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C1 indices/ surd question

Screenshot_2016-04-01-13-36-04.png can anyone go through this question step by step? I don't get how to solve it. Thanks.
Reply 1
=2x3×x232 \displaystyle = \sqrt 2x^3 \times \frac{\sqrt{x^2}}{\sqrt{32}} .

32=.. \sqrt{32}= ..

x2=... \sqrt{x^2}=... <---- be careful with this one.

Spoiler

:tongue:
(edited 7 years ago)
Reply 2
Original post by coconut64
Screenshot_2016-04-01-13-36-04.png can anyone go through this question step by step? I don't get how to solve it. Thanks.


Step 1: a÷bc=a×cba \div \frac{b}{c} = a \times \frac{c}{b}: this gives 2(x3)÷32x2=2(x3)×x232\sqrt{2}(x^3) \div \sqrt{\frac{32}{x^2}} = \sqrt{2}(x^3) \times \sqrt{\frac{x^2}{32}}

Step 2: ab=ab\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}, this gives: 2(x3)×x232=2(x3)×x232\sqrt{2}(x^3) \times \sqrt{\frac{x^2}{32}} = \sqrt{2}(x^3) \times \frac{\sqrt{x^2}}{\sqrt{32}}

Step 3: 32=16×2=162=42\sqrt{32} = \sqrt{16 \times 2} = \sqrt{16}\sqrt{2} = 4\sqrt{2} so that 2×132=242=14\sqrt{2} \times \frac{1}{\sqrt{32}} = \frac{\sqrt{2}}{4\sqrt{2}} = \frac{1}{4}

Step 4: 2(x3)×x232=14×x3×x2\sqrt{2}(x^3) \times \frac{\sqrt{x^2}}{\sqrt{32}} = \frac{1}{4} \times x^3 \times \sqrt{x^2}

Step 5: a=a1/2\sqrt{a} = a^{1/2} so x2=(x2)1/2\sqrt{x^2} = (x^{2})^{1/2}

Step 6: (ab)c=abc(a^b)^c = a^{bc} so x3=(x2)1/2=x2×1/2=x1=x\sqrt{x^3} = (x^2)^{1/2} = x^{2 \times 1/2} = x^{1} = x

Can you finish off from here?
(edited 7 years ago)
Reply 3
Original post by Zacken
Step 1: a÷bc=a×cba \div \frac{b}{c} = a \times \frac{c}{b}: this gives 2(x3)÷32x2=2(x3)×x232\sqrt{2}(x^3) \div \sqrt{\frac{32}{x^2}} = \sqrt{2}(x^3) \times \sqrt{\frac{x^2}{32}}

Step 2: ab=ab\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}, this gives: 2(x3)×x232=2(x3)×x232\sqrt{2}(x^3) \times \sqrt{\frac{x^2}{32}} = \sqrt{2}(x^3) \times \frac{\sqrt{x^2}}{\sqrt{32}}

Step 3: 32=16×2=162=42\sqrt{32} = \sqrt{16 \times 2} = \sqrt{16}\sqrt{2} = 4\sqrt{2} so that 2×132=242=14\sqrt{2} \times \frac{1}{\sqrt{32}} = \frac{\sqrt{2}}{4\sqrt{2}} = \frac{1}{4}

Step 4: 2(x3)×x232=14×x3×x2\sqrt{2}(x^3) \times \frac{\sqrt{x^2}}{\sqrt{32}} = \frac{1}{4} \times x^3 \times \sqrt{x^2}

Step 5: a=a1/2\sqrt{a} = a^{1/2} so x2=(x2)1/2\sqrt{x^2} = (x^{2})^{1/2}

Step 6: (ab)c=abc(a^b)^c = a^{bc} so x3=(x2)1/2=x2×1/2=x1=x\sqrt{x^3} = (x^2)^{1/2} = x^{2 \times 1/2} = x^{1} = x

Can you finish off from here?


1459515459784207341171.jpg I think this is right?
Reply 4
Original post by coconut64
1459515459784207341171.jpg I think this is right?


Perfect. First class work.
Reply 5
Original post by coconut64
1459515459784207341171.jpg I think this is right?


hmmm. try x=-1 and see if this works :wink:
Reply 6
Original post by B_9710
=2x3×x232 \displaystyle = \sqrt 2x^3 \times \frac{\sqrt{x^2}}{\sqrt{32}} .

32=.. \sqrt{32}= ..

x2=... \sqrt{x^2}=... <---- be careful with this one.

Spoiler

:tongue:


Thanks
Reply 7
Original post by coconut64
Thanks


I have to say, you did not get the right answer. As I said try any negative value of x and see if you get the same answer from the original expression and the final one you posted.
It is right if there is a condition stated somewhere that 0x 0 \leqslant x .
(edited 7 years ago)
Reply 8
Original post by coconut64
Thanks


Would the question happen to have a condition x>0x>0 that you haven't shown us? I'd think so. :smile:
Reply 9
Original post by Zacken
Would the question happen to have a condition x>0x>0 that you haven't shown us? I'd think so. :smile:


Not really. 1459516221392-1339212355.jpg
Reply 10
Original post by B_9710
I have to say, you did not get the right answer. As I said try any negative value of x and see if you get the same answer from the original expression and the final one you posted.
It is right if there is a condition stated somewhere that 0x 0 \leqslant x .

True but it works with a positive number tho. This is a non calculator exam so I would probs to go along with it. The question doesn't give other info BTW. Thanks.
Reply 11
Original post by coconut64
Not really. 1459516221392-1339212355.jpg


Oh well, it's just crappy question writing on Edexcel's part, nothing you need worry about at this stage.

Spoiler

Reply 12
I'm saying the answer is 14x3x \displaystyle \frac{1}{4}x^3|x| .
Reply 13
Original post by Zacken
Oh well, it's just crappy question writing on Edexcel's part, nothing you need worry about at this stage.

Spoiler



Oh that I have abosolutely no clue what it is. I don't do further maths. Thanks again.
Reply 14
Original post by coconut64
Oh that I have abosolutely no clue what it is. I don't do further maths. Thanks again.


No problem.

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