Question

a) by completing the square find in the form of the constant p the roots of the equation x^2+px+4=0
b) hence find the set of values of p for which the equation has no real roots ​

Answer 1

Answer:

a)

$\displaystyle x=-\frac{1}{2}p\pm\sqrt{\frac{1}{4}p^2-4}$

b) p is in the interval (-4,4)

Step-by-step explanation:

Quadratic Equation

It's given the following quadratic equation:

$x^2+px+4=0$

a)

It's required to complete squares and find the roots. This can be done by recalling the polynomial identity:

$a^2+2ab+b^2=(a+b)^2$

We already have the first term squared, and we need to find the second term. Rewriting the equation:

$\displaystyle x^2+2(\frac{1}{2}px)+4=0$

The second term of the binomial is 1/2p, thus completing the squares with $b^2$:

$\displaystyle x^2+2(\frac{1}{2}px)+\left(\frac{1}{2}p\right)^2+4-\left(\frac{1}{2}p\right)^2=0$

Factoring:

$\displaystyle \left(x+\frac{1}{2}p\right)^2+4-\frac{1}{4}p^2=0$

Moving the independent term to the right side:

$\displaystyle \left(x+\frac{1}{2}p\right)^2=\frac{1}{4}p^2-4$

Taking the square root:

$\displaystyle x+\frac{1}{2}p=\pm\sqrt{\frac{1}{4}p^2-4}$

Solving for x:

$\displaystyle x=-\frac{1}{2}p\pm\sqrt{\frac{1}{4}p^2-4}$

b) If the equation won't have real roots, then the radicand should be negative:

$\displaystyle \frac{1}{4}p^2-4<0$

Factoring:

$\displaystyle \left(\frac{1}{2}p-2\right)\left(\frac{1}{2}p+2\right)<0$

The zeros of the left-side polynomial are:

$\displaystyle \frac{1}{2}p-2=0$

$\displaystyle \frac{1}{2}p=2$

p = 4

$\displaystyle \frac{1}{2}p+2=0$

$\displaystyle \frac{1}{2}p=-2$

p = -4

The inequality:

$\displaystyle \left(\frac{1}{2}p-2\right)\left(\frac{1}{2}p+2\right)<0$

Is satisfied for values of p in the interval (-4,4)