Drag the tiles tot he correct boxes to complete the pairs. Not all tiles will be used... Match each quadratic equation with its solution set.

QUESTION 1

The given quadratic equation is

[tex]2 {x}^{2} - 8x + 5 = 0[/tex]

Comparing this to the general quadratic equation,

[tex]a {x}^{2} + bx + c = 0[/tex]

[tex]a=2, b=-8,c=5[/tex]

The solution to the above quadratic equation can be found using the quadratic formula,

[tex]x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} [/tex]

We substitute the values into the formula to obtain,

[tex]x = \frac{ - - 8 \pm \sqrt{ {( - 8)}^{2} - 4(2)(5)} }{2(2)} [/tex]

This will simplify to,

[tex]x = \frac{ 8 \pm \sqrt{ 64 -40} }{4} [/tex]

[tex]x = \frac{ 8 \pm \sqrt{ 24} }{4} [/tex]

[tex]x = \frac{ 8 \pm 2\sqrt{ 6} }{4} [/tex]

[tex]x = \frac{ 4 \pm \sqrt{ 6} }{2} [/tex]

[tex] \boxed {2 {x}^{2} - 8x + 5 = 0 \rightarrow \frac{ 4 \pm \sqrt{ 6} }{2} }[/tex]

QUESTION 2

The given quadratic equation is

[tex]2 {x}^{2} - 10x - 3 = 0[/tex]

This implies that,

[tex]a=2,b=-10,c=-3[/tex]

We use the formula to obtain,

[tex]x = \frac{ - - 10\pm \sqrt{ {( - 10)}^{2} - 4(2)( - 3) } }{2(2)} [/tex]

[tex]x = \frac{ 10\pm \sqrt{ 100 + 24 } }{4} [/tex]

[tex]x = \frac{ 10\pm \sqrt{ 124 } }{4} [/tex]

[tex]x = \frac{ 10\pm 2\sqrt{ 31 } }{4} [/tex]

[tex]x = \frac{ 5\pm \sqrt{ 31 } }{2} [/tex]

[tex] \boxed {2 {x}^{2} - 10x - 3 = 0 \rightarrow \frac{ 5\pm \sqrt{ 31 } }{2} }[/tex]

QUESTION 3

The given quadratic equation is

[tex]2 {x}^{2} - 8x - 3 = 0[/tex]

[tex]a=2,b=-8,c=-3[/tex]

[tex]x = \frac{ - - 8\pm \sqrt{ {( - 8)}^{2} - 4(2)( - 3) } }{2(2)} [/tex]

[tex]x = \frac{ 8\pm \sqrt{ 64 + 24 } }{4} [/tex]

[tex]x = \frac{ 8\pm \sqrt{ 88} }{4} [/tex]

[tex]x = \frac{ 8\pm 2\sqrt{ 22} }{4} [/tex]

[tex]x = \frac{ 4\pm \sqrt{ 22} }{2} [/tex]

[tex] \boxed {2 {x}^{2} - 8x - 3 = 0 \rightarrow \frac{ 4\pm \sqrt{ 22} }{2} }[/tex]

QUESTION 4

The given quadratic expression is

[tex]2 {x}^{2} - 9x - 1 = 0[/tex]

[tex]a=2,b=-9,c=-1[/tex]

We use the formula to get,

[tex]x = \frac{ - - 9\pm \sqrt{ {( - 9)}^{2} - 4(2)( - 1) } }{2(2)} [/tex]

[tex]x = \frac{ 9\pm \sqrt{ 81 + 8} }{4} [/tex]

[tex]x = \frac{ 9\pm \sqrt{ 89} }{4} [/tex]

[tex] \boxed {2 {x}^{2} - 9x - 1 = 0 \rightarrow \frac{ 9\pm \sqrt{ 89} }{4} }[/tex]

QUESTION 5

The given quadratic expression is

[tex]2 {x}^{2} - 9x + 6 = 0[/tex]

[tex]a=2,b=-9,c=6[/tex]

[tex]x = \frac{ - - 9\pm \sqrt{ {( - 9)}^{2} - 4(2)( 6) } }{2(2)} [/tex]

[tex]x = \frac{ 9\pm \sqrt{ 81 - 48 } }{4} [/tex]

[tex]x = \frac{ 9\pm \sqrt{ 33 } }{4} [/tex]

[tex] \boxed {2 {x}^{2} - 9x + 6= 0 \rightarrow \frac{ 9\pm \sqrt{ 33 } }{4} }[/tex]

NOT ALL THE TILES ARE USED