Completing the Square - Harder

Factor 3 out of the terms containing x before completing the square.

1. First take out the coefficient of x^2: 3x^2+12x+5=3(x^2+4x)+5.
2. Half of 4 is 2, so x^2+4x=(x+2)^2-4.
3. The subtraction is inside a bracket multiplied by 3, so subtract 12 from the constant 5.
4. The completed square form is 3(x+2)^2-7.

Factor out the coefficient of x^2 before halving the coefficient inside.

1. Take 5 out of the terms containing x: 5(x^2-4x)+6.
2. Inside the bracket, x^2-4x=(x-2)^2-4.
3. The correction is multiplied by 5, giving -20+6=-14.
4. The answer is 5(x-2)^2-14.

Check the outside coefficient as well as the terms produced by the squared bracket.

1. First write the expression as 3(x^2+2x)-2.
2. Inside the bracket, x^2+2x=(x+1)^2-1.
3. Multiplying by 3 and then subtracting 2 gives 3(x+1)^2-5.
4. 3(x+1)^2-3 forgets to multiply the -1 by 3. 3(x+2)^2-14 puts the whole coefficient 2 in the bracket instead of half of it: (x+2)^2=x^2+4x+4, which gives 4x, not 2x.

Take out the negative sign before completing the square.

1. Take out the negative sign: -(x^2-8x)+1.
2. Inside the bracket, x^2-8x=(x-4)^2-16.
3. Negating this gives -(x-4)^2+16, then add 1.
4. The answer is -(x-4)^2+17.

Remember to multiply the constant from the squared bracket by the outside coefficient.

1. The squared bracket contributes the constant 3^2=9.
2. The whole bracket is multiplied by 4.
3. Therefore k=4\times9=36.

Keep the outside multiplier when adjusting the constant.

1. Write the expression as 6(x^2+2x)+9.
2. Replace x^2+2x with (x+1)^2-1.
3. The constant becomes 9-6=3.
4. Thus the completed square form is 6(x+1)^2+3.

Expand the proposed squared expression to check its constant.

1. Expanding 2(x+3)^2 gives 2x^2+12x+18.
2. To leave the required constant 1, subtract 17.
3. The correct form is 2(x+3)^2-17, so the final constant must change.

After taking out 2, halve the coefficient of x inside the bracket.

1. Take 2 out of the first two terms: 2x^2-6x+1=2(x^2-3x)+1.
2. Half of -3 is -\frac{3}{2}, and its square is \frac{9}{4}.
3. So x^2-3x=(x-\frac{3}{2})^2-\frac{9}{4}.
4. Multiplying the correction by 2 gives -\frac{9}{2}; adding 1 gives -\frac{7}{2}.
5. The answer is 2(x-\frac{3}{2})^2-\frac{7}{2}.

After factoring out 4, the linear coefficient inside the bracket is 1.

1. Take out 4 from the first two terms: 4(x^2+x)+5.
2. Inside the bracket, x^2+x=(x+\frac{1}{2})^2-\frac{1}{4}.
3. Multiplying the correction by 4 subtracts 1.
4. The answer is 4(x+\frac{1}{2})^2+4.

Reorder the terms, then factor out the negative leading coefficient.

1. Reorder the terms and take out -3: -3(x^2+4x)+2.
2. Complete the square inside: x^2+4x=(x+2)^2-4.
3. Multiplying the correction by -3 adds 12.
4. The result is -3(x+2)^2+14.

Divide the linear coefficient by the outside multiplier before halving it.

Part (a)

1. The expression begins 8(x^2-3x).
2. The shift is half of 3, so a=\frac{3}{2}.

Part (b)

1. The square contributes 8\times(\frac{3}{2})^2=18 to the constant.
2. The required constant is 7, so 18+b=7.
3. Therefore b=-11.

Complete the square and check the least possible value of the expression.

1. Write the expression as 2(x^2+2x)+7.
2. Completing the square gives 2(x+1)^2+5.
3. A square is never negative, so the expression is at least 5 and cannot equal zero.
4. 2(x+1)^2+3 has the wrong constant: 2(x+1)^2=2x^2+4x+2, so +5 (not +3) is needed to make +7. The statement ending "at most 5" has the inequality the wrong way round.

Complete the square, isolate the squared bracket, then take both square roots.

