Question

Find the range of values of k for which the equation 4x² + 12x + k = 0 has no real roots.

Answer 1

we can approach this from the standpoint of the discriminant, if the discriminant is negative then we only get complex roots or namely "imaginary" solutions or values, well, let's check before that happens, when the discriminant is 0.

`\qquad \qquad \qquad \textit{discriminant of a quadratic} \\\\\\ y=\stackrel{\stackrel{a}{\downarrow }}{4}x^2\stackrel{\stackrel{b}{\downarrow }}{+12}x\stackrel{\stackrel{c}{\downarrow }}{+k} ~~~~~~~~ \stackrel{discriminant}{b^2-4ac}= \begin{cases} 0&\textit{one solution}\\ positive&\textit{two solutions}\\ negative&\textit{no solution} \end{cases}`

`0=(12)^2 - 4(4)(k)\implies 0=144-16k\implies 16k=144 \\\\\\ k=\cfrac{144}{16}\implies k=9\hspace{5em} \stackrel{if~k > 9\textit{, then we land on negative territory}}{{\Large \begin{array}{llll} k > 9 \end{array}}}`

Answer 2

Hello there!

Answer:

`(9, +\infty)`

Explanation:

It has no real roots only if

`\sqrt{b {}^(2) - 4ac }`

has no roots. That means that b² - 4ac is smaller than 0:

`b {}^(2) - 4ac < 0 \\ 12 {}^(2) - 4 * 4 * k < 0 \\ 144 - 16k < 0 \\ 144 - 144 - 16k < - 144 \\ - 16k < - 144 \\ k > 9 \\ k \: \: \: in \: \: \: (9, +\infty)`