The Quadratic Formula

Identify the three coefficients with their signs before substituting.

1. For ax^2+bx+c=0, the quadratic formula is x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.
2. Here a=2, b=3 and c=-4, so b^2-4ac=3^2-4\times2\times(-4)=9+32=41.
3. Substituting gives x=\dfrac{-3\pm\sqrt{41}}{4}, since 2a=4.
4. The other options each make a common slip: \dfrac{3\pm\ldots}{4} forgets the minus in -b, a denominator of 2 uses 2 instead of 2a, and \sqrt{-23} comes from treating c as +4.

Use both signs in the quadratic formula.

1. For ax^2+bx+c=0, the quadratic formula is x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.
2. Here a=1, b=-3 and c=-1, so b^2-4ac=(-3)^2-4\times1\times(-1)=9+4=13.
3. Substituting gives x=\dfrac{3\pm\sqrt{13}}{2}.
4. So x=\dfrac{3+\sqrt{13}}{2}=3.3027\ldots or x=\dfrac{3-\sqrt{13}}{2}=-0.3027\ldots
5. To 2 decimal places, x=3.30 or x=-0.30.

Keep the negative constant when calculating the discriminant.

1. For ax^2+bx+c=0, the quadratic formula is x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.
2. Here a=2, b=5 and c=-4, so b^2-4ac=25-4\times2\times(-4)=25+32=57.
3. Substituting gives x=\dfrac{-5\pm\sqrt{57}}{4}.
4. So x=\dfrac{-5+\sqrt{57}}{4}=0.6374\ldots or x=\dfrac{-5-\sqrt{57}}{4}=-3.1374\ldots
5. To 2 decimal places, x=0.64 or x=-3.14.

Simplify the square root in the formula before cancelling common factors.

1. For ax^2+bx+c=0, the quadratic formula is x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.
2. Here a=3, b=-2 and c=-2, so b^2-4ac=4+24=28 and x=\dfrac{2\pm\sqrt{28}}{6}.
3. Simplify the surd: \sqrt{28}=\sqrt{4\times7}=2\sqrt7, so x=\dfrac{2\pm2\sqrt7}{6}.
4. Divide the top and bottom by 2: x=\dfrac{1+\sqrt7}{3} or x=\dfrac{1-\sqrt7}{3}.

Apply the requested significant figures to each root separately.

1. For ax^2+bx+c=0, the quadratic formula is x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.
2. Here a=4, b=2 and c=-1, so b^2-4ac=4+16=20 and x=\dfrac{-2\pm\sqrt{20}}{8}.
3. So x=\dfrac{-2+\sqrt{20}}{8}=0.30901\ldots or x=\dfrac{-2-\sqrt{20}}{8}=-0.80901\ldots
4. To 3 significant figures (counting from the first non-zero digit), x=0.309 or x=-0.809.

Rearrange the equation so that one side is zero before identifying the coefficients.

1. The formula needs the equation in the form ax^2+bx+c=0, so subtract 7: x^2+4x-7=0.
2. For ax^2+bx+c=0, the quadratic formula is x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.
3. Here a=1, b=4 and c=-7, so b^2-4ac=16+28=44 and x=\dfrac{-4\pm\sqrt{44}}{2}.
4. So x=1.3166\ldots or x=-5.3166\ldots
5. To 2 decimal places, x=1.32 or x=-5.32.

Multiplying the whole equation by minus one can make the leading coefficient positive.

1. Write the terms in order and multiply every term by -1: 3x^2-5x-2=0. This has the same solutions.
2. Use x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} with a=3, b=-5 and c=-2: b^2-4ac=25+24=49.
3. So x=\dfrac{5\pm\sqrt{49}}{6}, which is \dfrac{5\pm7}{6}.
4. x=\dfrac{5+7}{6}=2 or x=\dfrac{5-7}{6}=-\dfrac13.
5. Because 49 is a square number the solutions are not surds: the quadratic also factorises as (3x+1)(x-2)=0.

Clear the fractional coefficient first.

1. Multiply every term by 2 to clear the fraction: x^2-6x+2=0. This has the same solutions.
2. Use x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} with a=1, b=-6 and c=2: b^2-4ac=36-8=28.
3. So x=\dfrac{6\pm\sqrt{28}}{2}. Since \sqrt{28}=2\sqrt7, this is x=\dfrac{6\pm2\sqrt7}{2}.
4. Divide each term on top by 2: x=3+\sqrt7 or x=3-\sqrt7.

Calculate the discriminant before deciding how many distinct roots there are.

1. Use x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} with a=9, b=-12 and c=4.
2. b^2-4ac=(-12)^2-4\times9\times4=144-144=0.
3. Since \sqrt0=0, both signs give the same value: x=\dfrac{12}{18}=\dfrac23.
4. There is just one solution because 9x^2-12x+4=(3x-2)^2 is a perfect square.

Look at the sign of the discriminant.

1. The number of real solutions depends on b^2-4ac, the part under the square root in the quadratic formula.
2. Here a=2, b=3 and c=5, so b^2-4ac=3^2-4\times2\times5=9-40=-31.
3. A negative number has no real square root, so there are no real solutions.
4. The other options all need b^2-4ac to be 0 or more.

Set the height equal to zero and choose the root allowed by the time domain.

