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Surds C1

Expand and simplify (7 + 3ROOT 2)(5 2 ROOT2).


I got the answer to be -11 but i'm not tooo sure !



Also, does anyone know how to explain (iv) ,the last bit of this question, from the JAN2012 C1 OCREMEI past paper? Thaks everybody! :smile:

(iii) Show that, where the line y = 2x + k intersects the circle,
5x^2 + (4k 4) x + k^2 16 = 0. [3]
(iv) Hence find the values of k for which the line y = 2x + k is a tangent to the circle.
Reply 1
Original post by alex7892
Expand and simplify (7 + 3ROOT 2)(5 2 ROOT2).


I got the answer to be -11 but i'm not tooo sure !



Also, does anyone know how to explain (iv) ,the last bit of this question, from the JAN2012 C1 OCREMEI past paper? Thaks everybody! :smile:

(iii) Show that, where the line y = 2x + k intersects the circle,
5x^2 + (4k 4) x + k^2 16 = 0. [3]
(iv) Hence find the values of k for which the line y = 2x + k is a tangent to the circle.


(7+32)(522)11 (7+3\sqrt2)(5-2\sqrt2) \not= -11

Show your working.
Reply 2
Sorry looks like I made a mistake. Would the answer be:

35-14root5+15root2-6root10 ?
Reply 3
Original post by alex7892

Also, does anyone know how to explain (iv) ,the last bit of this question, from the JAN2012 C1 OCREMEI past paper? Thaks everybody! :smile:

(iii) Show that, where the line y = 2x + k intersects the circle,
5x^2 + (4k 4) x + k^2 16 = 0. [3]
(iv) Hence find the values of k for which the line y = 2x + k is a tangent to the circle.



We know a tangent intersects the circle at only one point, hence the quadratic, 5x2+(4k4)x+k216=0. 5x^2 + (4k - 4) x + k^2 -16 = 0. , will have repeated roots.
So solve, b24ac=0 b^2 - 4ac =0
Reply 4
..Ok, thanks Raheem, but what would be A, B and C? :s-smilie:
Reply 5
Original post by alex7892
Sorry looks like I made a mistake. Would the answer be:

35-14root5+15root2-6root10 ?


No.

(7+32)(522)=7×57×22+5×3232×22 (7+3\sqrt2)(5-2\sqrt2) = 7\times 5 - 7 \times 2\sqrt2 +5 \times 3 \sqrt2 - 3\sqrt2 \times 2\sqrt2

Remember, 2×2=2 \sqrt2 \times \sqrt2 = 2
Reply 6
Original post by alex7892
..Ok, thanks Raheem, but what would be A, B and C? :s-smilie:


Do you know about the discriminant?

For repeated roots it is, b24ac=0 b^2 -4ac =0

5x2+(4k4)x+k216=0 5x^2 + (4k - 4) x + k^2 -16 = 0
a=5   b=4k4   c=k216 a=5 \ \ \ b=4k-4 \ \ \ c=k^2-16
Reply 7
Original post by raheem94
No.

(7+32)(522)=7×57×22+5×3232×22 (7+3\sqrt2)(5-2\sqrt2) = 7\times 5 - 7 \times 2\sqrt2 +5 \times 3 \sqrt2 - 3\sqrt2 \times 2\sqrt2

Remember, 2×2=2 \sqrt2 \times \sqrt2 = 2


Oh so sorry, I posted the wrong question, and my answer was for the other question haha I mean (7+3root2) (5-2root5)
Sorry!! And what would A, and C be then that is confusing :s-smilie:
Reply 8
Okay thanks for the discriminant thing, just quite weired using k2-16 for c etc... never seen that before...
Reply 9
Original post by alex7892
..but what would be A, B and C? :s-smilie:


Quadratics have the form ax2+bx+c=0ax^2 + bx + c = 0

a,b and c are the coefficients of each term in the general quadratic equation. So, 'a' is whatever comes before an x^2, 'b' is whatever comes before an 'x', and 'c' is anything that doesn't have an 'x' in it.

(extra: technically, c is the coefficient of x^0, but as x^0 = 1 theres no need to write it.)
(edited 11 years ago)
Reply 10
Thaaanks everyone :smile:
Reply 11
Original post by alex7892
Oh so sorry, I posted the wrong question, and my answer was for the other question haha I mean (7+3root2) (5-2root5)
Sorry!! And what would A, and C be then that is confusing :s-smilie:


(7+32)(525)11 (7+3 \sqrt2) (5-2\sqrt5) \not= -11

By the way, are you sure you have typed out the right question now?
Reply 12
Okay was it the second answer I gave? and yesss... :smile:
Reply 13
Original post by alex7892
Sorry looks like I made a mistake. Would the answer be:

35-14root5+15root2-6root10 ?


Yes, this is correct :smile:
Reply 14
Thanks a lot Raheem going to personal rep you in a minute, always sorting out my maths problems aren't you! lol :biggrin:
Reply 15
Original post by alex7892
Thanks a lot Raheem going to personal rep you in a minute, always sorting out my maths problems aren't you! lol :biggrin:


I answer a lot of questions on the maths forum, so it is difficult to keep track of the people, except very active members.

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