Question

What are the roots of f(x) = x2 – 48? –48 and 48 –24 and 24 Negative 8 StartRoot 3 EndRoot and 8 StartRoot 3 EndRoot Negative 4 StartRoot 3 EndRoot and 4 StartRoot 3 EndRoot

Answer 1

Answer:

$x=4\sqrt{3},\:x=-4\sqrt{3}$ are the roots.

Step-by-step explanation:

$x=4\sqrt{3},\:x=-4\sqrt{3}$

Considering the expression

$x^2\:-\:48$

Solving

$x^2-48=0$

$x^2-48+48=0+48$

$x^2=48$

$\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$x=\sqrt{48},\:x=-\sqrt{48}$

Solving

$x=\sqrt{48}$

  = $\sqrt{2^4\cdot \:3}$

  $\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}$

  = $\sqrt{3}\sqrt{2^4}$

  $\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a^m}=a^{\frac{m}{n}}$

  $\sqrt{2^4}=2^{\frac{4}{2}}=2^2$

  $=2^2\sqrt{3}$

  $=4\sqrt{3}$

So,

$x=4\sqrt{3}$

Similarly,

$x=-\sqrt{48}=-4\sqrt{3}$

Therefore, $x=4\sqrt{3},\:x=-4\sqrt{3}$ are the roots.

Keywords: roots, expression

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Answer 2

Answer:

The solutions are  $x=4\sqrt{3}$ an $x=-4\sqrt{3}$

Step-by-step explanation:

We want to find the root of the quadratic function $f(x)=x^2-48$

To find the roots of this function, we set f(x)=0

This implies that:

$x^2-48=0$

This implies that:

$x^2=48$

Take square root to get;

$x=\pm \sqrt{48}$

$x=\pm4\sqrt{3}$

Therefore the solutions are  $x=4\sqrt{3}$ an $x=-4\sqrt{3}$