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

1 Square root method / 3 pts

a) \(x^2=25\), so \(x=\pm 5\). (1 pt)

b) \(x+2=\pm\sqrt{20}=\pm 2\sqrt{5}\) (1 pt), so \(x=-2\pm 2\sqrt{5}\), about \(2.47\) or \(-6.47\). (1 pt)

2 Completing the square / 4 pts

a) \(x^2-12x=-20\), add \(36\): \((x-6)^2=16\) (1 pt), so \(x=10\) or \(x=2\). (1 pt)

b) Divide by \(3\): \(x^2+2x=\dfrac{1}{3}\); add \(1\): \((x+1)^2=\dfrac{4}{3}\) (1 pt); \(x=-1\pm\dfrac{2\sqrt{3}}{3}\), about \(0.15\) or \(-2.15\). (1 pt)

3 Quadratic formula and discriminant / 4 pts

a) \(\Delta=49-24=25\) (1 pt). \(x=\dfrac{7\pm 5}{4}\) (1 pt), so \(x=3\) or \(x=\dfrac{1}{2}\). (1 pt)

b) \(\Delta=64-68=-4<0\): no real solution. (1 pt)

4 A parameter / 3 pts

One solution means \(\Delta=0\): \(100-4k=0\) (1 pt), so \(k=25\) (1 pt).

Then \(x^2-10x+25=(x-5)^2=0\), so \(x=5\). (1 pt)

5 A poster / 3 pts

Let \(w\) be the width in inches: \(w(w+6)=187\), so \(w^2+6w-187=0\). (1 pt)

\(\Delta=36+748=784=28^2\), \(w=\dfrac{-6\pm 28}{2}\), so \(w=11\) or \(w=-17\). (1 pt)

The width is positive: the poster is \(11\) in by \(17\) in. Check: \(11\times 17=187\). (1 pt)

6 Line and parabola / 3 pts

\(x^2-1=x+1\), so \(x^2-x-2=0\) (1 pt), \((x-2)(x+1)=0\), so \(x=2\) or \(x=-1\). (1 pt)

Then \(y=3\) or \(y=0\): the solutions are \((2,3)\) and \((-1,0)\). (1 pt)