simplify the expression $3\sqrt{7\div\sqrt{14-4\sqrt{7}}}$3√7÷√14−4√7 by rationalizing th denominator
6 years ago
Answered By Erin A
Rationalizing the denominator just means make sure there are no square root signs in the denominator of a fraction - they're only allowed in the numerator!
In order to do this, we can multiply a square root denominator by itself, because we know that $\sqrt{x}$√x multiplied by $\sqrt{x}$√x is just x, which gets rid of the square root sign. Then, so that we don't change the actual value of the expression, whatever we do to the denominator we also do to the numerator.
This problem looks easier if you apply the largest square root separately to the numerator and denominator of the fraction (using the principle $\sqrt{\frac{x}{y}}=\frac{\sqrt{x}}{\sqrt{y}}$√xy=√x√y ) and then work from the outside-in.
Now, we start by getting rid of the largest square root in our denominator
$\frac{3\sqrt{7}}{\sqrt{\sqrt{14-4\sqrt{7}}}}\times\frac{\sqrt{\sqrt{14-4\sqrt{7}}}}{\sqrt{\sqrt{14-4\sqrt{7}}}}$3√7√√14−4√7×√√14−4√7√√14−4√7 --> multiply the numerators together, then the denominators together
$\frac{3\sqrt{7}\sqrt{\sqrt{14-4\sqrt{7}}}}{\sqrt{14-4\sqrt{7}}}$3√7√√14−4√7√14−4√7 --> now we will rationalize the denominator a second time
*since our denominator is now a binomial (its not one big square root anymore), to rationalize this we multiply the numerator and denominator by the conjugate binomal of the denominator - see below
$\frac{3\sqrt{7}\sqrt{\sqrt{14-4\sqrt{7}}}\sqrt{14-4\sqrt{7}}}{14-4\sqrt{7}}\times\frac{14+4\sqrt{7}}{14+4\sqrt{7}}$3√7√√14−4√7√14−4√714−4√7×14+4√714+4√7 --> multiply, and expand the denominator
$\frac{\left(3\sqrt{7}\sqrt{\sqrt{14-4\sqrt{7}}}\sqrt{14-4\sqrt{7}}\right)\left(14+4\sqrt{7}\right)}{14^2+14\left(4\sqrt{7}\right)-14\left(4\sqrt{7}\right)-\left(4\sqrt{7}\right)^2}$(3√7√√14−4√7√14−4√7)(14+4√7)142+14(4√7)−14(4√7)−(4√7)2 --> the middle terms in the denominator cancel out
$\frac{\left(3\sqrt{7}\sqrt{\sqrt{14-4\sqrt{7}}}\sqrt{14-4\sqrt{7}}\right)\left(14+4\sqrt{7}\right)}{14^2-16\left(7\right)}$(3√7√√14−4√7√14−4√7)(14+4√7)142−16(7) --> and the denominator simplifies to 84
You're left with a whole number in the denominator. The numerator is one big monomial multiplied by a binomial - you can simplify this as well (exponent laws may help you with that) but if the question only needs you to rationalize the denominator, then you're done!
6 years ago
Answered By Erin A
Rationalizing the denominator just means make sure there are no square root signs in the denominator of a fraction - they're only allowed in the numerator!
In order to do this, we can multiply a square root denominator by itself, because we know that $\sqrt{x}$√x multiplied by $\sqrt{x}$√x is just x, which gets rid of the square root sign. Then, so that we don't change the actual value of the expression, whatever we do to the denominator we also do to the numerator.
This problem looks easier if you apply the largest square root separately to the numerator and denominator of the fraction (using the principle $\sqrt{\frac{x}{y}}=\frac{\sqrt{x}}{\sqrt{y}}$√xy =√x√y ) and then work from the outside-in.
$3\sqrt{\frac{7}{\sqrt{14-4\sqrt{7}}}}=3\frac{\sqrt{7}}{\sqrt{\sqrt{14-4\sqrt{7}}}}=\frac{3\sqrt{7}}{\sqrt{\sqrt{14-4\sqrt{7}}}}$3√7√14−4√7 =3√7√√14−4√7 =3√7√√14−4√7
Now, we start by getting rid of the largest square root in our denominator
$\frac{3\sqrt{7}}{\sqrt{\sqrt{14-4\sqrt{7}}}}\times\frac{\sqrt{\sqrt{14-4\sqrt{7}}}}{\sqrt{\sqrt{14-4\sqrt{7}}}}$3√7√√14−4√7 ×√√14−4√7√√14−4√7 --> multiply the numerators together, then the denominators together
$\frac{3\sqrt{7}\sqrt{\sqrt{14-4\sqrt{7}}}}{\sqrt{14-4\sqrt{7}}}$3√7√√14−4√7√14−4√7 --> now we will rationalize the denominator a second time
$\frac{3\sqrt{7}\sqrt{\sqrt{14-4\sqrt{7}}}}{\sqrt{14-4\sqrt{7}}}\times\frac{\sqrt{14-4\sqrt{7}}}{\sqrt{14-4\sqrt{7}}}$3√7√√14−4√7√14−4√7 ×√14−4√7√14−4√7
$\frac{3\sqrt{7}\sqrt{\sqrt{14-4\sqrt{7}}}\sqrt{14-4\sqrt{7}}}{14-4\sqrt{7}}$3√7√√14−4√7√14−4√714−4√7
*since our denominator is now a binomial (its not one big square root anymore), to rationalize this we multiply the numerator and denominator by the conjugate binomal of the denominator - see below
$\frac{3\sqrt{7}\sqrt{\sqrt{14-4\sqrt{7}}}\sqrt{14-4\sqrt{7}}}{14-4\sqrt{7}}\times\frac{14+4\sqrt{7}}{14+4\sqrt{7}}$3√7√√14−4√7√14−4√714−4√7 ×14+4√714+4√7 --> multiply, and expand the denominator
$\frac{\left(3\sqrt{7}\sqrt{\sqrt{14-4\sqrt{7}}}\sqrt{14-4\sqrt{7}}\right)\left(14+4\sqrt{7}\right)}{14^2+14\left(4\sqrt{7}\right)-14\left(4\sqrt{7}\right)-\left(4\sqrt{7}\right)^2}$(3√7√√14−4√7√14−4√7)(14+4√7)142+14(4√7)−14(4√7)−(4√7)2 --> the middle terms in the denominator cancel out
$\frac{\left(3\sqrt{7}\sqrt{\sqrt{14-4\sqrt{7}}}\sqrt{14-4\sqrt{7}}\right)\left(14+4\sqrt{7}\right)}{14^2-16\left(7\right)}$(3√7√√14−4√7√14−4√7)(14+4√7)142−16(7) --> and the denominator simplifies to 84
You're left with a whole number in the denominator. The numerator is one big monomial multiplied by a binomial - you can simplify this as well (exponent laws may help you with that) but if the question only needs you to rationalize the denominator, then you're done!