Question

Given that x² + px+q and 4x²+q have a factor (x-r), where p, q and r are non-zero.
Show that 4p² +9q =0.

Answer 1

Answer:

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Answer 2

SOLUTION

GIVEN

x² + px + q and 4x² + q have a factor (x-r), where p, q and r are non-zero.

TO PROVE

4p² + 9q = 0

EVALUATION

Here it is given that x² + px + q have a factor (x-r)

So we have

$\sf {r}^{2} + pr + q = 0 \: \: \: - - - (1)$

Again 4x² + q have a factor (x-r)

So we have

$\sf 4{r}^{2} + q = 0$

$\sf \implies \: q = - 4{r}^{2} \: \: \: - - - (2)$

From Equation 1 and Equation 2 we get

$\sf {r}^{2} + pr - 4 {r}^{2} = 0$

$\sf \implies pr - 3 {r}^{2} = 0$

$\sf \implies pr = 3 {r}^{2}$

$\sf \implies p = 3 r$

Now

$\sf 4{p}^{2} + 9q$

$\sf = 4 \times {(3r)}^{2} + 9 \times ( - 4 {r}^{2} )$

$\sf =36 {r}^{2} - 36{r}^{2}$

$= 0$

Hence proved

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