Question

f(x) = x² + (k-6) x +9, k * 0. The roots of the equation f(x) = 0 are a and B. (a) Find, in terms of k, the value of (i) a² + ß² (ii) a² ß² Given that 9(a²+ B²) = 2a²p². find the value of k. (b) (c) Using your value of k, and without solving the equation f(x) = 0. form a quadratic equation, with integer coefficients, which has roots and 33² f ( x ) = x² + ( k - 6 ) x +9 , k * 0 . The roots of the equation f ( x ) = 0 are a and B. ( a ) Find , in terms of k , the value of ( i ) a² + ß² ( ii ) a² ß² Given that 9 ( a² + B² ) = 2a²p² . find the value of k . ( b ) ( c ) Using your value of k , and without solving the equation f ( x ) = 0 . form a quadratic equation , with integer coefficients , which has roots and 33²​

Answer 1

(a) If
`\alpha` and
`\beta` are roots of
`f(x)`, then we can factorize
`f` as

`f(x) = x^2 + (k - 6) x + 9 = (x - \alpha) (x - \beta)`

Expand the right side and match up coefficients:

`x^2 + (k-6) x + 9 = x^2 - (\alpha + \beta) x + \alpha \beta \implies \begin{cases} \alpha + \beta = -(k-6) \\ \alpha \beta = 9 \end{cases}`

Now, recall that
`(x+y)^2 = x^2 + 2xy + y^2`. It follows that

`\boxed{\alpha^2 + \beta^2} = (\alpha + \beta)^2 - 2\alpha\beta = (-(k-6))^2 - 2*9 = \boxed{k^2 - 12k + 18}`

and

`\boxed{\alpha^2\beta^2} = 9^2 = \boxed{81}`

(b) If
`9(\alpha^2+\beta^2) = 2\alpha^2\beta^2`, then

`9 (k^2 - 12k + 18) = 2*81 \implies 9k^2 - 108k = 0 \implies 9k (k - 12) = 0`

Since
`k\\eq0`, it follows that
`\boxed{k=12}`.

(c) The simplest quadratic expression with roots
`\frac1{\alpha^2}` and
`\frac1{\beta^2}` is

`\left(x - \frac1{\alpha^2}\right) \left(x - \frac1{\beta^2}\right)`

which expands to

`x^2 - \left(\frac1{\alpha^2} + \frac1{\beta^2}\right) x + \frac1{\alpha^2\beta^2}`

Reusing the identity from (a-i) and the result from part (b), we have

`\left(\frac1\alpha + \frac1\beta\right)^2 = \frac1{\alpha^2} + \frac2{\alpha\beta} + \frac1{\beta^2} \\\\ \implies \frac1{\alpha^2} + \frac1{\beta^2} = \left((\alpha + \beta)/(\alpha\beta)\right)^2 - \frac2{\alpha\beta} = \left(\frac{-(k-6)}9\right)^2 - \frac29 = \frac29`

We also know from part (a-ii) that
`\alpha^2\beta^2=81`.

So, the simplest quadratic that fits the description is

`x^2 - \frac29 x + \frac1{81}`

To get one with integer coefficients, we multiply the whole expression by 81 to get
`\boxed{81x^2 - 18x + 1}`.