If 5+35−3=a+15b for integers a and b then what are a and b?

Step 1 Assumption: The question should be: 5+35−3=a+15b Consider the left hand side (LHS) only 5+35−3×1 5+35−3×5+35+3 (5+3)(5+3)5−3 5+23×5+32 82+2152 4+15 Step 2 4+15=a+15b Thus a=4; b=1 Step 3 If the RHS →a+15b The b is still 1

"If 5+35−3=a+15b for integers a and b then..." was answered by students like you

High School
Algebra
Pre-Algebra
Equations, expressions, and inequalitie

Iris Vaughan

Answered question

2022-02-03

\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} = a + \sqrt{15} b

for integers a and b then what are a and b?

Answer & Explanation

Michaela Boyle

Beginner2022-02-04Added 10 answers

Step 1
Assumption: The question should be:

\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} = a + \sqrt{15} b

Consider the left hand side (LHS) only

\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} \times 1

\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} \times \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} + \sqrt{3}}

\frac{(\sqrt{5} + \sqrt{3}) (\sqrt{5} + \sqrt{3})}{5 - 3}

\frac{5 + 2 \sqrt{3 \times 5} + 3}{2}

\frac{8}{2} + \frac{2 \sqrt{15}}{2}

4 + \sqrt{15}

Step 2

4 + \sqrt{15} = a + \sqrt{15} b

Thus

a = 4; b = 1

Step 3
If the RHS

\to a + \sqrt{15 b}

The b is still 1

Iacopelli5co

Beginner2022-02-05Added 15 answers

Step 1
To rational the denominator

\sqrt{5} - \sqrt{3}

we can multiply both numerator and denominator by the radical conjugate

\sqrt{5} + \sqrt{3}

\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} = \frac{(\sqrt{5} + \sqrt{3}) (\sqrt{5} + \sqrt{3})}{(\sqrt{5} - \sqrt{3}) (\sqrt{5} + \sqrt{3})}

= \frac{{(\sqrt{5})}^{2} + 2 \sqrt{5} \sqrt{3} + {(\sqrt{3})}^{2}}{{(\sqrt{5})}^{2} - {(\sqrt{3})}^{2}}

= \frac{5 + 2 \sqrt{15} + 3}{5 - 3}

= \frac{8 + 2 \sqrt{15}}{2}

= 4 + \sqrt{15}

So,

a = 4

and

b = 1

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