1. Take 2 out of the x terms: 2(x^2-4x)+1=0.
2. Half of -4 is -2, so x^2-4x=(x-2)^2-4. The equation becomes 2(x-2)^2-8+1=0, which is 2(x-2)^2-7=0.
3. Isolate the square: (x-2)^2=\dfrac72.
4. Take both square roots: x-2=\pm\sqrt{\dfrac72}. Multiplying top and bottom inside the root by 2, \sqrt{\dfrac72}=\sqrt{\dfrac{14}{4}}=\dfrac{\sqrt{14}}{2}.
5. So x=2-\dfrac{\sqrt{14}}{2} or x=2+\dfrac{\sqrt{14}}{2}.

Simplify the equation by dividing through before completing the square.

1. Divide the equation by 5: x^2+2x=4.
2. Add 1 to both sides: (x+1)^2=5.
3. Take both square roots and subtract 1.
4. Thus x=-1-\sqrt{5} or x=-1+\sqrt{5}.

Complete the square after taking 3 out of the first two terms, then isolate the squared bracket.

1. Write the left side as 3(x^2+4x)+7.
2. Since x^2+4x=(x+2)^2-4, the equation becomes 3(x+2)^2-5=0.
3. Isolate the square: (x+2)^2=\frac{5}{3}.
4. Take both square roots: x+2=\pm\frac{\sqrt{15}}{3}.
5. So x=-2-\frac{\sqrt{15}}{3} or x=-2+\frac{\sqrt{15}}{3}.

Expand first, then divide by the leading coefficient before completing the square.

1. Expand the product: 2x^2+7x-4=6.
2. Divide the resulting equation by 2: x^2+\frac{7}{2}x=5.
3. Add \frac{49}{16} to both sides: (x+\frac{7}{4})^2=\frac{129}{16}.
4. Take both square roots: x+\frac{7}{4}=\pm\frac{\sqrt{129}}{4}.
5. Hence x=\frac{-7-\sqrt{129}}{4} or x=\frac{-7+\sqrt{129}}{4}.

A square is never negative, so think about what multiplying it by a negative number does.

1. Factor out -4 from the first two terms: -4(x^2-2x)-9.
2. Completing the square gives -4(x-1)^2-5.
3. Since (x-1)^2 is never negative, -4(x-1)^2 is zero or negative.
4. Subtracting 5 makes the expression at most -5, so it is always negative.
5. The statement saying the square is never positive is wrong: a square is never negative. The statements using -4(x+1)^2-5 and -4(x-1)^2+5 have the wrong bracket or constant.

Factor out -2 carefully, then remember that the correction term is also multiplied by -2.

Part (a)

1. Factor out -2 from the terms containing x: -2(x^2-5x)+3.
2. Complete the square inside the bracket: x^2-5x=(x-\frac{5}{2})^2-\frac{25}{4}.
3. The correction contributes (-2)\times(-\frac{25}{4})=\frac{25}{2}.
4. Adding 3 gives \frac{31}{2}, so the answer is -2(x-\frac{5}{2})^2+\frac{31}{2}.

Part (b)

1. Using part (a), -2(x-\frac{5}{2})^2+\frac{31}{2}=9.
2. Subtract \frac{31}{2} and divide by -2: (x-\frac{5}{2})^2=\frac{13}{4}.
3. Both signs are needed when taking square roots: x-\frac{5}{2}=\pm\frac{\sqrt{13}}{2}.
4. So x=\frac{5-\sqrt{13}}{2} or x=\frac{5+\sqrt{13}}{2}.

Use the distance of each root from the centre of the completed square.

1. Complete the square on the left: 2(x-1)^2+7.
2. For either given solution, x-1=\pm\sqrt{3}, so (x-1)^2=3.
3. Therefore c=2\times3+7=13.

Take 2 out of the x terms; the number in the bracket is half of the new coefficient of x, in terms of p.

Part (a)

1. Take 2 out of the terms in x: 2x^2+8px+5=2(x^2+4px)+5.
2. Half of 4p is 2p, so x^2+4px=(x+2p)^2-4p^2.
3. Multiply the correction by 2: 2(x+2p)^2-8p^2+5.
4. So the expression is 2(x+2p)^2+5-8p^2, with a=2, b=2p and c=5-8p^2.

Part (b)

1. From part (a), 2(x+2p)^2 is never negative and is 0 when x=-2p, so the minimum value is 5-8p^2.
2. Set 5-8p^2=-13. Then 8p^2=18, so p^2=\dfrac94.
3. p is positive, so p=\dfrac32.