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# A few minor problems. watch

1. Three questions I am not totally sure how to solve I could do with some help on.

1) Solve the equation 2x^2+32x+119=0
write you answers in the form p+q√2 where p and q are rational numbers.

I completed the square (not totally sure that this is right) and got 2(x+8)^2-9=0. Then 2x+16=3 so x+8=3/2 but don't know if this is right or what to do next.

Secondly, I am stuck on this

1)Express x^2-12x+40 in the form (x-p)^2+q. I did this and got (x-6)^2+4.
but then, in part b it says find the least value of x^2-12x+40 and I am not totally sure of what this means.

3)Lastly, I am having trouble with this. Find the constant a,b and c such that, for all values of x,
3x^2-6x+10=a(x+b)^2+c.

Now I completed the square on it but not sure what to do next.

Any helo will be much appreciated.
2. (Original post by stefl14)
Three questions I am not totally sure how to solve I could do with some help on.

1) Solve the equation 2x^2+32x+119=0
write you answers in the form p+q√2 where p and q are rational numbers.

I completed the square (not totally sure that this is right) and got 2(x+8)^2-9=0. Then 2x+16=3 so x+8=3/2 but don't know if this is right or what to do next.
TSR is horrendously slow at present; perhaps they're doing their upgrade.

You're correct up to "2(x+8)^2-9=0"

Now you need to find x.

What you had wasn't correct, although it had some good elements in it. Post your working there after and we can see what you were doing.
3. (Original post by stefl14)
Three questions I am not totally sure how to solve I could do with some help on.

1) Solve the equation 2x^2+32x+119=0
write you answers in the form p+q√2 where p and q are rational numbers.

I completed the square (not totally sure that this is right) and got 2(x+8)^2-9=0. Then 2x+16=3 so x+8=3/2 but don't know if this is right or what to do next.

Secondly, I am stuck on this

1)Express x^2-12x+40 in the form (x-p)^2+q. I did this and got (x-6)^2+4.
but then, in part b it says find the least value of x^2-12x+40 and I am not totally sure of what this means.

3)Lastly, I am having trouble with this. Find the constant a,b and c such that, for all values of x,
3x^2-6x+10=a(x+b)^2+c.

Now I completed the square on it but not sure what to do next.

Any helo will be much appreciated.
Hello,

1)Your completing is right, then add both side 9 then divide by 2
Taking the root rationalize
the denominator at the right side
Solve if a) x>=-8 b) x<-8
2)You are right: (x-6)^2+4
Consider that it is a function, and taking any value into x and calculating, you get the value of function at that x location. This type of quaratic function is a parabola, which has so called stationary point.
It is that point where the parabola change direction. It is minimum (least value) or maximum (greatest value) depending on the sign of the quadratic term.
In you example the sign is postive, so this function has minimum.
The equation if the parabola: Y-y0=1/(2p)(X-x0)^2 where
(x0,y0) is cooordinates of the vertex (the stationary point).
So Y=(x-6)^2+4 -> x0=6 y0=4
Your least value is 4, the location is 6.
3. Complete the left side of the equation an compare coefficients of
both side and read the values.

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