beta+1)) = (\alpha\beta)/(\alpha\beta + \alpha + \beta + 1) \\\\ ~~~~~~~~~~~~~~~~~~~~~ = (\frac83)/(\frac83 + 2 + 1) \\\\ ~~~~~~~~~~~~~~~~~~~~~ = \frac8{17}`

`\implies p_(\rm iv)(x) = x^2 - (22)/(17)x + \frac8{17}`

`\implies p_(\rm iv)(x) = \boxed{17x^2 - 22x + 8}`