Solving Quadratic Equations: Algebra 1 Math Worksheet Answer Key – PDF Download

Algebra 1 Solving Worksheet Answer Key: problems 1–5

Problem 1: Standard form

1. Subtract \(5x\) and add 2: \(3x^{2}-5x+2=0\), so \(a=3\), \(b=-5\), \(c=2\).
2. Distribute: \(x^{2}+4x=21\). Subtract 21: \(x^{2}+4x-21=0\), so \(a=1\), \(b=4\), \(c=-21\).

Problem 2: Perfect-square roots

1. \(x=\pm\sqrt{81}\). \(x=9\) or \(x=-9\).
2. Add 11: \(x^{2}=36\). \(x=6\) or \(x=-6\). Check: \(36-11=25\).

Problem 3: Square roots, not perfect

1. \(x=\pm\sqrt{20}=\pm 2\sqrt{5}\approx\pm 4.47\). \(x=\pm 2\sqrt{5}\).
2. Divide by 3: \(x^{2}=25\). \(x=\pm 5\).

Problem 4: Square both sides back

Take square roots: \(x-3=\pm 4\). Then \(x=3+4=7\) or \(x=3-4=-1\). \(x=7\) or \(x=-1\).

Check: \((7-3)^{2}=16\) and \((-1-3)^{2}=16\). Both points are 4 units from 3.

Problem 5: Square roots with a shift

\(x+2=\pm\sqrt{10}\), so \(x=-2\pm\sqrt{10}\).

With \(\sqrt{10}\approx 3.162\): \(x\approx 1.16\) or \(x\approx -5.16\).

Algebra 1 Solving Worksheet Answer Key: problems 6–10

Problem 6: Zero-product property

Set each factor to 0: \(x-4=0\) or \(x+7=0\). \(x=4\) or \(x=-7\).

Problem 7: Factor a trinomial

Find two numbers with product 20 and sum 9: 4 and 5. So \((x+4)(x+5)=0\).

\(x=-4\) or \(x=-5\). Check: \(16-36+20=0\).

Problem 8: Negative constant

Two numbers with product \(-28\) and sum \(-3\): \(-7\) and 4. So \((x-7)(x+4)=0\).

\(x=7\) or \(x=-4\). Check: \(49-21-28=0\).

Problem 9: Common factor

1. Factor out \(x\): \(x(x-6)=0\). \(x=0\) or \(x=6\).
2. \(x(x+8)=0\). \(x=0\) or \(x=-8\).

Do not divide by \(x\): that would lose the solution \(x=0\).

Problem 10: Rearrange first

Move every term to the left: \(x^{2}-2x-15=0\). Factor: \((x-5)(x+3)=0\).

\(x=5\) or \(x=-3\). Check \(x=-3\): \(9=-6+15\).

Algebra 1 Solving Worksheet Answer Key: problems 11–15

Problem 11: Leading coefficient 2

Split the middle term: \(2x^{2}+6x+x+3=2x(x+3)+1(x+3)=(2x+1)(x+3)\).

\(2x+1=0\) gives \(x=-\frac{1}{2}\), and \(x+3=0\) gives \(x=-3\). \(x=-\frac{1}{2}\) or \(x=-3\).

Problem 12: Solve from a graph

The solutions are the x-intercepts: \(x=-2\) and \(x=4\).

Factoring: \((x-4)(x+2)=0\), so \(x=4\) or \(x=-2\). The graph and the algebra agree. The lowest point is \((1, -9)\), halfway between the zeros.

Problem 13: Complete the square

Subtract 5: \(x^{2}+6x=-5\). Add \(3^{2}=9\): \(x^{2}+6x+9=4\), so \((x+3)^{2}=4\).

\(x+3=\pm 2\). \(x=-1\) or \(x=-5\).

Problem 14: Irrational solutions

\(x^{2}-8x=-3\). Add \((-4)^{2}=16\): \((x-4)^{2}=13\).

\(x=4\pm\sqrt{13}\), so \(x\approx 7.61\) or \(x\approx 0.39\).

Problem 15: Odd middle coefficient

\(x^{2}+5x=2\). Half of 5 is \(\frac{5}{2}\), and its square is \(\frac{25}{4}\). So \(x^{2}+5x+\frac{25}{4}=2+\frac{25}{4}=\frac{33}{4}\).

