Solving Quadratic Equations: practice solutions, Grade 9 – download the PDF

1 Square root method, first steps ★★★

a) \(x=\pm\sqrt{49}=\pm 7\). The solutions are \(7\) and \(-7\).

b) Divide by \(5\): \(x^2=9\), so \(x=\pm 3\). Check: \(5\cdot 9=45\). The solutions are \(3\) and \(-3\).

2 A square plus a shift ★★★

Take square roots: \(x+4=\pm 5\).

If \(x+4=5\) then \(x=1\); if \(x+4=-5\) then \(x=-9\). Check \(x=-9\): \((-5)^2=25\). The solutions are \(1\) and \(-9\).

3 Answer as a radical ★★★

\(x^2=20\), so \(x=\pm\sqrt{20}=\pm 2\sqrt{5}\).

Since \(\sqrt{5}\approx 2.236\), \(x\approx 4.47\) or \(x\approx -4.47\).

4 True or false? ★★★

She is wrong. Both \(3^2=9\) and \((-3)^2=9\) are positive, so no real number squared equals \(-9\).

The equation \(x^2=-9\) has no real solution.

5 Standard form ★★★

a) Subtract \(2x+7\): \(3x^2-2x-7=0\), so \(a=3\), \(b=-2\), \(c=-7\).

b) Expand: \(x^2+2x-3=5\). Subtract \(5\): \(x^2+2x-8=0\), so \(a=1\), \(b=2\), \(c=-8\).

6 Counting solutions ★★★

a) \(\Delta=25-8=17>0\): two solutions.

b) \(\Delta=144-4\cdot 4\cdot 9=144-144=0\): one solution.

c) \(\Delta=1-4\cdot 2\cdot 3=-23<0\): no real solution.

7 Zero-product warm-up ★★★

Either \(x-6=0\) or \(x+2=0\), so \(x=6\) or \(x=-2\).

Expanding: \(x^2+2x-6x-12=x^2-4x-12\). Check \(x=6\): \(36-24-12=0\).

8 Complete the square ★★★

\(x^2+8x=-7\). Half of \(8\) is \(4\), and \(4^2=16\): \(x^2+8x+16=9\), so \((x+4)^2=9\).

Then \(x+4=\pm 3\), so \(x=-1\) or \(x=-7\). Check \(x=-1\): \(1-8+7=0\).

9 Completing the square with a radical ★★★

\(x^2-10x=-18\). Add \(25\): \((x-5)^2=7\).

So \(x=5\pm\sqrt{7}\), which gives \(x\approx 7.65\) or \(x\approx 2.35\).

10 Quadratic formula with a leading coefficient ★★★

\(a=3\), \(b=1\), \(c=-10\). \(\Delta=1+120=121=11^2\).

\(x=\dfrac{-1\pm 11}{6}\), so \(x=\dfrac{10}{6}=\dfrac{5}{3}\) or \(x=\dfrac{-12}{6}=-2\).

11 Irrational solutions ★★★

\(\Delta=36-28=8\), so \(x=\dfrac{6\pm 2\sqrt{2}}{2}=3\pm\sqrt{2}\).

Numerically, \(x\approx 4.41\) or \(x\approx 1.59\).

12 Which method? ★★★

a) Square root method: \(x^2=9\), so \(x=\pm 3\).

b) Factoring: \((x+3)(x+4)=0\), so \(x=-3\) or \(x=-4\).

c) The quadratic formula: \(\Delta=9+20=29\), so \(x=\dfrac{3\pm\sqrt{29}}{2}\), that is \(x\approx 4.19\) or \(x\approx -1.19\).

13 A rectangular patio ★★★

Let \(w\) be the width in feet. Then \(w(w+5)=84\), so \(w^2+5w-84=0\).

\(\Delta=25+336=361=19^2\), so \(w=\dfrac{-5\pm 19}{2}\), that is \(w=7\) or \(w=-12\).

A width cannot be negative, so the patio is 7 ft wide and 12 ft long. Check: \(7\times 12=84\).

14 Find the error ★★★

The zero-product property only works when the product equals 0, not \(10\).

Correct work: \(x^2+3x-10=0\), so \((x+5)(x-2)=0\) and \(x=-5\) or \(x=2\). Check \(x=2\): \(4+6=10\).

15 Reading a table of values ★★★

a) The solutions are the values of \(x\) where \(y=0\): \(x=1\) and \(x=3\). Check: \(9-12+3=0\).

b) The lowest value in the table is \(-1\), reached at \(x=2\), halfway between the two zeros. The lowest point is \((2,-1)\).

16 One solution exactly ★★★

We need \(\Delta=k^2-4\cdot 1\cdot 16=k^2-64=0\), so \(k=8\) or \(k=-8\).

For \(k=8\): \(x^2+8x+16=(x+4)^2=0\), so \(x=-4\).

17 Completing the square when a is not 1 ★★★

Divide by \(2\): \(x^2-6x=-\dfrac{5}{2}\). Add \(9\): \((x-3)^2=\dfrac{13}{2}\).

So \(x=3\pm\sqrt{\dfrac{13}{2}}=3\pm\dfrac{\sqrt{26}}{2}\), about \(5.55\) or \(0.45\).

Formula check: \(\Delta=144-40=104\), \(x=\dfrac{12\pm\sqrt{104}}{4}=\dfrac{6\pm\sqrt{26}}{2}\), the same values.

18 The falling tool ★★★

It hits the ground when \(h=0\): \(144-16t^2=0\), so \(t^2=9\) and \(t=3\) (the value \(-3\) is rejected because time cannot be negative).

After 1 second: \(h=144-16=128\) feet. The tool lands after 3 seconds.

19 A ball thrown upward ★★★

a) Solve \(-16t^2+32t+48=0\). Divide by \(-16\): \(t^2-2t-3=0\), so \((t-3)(t+1)=0\). Since \(t\ge 0\), the ball lands at \(t=3\) seconds.

b) The ledge is at \(48\) ft: \(-16t^2+32t=0\), so \(-16t(t-2)=0\) and \(t=0\) or \(t=2\). The ball is back at ledge height after 2 seconds.

20 A line cuts a parabola ★★★

Set equal: \(x^2-3x=2x-6\), so \(x^2-5x+6=0\), \((x-2)(x-3)=0\), and \(x=2\) or \(x=3\).

Using \(y=2x-6\): \(y=-2\) and \(y=0\). The solutions are \((2,-2)\) and \((3,0)\). Check \((2,-2)\): \(4-6=-2\).

21 Tangent line or no intersection ★★★

a) \(x^2+2x+1=4x\), so \(x^2-2x+1=0\), \((x-1)^2=0\), \(x=1\). One point: \((1,4)\); the line is tangent to the parabola.

b) \(x^2+1=2x-3\), so \(x^2-2x+4=0\). \(\Delta=4-16=-12<0\): no solution, so the graphs never meet.

22 Reading a graph ★★★

a) The x-intercepts are \((-4,0)\) and \((2,0)\), so \(x=-4\) or \(x=2\). Check: \(16-8-8=0\).

b) \(x^2+2x=0\), so \(x(x+2)=0\) and \(x=0\) or \(x=-2\). These are the x-coordinates where the parabola meets the horizontal line \(y=-8\).