Help! Simultaneous quadratic equation

How do I solve that? Please show working out.
I havent studied maths for over 30 years!

Im trying to explain this to my daughter who thinks I know

Reply 1

gullyyt

14

So you complete the square. Basically half the ten and put it inside brackets like so.
(x-5)^2+33

you square the 5 inside the bracket to get 25.However, you want to get rid of it so you subtract 25 rather than adding it

(x-5)^2-25+33
Simplify
(x-5)^2+8

a=-5
b=8

The minimum value would be the opposite of whats inside the bracket and whats on the outside
(5,8)

Sorry if my explanation is horrible xD

(edited 6 years ago)

Original post by markmath

How do I solve that? Please show working out.
I havent studied maths for over 30 years!

Im trying to explain this to my daughter who thinks I know

Google 'completing the square'.

Reply 3

gullyyt

14

Original post by markmath

How do I solve that? Please show working out.
I havent studied maths for over 30 years!

Im trying to explain this to my daughter who thinks I know

Oh sorry, I may have answered the second part of the question incorrectly, does it mean the minimum point or the actual value of X , :\

Reply 4

markmath

OP

Original post by Gulled Bulhan

Oh sorry, I may have answered the second part of the question incorrectly, does it mean the minimum point or the actual value of X , :\

I really couldnt tell you. You can see the question as asked by gsce paper.

Reply 5

markmath

OP

Original post by Gulled Bulhan

So you complete the square. Basically half the ten and put it inside brackets like so.
(x-5)^2+33

you square the 5 inside the bracket to get 25.However, you want to get rid of it so you subtract 25 rather than adding it

(x-5)^2-25+33
Simplify
(x-5)^2+8

a=-5
b=8

The minimum value would be the opposite of whats inside the bracket and whats on the outside
(5,8)

Sorry if my explanation is horrible xD

Your explanation is good thankyou.

Reply 6

ManLike007

19

Original post by markmath

How do I solve that? Please show working out.
I havent studied maths for over 30 years!

Im trying to explain this to my daughter who thinks I know

Just a side note, the

\equiv

sign means equivalent, so in the question it's saying this 'equation' is equivalent to this 'equation' just expressed differently.

Part a is explained well enough however there is a mistake from what the other person said (Gulled Bulhan user).

We've established that

(x-a)^{2}+b \equiv (x-5)^{2} + 8

.

Therefore a = 5 because a is already negative. If you say

a=-5

then you'll get

(x-(-5))^{2} + 8 \Rightarrow (x+5)^{2} + 8

which is not the correct format.

As for part b, you need the minimum x coordinate. The minimum point is when the part of the equation which varies is minimised to zero so you get

(x-5)^{2}=0

Then square root both sides.

(x-5)=\sqrt {0}

(x-5)=0

. Then add 5 to both sides so you get *drumroll*

x=5

.

This

x=5

gives you the minimum

x

coordinate.

Now substitute this back into the equation of

x^2 - 10x + 33

Therefore

(5)^2 - 10(5) + 33

equals to

8

.

(edited 6 years ago)

Help! Simultaneous quadratic equation

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