\((x+\frac{5}{2})^{2}=\frac{33}{4}\), so \(x=\frac{-5\pm\sqrt{33}}{2}\). \(x\approx 0.37\) or \(x\approx -5.37\).

Algebra 1 Solving Worksheet Answer Key: problems 16–20

Problem 16: Divide first

Divide by 2: \(x^{2}-6x+5=0\), so \(x^{2}-6x=-5\). Add 9: \((x-3)^{2}=4\).

\(x-3=\pm 2\). \(x=5\) or \(x=1\). Check: \(50-60+10=0\).

Problem 17: Use the formula

\(a=2\), \(b=-5\), \(c=-3\). \(b^{2}-4ac=25+24=49\).

\(x=\frac{5\pm 7}{4}\). \(x=3\) or \(x=-\frac{1}{2}\).

Problem 18: Formula with a radical

\(b^{2}-4ac=16+12=28\), and \(\sqrt{28}=2\sqrt{7}\).

\(x=\frac{4\pm 2\sqrt{7}}{2}=2\pm\sqrt{7}\). \(x\approx 4.65\) or \(x\approx -0.65\).

Problem 19: Formula, a not 1

\(a=3\), \(b=-4\), \(c=-2\). \(b^{2}-4ac=16+24=40\).

\(x=\frac{4\pm\sqrt{40}}{6}=\frac{2\pm\sqrt{10}}{3}\). \(x\approx 1.72\) or \(x\approx -0.39\).

Problem 20: Discriminant and graph

The parabola never touches the x-axis, so there are no real solutions.

Discriminant: \(b^{2}-4ac=(-2)^{2}-4(1)(3)=4-12=-8 < 0\). A negative discriminant means no real solutions.

Algebra 1 Solving Worksheet Answer Key: problems 21–25

Problem 21: Find k

One solution means \(b^{2}-4ac=0\): \(36-4k=0\), so \(k=9\).

Then \(x^{2}+6x+9=(x+3)^{2}=0\) gives \(x=-3\).

Problem 22: Rectangle area

\(w(w+3)=70\), so \(w^{2}+3w-70=0\). Factor: \((w+10)(w-7)=0\).

\(w=-10\) is rejected because a width cannot be negative. So \(w=7\) and the length is \(7+3=10\).

7 ft by 10 ft. Check: \(7\times 10=70\).

Problem 23: Stone from a cliff

1. Set \(h=0\): \(-16(t^{2}-2t-3)=0\), so \(-16(t-3)(t+1)=0\). \(t=-1\) is rejected because time cannot be negative. \(t=3\) seconds.
2. Set \(h=48\): \(-16t^{2}+32t=0\), so \(-16t(t-2)=0\). \(t=0\) and \(t=2\) seconds: the launch and the return to ledge height.

Problem 24: Consecutive integers

Let the integers be \(n\) and \(n+1\). \(n(n+1)=156\), so \(n^{2}+n-156=0\). Factor: \((n+13)(n-12)=0\).

\(n=12\) gives 12 and 13. \(n=-13\) gives \(-13\) and \(-12\), whose product is also 156, so \(-13\) and \(-12\) is the other pair, but it is not positive.

Problem 25: Garden border

The outer rectangle is \((12+2x)\) by \((8+2x)\). \((12+2x)(8+2x)=192\), so \(4x^{2}+40x+96=192\) and \(4x^{2}+40x-96=0\). Divide by 4: \(x^{2}+10x-24=0\), so \((x+12)(x-2)=0\).

\(x=-12\) is rejected. \(x=2\) feet. Check: \(16\times 12=192\).

Algebra 1 Solving Worksheet Answer Key: problems 26–27

Problem 26: School play tickets

1. \(200x-10x^{2}=1000\), so \(x^{2}-20x+100=0\), or \((x-10)^{2}=0\). \(x=10\): a price of \(\$10\), with 100 tickets sold. This is the only price, because it is the maximum.
2. \(200x-10x^{2}=1200\) gives \(x^{2}-20x+120=0\). \(D=400-480=-80 < 0\). No: \(\$1{,}200\) is never reached.

Problem 27: Find the error

The zero-product property only applies when the product is 0. Here the product is 14, so the factors do not have to equal 14.

Correct work: \(x^{2}-5x-14=0\), so \((x-7)(x+2)=0\). \(x=7\) or \(x=-2\). Check \(x=7\): \(49-35=14\).

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