Completing the Square

Halve the coefficient of the linear term, then subtract the square of that half.

1. Half of 8 is 4, so start with (x+4)^2.
2. This square expands to x^2+8x+16, which has 13 too much.
3. Therefore x^2+8x+3=(x+4)^2-13.

Halve the coefficient of the linear term, then subtract the square of that half.

1. Half of -10 is -5.
2. The square (x-5)^2 expands to x^2-10x+25.
3. Subtract the extra 25: (x-5)^2-25.

Expand the squared brackets and check both the linear term and the constant.

1. The 6x term comes from twice the number 3 in the square.
2. (x+3)^2=x^2+6x+9.
3. Subtract 5 to leave the required constant 4, giving (x+3)^2-5.
4. (x+3)^2+4 forgets to take away the 9 that the square adds, and (x-3)^2-5 has the wrong sign inside the bracket: (x-3)^2 gives -6x.

Halve the coefficient of the linear term, then subtract the square of that half.

1. Half of -4 is -2, so use (x-2)^2.
2. This adds a constant 4; the required constant is 11.
3. Add the remaining 7: (x-2)^2+7.

The constant in a perfect square is the square of half the linear coefficient.

1. In (x+a)^2, the coefficient of x is 2a.
2. So 2a=12 and a=6.
3. The constant must be a^2, so k=36.

Halve the coefficient of the linear term, then subtract the square of that half.

1. Half of 5 is \frac{5}{2}, so use (x+\frac{5}{2})^2=x^2+5x+\frac{25}{4}.
2. This is \frac{25}{4} more than x^2+5x, so x^2+5x+1=(x+\frac{5}{2})^2-\frac{25}{4}+1.
3. Combine the constants: -\frac{25}{4}+\frac{4}{4}=-\frac{21}{4}, so the answer is (x+\frac{5}{2})^2-\frac{21}{4}.

Halve the coefficient of the linear term, then subtract the square of that half.

1. Half of -3 is -\frac{3}{2}, whose square is \frac{9}{4}.
2. So u^2-3u=(u-\frac{3}{2})^2-\frac{9}{4}.
3. Adding 8 gives (u-\frac{3}{2})^2+\frac{23}{4}.

Work out the constant term produced by the squared bracket.

1. Expanding (x+5)^2 introduces the constant 25, not 5.
2. To keep the expression unchanged, subtract that whole constant.
3. The correct step is (x+5)^2-25+9, so replace -5 with -25.

Compare the linear term first, then the constants.

Part (a)

1. Expanding the square gives a linear term of -2at.
2. Compare -2a=-14, giving a=7.

Part (b)

1. The square contributes the constant 7^2=49.
2. The constant on the left is 20, so 49-b=20.
3. Therefore b=29.

Halve the coefficient of the linear term, then subtract the square of that half.

1. Put the terms in order: x^2+2x+9.
2. The square (x+1)^2 contains x^2+2x+1.
3. Add the remaining 8: (x+1)^2+8.

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

1. Complete the square: x^2+4x-3=(x+2)^2-7.
2. The equation is therefore (x+2)^2=7.
3. Taking both square roots gives x+2=\pm\sqrt{7}.
4. So x=-2-\sqrt{7} or x=-2+\sqrt{7}.

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

1. The left side is (x-3)^2-7.
2. So (x-3)^2-7=7, giving (x-3)^2=14.
3. Take both square roots and add 3.
4. The solutions are x=3-\sqrt{14} and x=3+\sqrt{14}.

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

1. Move the linear term to the left: x^2+2x=5.
2. Add 1 to both sides to make a square: (x+1)^2=6.
3. Take both square roots and subtract 1.
4. Thus x=-1-\sqrt{6} or x=-1+\sqrt{6}.

Complete the square, then use the condition that the number is positive.

1. Complete the square by adding 9 to both sides: (x+3)^2=29.
2. Hence x=-3\pm\sqrt{29}.
3. The root -3-\sqrt{29} is negative, so reject it.
4. The required positive number is -3+\sqrt{29}.

Let the width be w cm and write the length in terms of w.

1. Let the width be w cm. The length is w+8 cm, so w(w+8)=50, which is w^2+8w=50.
2. Half of 8 is 4, so add 4^2=16 to both sides: w^2+8w+16=66.
3. The left side is a perfect square: (w+4)^2=66.
4. Take both square roots: w+4=\pm\sqrt{66}, so w=-4\pm\sqrt{66}.
5. A width must be positive, so the width is -4+\sqrt{66} cm (about 4.12 cm).

Expand the product before looking for a completed square.

1. Expand the product: x^2+6x-7=10.
2. Move the constant to give x^2+6x=17.
3. Add 9 to both sides: (x+3)^2=26.
4. Taking both square roots gives x=-3-\sqrt{26} or x=-3+\sqrt{26}.

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

1. Move the constant: x^2-3x=1.
2. Half of -3 is -\frac{3}{2}, so add \frac{9}{4} to both sides.
3. This gives (x-\frac{3}{2})^2=\frac{13}{4}.
4. Take both square roots: x-\frac{3}{2}=\pm\frac{\sqrt{13}}{2}.
5. Therefore x=\frac{3-\sqrt{13}}{2} or x=\frac{3+\sqrt{13}}{2}.

Complete the square and use the fact that a square is never negative.

1. Complete the square: x^2+6x+13=(x+3)^2+4.
2. The square of a real number cannot be negative, so this expression is at least 4.
3. It can never equal zero, which proves that the equation has no real solutions.
4. The statement that says "at most 4" has the inequality the wrong way round: because the square is at least 0, the expression is at least 4.

For exactly one solution, the left side must be a perfect square (x+a)^2.

Part (a)

1. Exactly one solution means the left side is a perfect square, (x+a)^2=x^2+2ax+a^2, which is zero only when x=-a.
2. Compare constants: a^2=49, so a=7 or a=-7.
3. Compare the x terms: k=2a. As k is negative, a=-7 and k=-14.

Part (b)

1. With k=-14, the equation is x^2-14x+49=0, which is (x-7)^2=0.
2. So x-7=0 and x=7.

Express both numbers using one unknown before forming the equation.

1. Let the smaller number be x, so the larger is x+2.
2. The sum of squares gives x^2+(x+2)^2=22.
3. Expand and simplify: 2x^2+4x+4=22, so x^2+2x=9.
4. Complete the square: (x+1)^2=10.
5. The two algebraic roots are -1\pm\sqrt{10}.
6. Since the number is positive, the smaller number is -1+\sqrt{10}.