Question

It is given the quadratic equation (q + 1 )x² - 8x + p = 0,where p and q are constants,has two equal real roots.Express p in terms of q.
Do help meeee!!!
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Answer 1

Step-by-step explanation:

answer is in the photo above

Answer 2

Answer:

$\displaystyle p = \frac{16}{q + 1}$.

Step-by-step explanation:

Factor out the coefficient of $x^{2}$:

$\displaystyle \frac{1}{q + 1}\, [(q + 1)\, x^2 - 8\, x + p] = 0$.

$\displaystyle x^2 - \frac{8}{q + 1} \, x + \frac{p}{q + 1} = 0$.

Let $m$ denote the real root of this equation. By the Factor Theorem, this equation would have the factor $(x - m)$ repeated for two times in total:

$(x - m)^{2} = 0$.

Expand to obtain: $x^2 - 2\, m\cdot x + m^{2} =0$.

Compare this expression with $\displaystyle x^2 - \frac{8}{q + 1} \, x + \frac{p}{q + 1} = 0$:

$\displaystyle -\frac{8}{q + 1} = -2\, m$ (for the coefficient of $x$.)

$\displaystyle \frac{p}{q + 1} = m^2$ (for the constant.)

Rewrite the first equation to find an expression for $m$ in terms of $q$:

$\displaystyle m = \frac{4}{q + 1}$.

Substitute this expression for $m$ into the second equation to find an expression for $p$ in terms of $q$:

$\displaystyle \frac{p}{q + 1} = m^2$.

$\displaystyle \frac{p}{q + 1} = \frac{4^2}{(q+1)^{2}}$.

$\displaystyle p = \frac{16}{q + 1}$.

Verify that this expression for $p$ satisfies the requirements:

$\displaystyle (q + 1) \, x^{2} - 8\, x + \frac{16}{q + 1} = 0$.

$\displaystyle x^2 - \frac{8}{q + 1} + \frac{16}{(q+1)^2} = 0$.

$\displaystyle x = \frac{4}{q + 1}$ is the (repeated) real root of this quadratic equation.