Hard A-Level Indices and surds Questions

Plan the approach

Both surds are nested. If each equals a perfect square, the outer roots simplify to simple surd expressions.

Try a square for the first surd

Test $(\sqrt5+2)^2$ using $(p+q)^2$.

Confirm the first square

The numbers give $5+4=9$, which matches.

Take the first square root

Since $\sqrt5+2>0$, this is the correct root.

Try a square for the second surd

Test $(\sqrt5-2)^2$ using $(p-q)^2$.

Confirm the second square

The numbers give $5+4=9$, which matches.

Check the sign of √5 − 2

Since $\sqrt5$ is bigger than 2, $\sqrt5-2$ is positive, so it is the correct root.

Take the second square root

The positive root is $\sqrt5-2$.

Set up the subtraction

Now carry out the subtraction from the original expression.

Remove the brackets

The minus sign flips the signs of the second bracket.

Simplify

The $\sqrt5$ terms cancel and $2+2=4$.

Recall the expansion identities

Recognising perfect squares of the form $(\sqrt5\pm2)^2$ was the key idea for de-nesting these surds.

Note both roots are positive

Both expressions under the outer roots are positive, so each square root is a positive surd expression.

Explain why the answer is whole

Because the two roots differ only in the sign of the constant, subtracting them cancels the surds and leaves a whole number.

Check numerically

The two roots are about 4.236 and 0.236, and their difference is 4, confirming the answer.

Understand the target

We know $2^a=5$. The plan is to rewrite $2^{3a+1}$ using $2^a$ so we can substitute the 5.

Split the index using the addition law

Adding indices corresponds to multiplying powers, so we separate the $+1$.

Rewrite 2^{3a} as a cube

The index $3a$ means $2^a$ cubed, using power of a power in reverse.

Substitute the given value

Replace $2^a$ with 5.

Cube the 5

$5\times5\times5=125$.

Put the pieces together

Multiply by the leftover factor of $2^1=2$.

State the answer

$125\times2=250$.

Recall the addition law

Adding indices means multiplying powers; this is the law that let us peel off the $+1$.

Recall the power-of-a-power law

Multiplying indices means a power raised to a power; this turned $2^{3a}$ into $\left(2^a\right)^3$.

Notice we never need a itself

We do not have to find the exact value of $a$; knowing $2^a=5$ is all that is needed to substitute.

Interpret the leftover factor

The extra $2^1$ is just 2; it simply doubles the result of $5^3$.

Recombine to check the split

Multiplying $2^{3a}$ by $2^1$ rebuilds the original index $3a+1$, so the split was valid.

View it all at once

Reading the whole thing as twice $(2^a)^3$ gives the same route to the answer.

Check numerically

Using the approximate value $a\approx2.322$, the power comes out to about 250, matching the exact answer.

State the conclusion

So the exact value is 250.

Plan the method

The two denominators are conjugates, so their product is surd-free. We combine into one fraction.

Find the common denominator

The common denominator is the product of the two brackets.

Write as a single fraction

Each numerator is multiplied by the other fraction's denominator.

Set up the denominator

Difference of two squares.

Simplify the denominator

This gives a denominator of 3.

Expand the numerator

Distribute the 3 across each bracket, being careful with the subtraction.

Simplify the numerator

The $3\sqrt5$ terms cancel and the $3\sqrt2$ terms add to $6\sqrt2$.

Write the fraction

Numerator over the denominator 3.

Simplify

Divide $6\sqrt2$ by 3 to get $2\sqrt2$.

Recall the difference of two squares

This identity is why the product of the two conjugate denominators has no surds.

Check the first numerator part

Expanding the first bracket carefully avoids sign slips.

Check the second numerator part

The leading minus sign flips both signs inside the second bracket.

Explain why √5 cancels

The two $\sqrt5$ terms are equal and opposite, so they cancel completely.

Check numerically

The original difference is about 2.83 and $2\sqrt2\approx2.83$, confirming the answer.

State the conclusion

So the whole expression simplifies to $2\sqrt2$.

Plan: simplify inside first

It is easier to simplify the fraction inside the bracket before dealing with the outer power.

Subtract the a indices

Divide by subtracting the bottom index from the top.

Finish the a index

Subtracting $-1$ is the same as adding 1: $\tfrac12+1=\tfrac32$.

Subtract the b indices

Again subtract the bottom index from the top.

Finish the b index

$-2-\tfrac12=-\tfrac52$.

Rewrite the bracket

The inside has simplified to $a^{3/2}b^{-5/2}$.

Apply the outer power to a

Multiply the indices: $\tfrac32\times(-2)=-3$.

Apply the outer power to b

Multiply the indices: $-\tfrac52\times(-2)=5$.

Combine

Put the two results together.

Make indices positive

The negative index on $a$ moves it to the denominator.

Recall the division law used

Dividing powers of the same base means subtracting indices; we used this for both $a$ and $b$ inside the bracket.

Recall the power-of-a-power law

Raising a power to another power means multiplying indices; this handled the outer $-2$.

Explain why b ends up on top

The two negative signs multiply to a positive, so $b$ finishes on top with a positive index.

Sense-check the form

The final answer uses only positive indices, exactly as the question required.

Spot-check with numbers

Testing $a=4,b=1$: the bracket is 8, so the whole thing is $\tfrac1{64}$, matching $\tfrac{b^5}{a^3}=\tfrac1{64}$.

Study the equation

There is a $5^{2x}$ term and a $5^x$ term, so a quadratic is hiding inside.

Recognise the hidden quadratic

By the power law, $5^{2x}$ is the square of $5^x$.

Substitute

Let $y=5^x$ to form a standard quadratic.

Find the factor pair

We need numbers multiplying to 5 and adding to $-6$: they are $-1$ and $-5$.

Factorise

Write the quadratic as a product of two brackets.

Solve for y

Set each bracket to zero.

Return to the first root

Any base to the power 0 is 1.

Solve the first case

Matching bases gives $x=0$.

Return to the second root

Rewrite the second value with base 5.

Solve the second case

Matching bases gives $x=1$.

State the solutions

Both solutions are valid.

Recall the substitution idea

Naming $5^x$ as $y$ is what turned an index equation into a quadratic we could factorise.

Note the positivity condition

Both $y$-values (1 and 5) are positive, so neither is rejected and no solutions are lost.

Check the first solution

Substituting $x=0$ gives 0, so it is correct.

Check the second solution

Substituting $x=1$ also gives 0, confirming both answers.