1. The rocket is on the ground when h=0: 8+12t-5t^2=0, which is 5t^2-12t-8=0.
2. Use t=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} with a=5, b=-12 and c=-8: b^2-4ac=144+160=304.
3. So t=\dfrac{12\pm\sqrt{304}}{10}, giving t=2.9435\ldots or t=-0.5435\ldots
4. Time after launch can't be negative, so the rocket reaches the ground after 2.94 seconds.

Represent the two numbers using one unknown, then use their product.

1. Let the smaller number be x. The other number is x+4.
2. Their product is 22: x(x+4)=22, so x^2+4x-22=0.
3. Use x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} with a=1, b=4 and c=-22: b^2-4ac=16+88=104.
4. So x=\dfrac{-4\pm\sqrt{104}}{2}, giving x=3.0990\ldots or x=-7.0990\ldots
5. The numbers are positive, so reject the negative value. The smaller number is 3.10 to 2 decimal places.

Calculate both roots, then use the stated interval to select one.

1. Use x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} with a=1, b=-5 and c=2: b^2-4ac=25-8=17.
2. So x=\dfrac{5\pm\sqrt{17}}{2}, giving x=4.5615\ldots or x=0.43844\ldots
3. Only 0.43844\ldots lies between 0 and 1. To 3 decimal places it is 0.438.

Check the sign of the linear coefficient and the minus sign in front of it in the formula.

1. In 3x^2-5x-4=0, a=3, b=-5 and c=-4.
2. The formula starts with -b, and -b=-(-5)=5, so the top must start with 5, not -5.
3. The rest is right: b^2-4ac=25+48=73 and 2a=6. The \pm must stay, because it gives the two solutions.
4. The corrected expression is x=\dfrac{5\pm\sqrt{73}}{6}.

Find both roots in exact form so that their difference can be simplified.

1. Use x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} with a=1, b=6 and c=-8: b^2-4ac=36+32=68.
2. So x=\dfrac{-6\pm\sqrt{68}}{2}. Since \sqrt{68}=2\sqrt{17}, this is x=-3\pm\sqrt{17}.
3. The larger root is -3+\sqrt{17} and the smaller is -3-\sqrt{17}.
4. Larger minus smaller: (-3+\sqrt{17})-(-3-\sqrt{17})=2\sqrt{17}.

Expand and collect all terms on one side before using the quadratic formula.

1. Expand the brackets: (2x+1)(x-3)=2x^2-6x+x-3=2x^2-5x-3.
2. So 2x^2-5x-3=5. Subtract 5 to make one side zero: 2x^2-5x-8=0.
3. Use x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} with a=2, b=-5 and c=-8: b^2-4ac=25+64=89.
4. So x=\dfrac{5+\sqrt{89}}{4} or x=\dfrac{5-\sqrt{89}}{4}.

Clear the denominators before collecting the quadratic terms.

1. Multiply both sides by 6, the lowest common multiple of 2 and 3: 3(x^2+3x)=2(2x+5).
2. Expand: 3x^2+9x=4x+10. Rearrange to make one side zero: 3x^2+5x-10=0.
3. Use x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} with a=3, b=5 and c=-10: b^2-4ac=25+120=145.
4. So x=\dfrac{-5\pm\sqrt{145}}{6}, giving x=1.1735\ldots or x=-2.8402\ldots
5. To 3 significant figures, x=1.17 or x=-2.84.

Set the model equal to the required income and retain every root inside its domain.

1. Set the income equal to 200: p(60-2p)=200, so 60p-2p^2=200.
2. Rearrange to 2p^2-60p+200=0 and divide every term by 2: p^2-30p+100=0.
3. Use p=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} with a=1, b=-30 and c=100: b^2-4ac=900-400=500.
4. So p=\dfrac{30\pm\sqrt{500}}{2}, giving p=26.1803\ldots or p=3.8196\ldots, both between 0 and 30.
5. To the nearest penny, the prices are £3.82 and £26.18.

Multiply every term by x(x+1) to clear both fractions.

1. Multiply every term by x(x+1): 2(x+1)+3x=x(x+1).
2. Expand: 2x+2+3x=x^2+x, so 5x+2=x^2+x.
3. Collect everything on one side: x^2-4x-2=0.
4. Use x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} with a=1, b=-4 and c=-2: b^2-4ac=16+8=24.
5. So x=\dfrac{4\pm\sqrt{24}}{2}. Since \sqrt{24}=2\sqrt6, this is x=\dfrac{4\pm2\sqrt6}{2}=2\pm\sqrt6.
6. Neither value is 0 or -1 (which would make a denominator zero), so x=2+\sqrt6 or x=2-\sqrt6.

Express each outside dimension in terms of the lawn dimension and the path width.

1. The path adds x m at each end of each side, so the outside measures 10+2x m by 6+2x m.
2. The total area gives (10+2x)(6+2x)=160. Expand: 60+20x+12x+4x^2=160.
3. Rearrange: 4x^2+32x-100=0, and divide every term by 4: x^2+8x-25=0.
4. Use x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} with a=1, b=8 and c=-25: b^2-4ac=64+100=164.
5. So x=\dfrac{-8\pm\sqrt{164}}{2}, giving x=2.4031\ldots or x=-10.403\ldots
6. A width must be positive, so the path is 2.40 m wide to 2 decimal places.