The quadratic formula is given by:
x = − b ± b 2 − 4 a c 2 a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} x = 2 a − b ± b 2 − 4 a c
a) 2 x 2 − 7 x + 1 = 0 2x^2 - 7x + 1 = 0 2 x 2 − 7 x + 1 = 0 Here, a = 2 a = 2 a = 2 , b = − 7 b = -7 b = − 7 , and c = 1 c = 1 c = 1 . Substituting these values into the quadratic formula:
x = − ( − 7 ) ± ( − 7 ) 2 − 4 ( 2 ) ( 1 ) 2 ( 2 ) x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(2)(1)}}{2(2)} x = 2 ( 2 ) − ( − 7 ) ± ( − 7 ) 2 − 4 ( 2 ) ( 1 ) x = 7 ± 49 − 8 4 x = \frac{7 \pm \sqrt{49 - 8}}{4} x = 4 7 ± 49 − 8 x = 7 ± 41 4 x = \frac{7 \pm \sqrt{41}}{4} x = 4 7 ± 41 The solution set is { 7 + 41 4 , 7 − 41 4 } \{\frac{7 + \sqrt{41}}{4}, \frac{7 - \sqrt{41}}{4}\} { 4 7 + 41 , 4 7 − 41 } .
b) 3 x 2 = − 9 x + 2 3x^2 = -9x + 2 3 x 2 = − 9 x + 2 . First, rearrange the equation to standard form: 3 x 2 + 9 x − 2 = 0 3x^2 + 9x - 2 = 0 3 x 2 + 9 x − 2 = 0 Here, a = 3 a = 3 a = 3 , b = 9 b = 9 b = 9 , and c = − 2 c = -2 c = − 2 . Substituting these values into the quadratic formula:
x = − 9 ± 9 2 − 4 ( 3 ) ( − 2 ) 2 ( 3 ) x = \frac{-9 \pm \sqrt{9^2 - 4(3)(-2)}}{2(3)} x = 2 ( 3 ) − 9 ± 9 2 − 4 ( 3 ) ( − 2 ) x = − 9 ± 81 + 24 6 x = \frac{-9 \pm \sqrt{81 + 24}}{6} x = 6 − 9 ± 81 + 24 x = − 9 ± 105 6 x = \frac{-9 \pm \sqrt{105}}{6} x = 6 − 9 ± 105 The solution set is { − 9 + 105 6 , − 9 − 105 6 } \{\frac{-9 + \sqrt{105}}{6}, \frac{-9 - \sqrt{105}}{6}\} { 6 − 9 + 105 , 6 − 9 − 105 } .
c) x 2 − 4 x = 3 x^2 - 4x = 3 x 2 − 4 x = 3 . First, rearrange the equation to standard form: x 2 − 4 x − 3 = 0 x^2 - 4x - 3 = 0 x 2 − 4 x − 3 = 0 Here, a = 1 a = 1 a = 1 , b = − 4 b = -4 b = − 4 , and c = − 3 c = -3 c = − 3 . Substituting these values into the quadratic formula:
x = − ( − 4 ) ± ( − 4 ) 2 − 4 ( 1 ) ( − 3 ) 2 ( 1 ) x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-3)}}{2(1)} x = 2 ( 1 ) − ( − 4 ) ± ( − 4 ) 2 − 4 ( 1 ) ( − 3 ) x = 4 ± 16 + 12 2 x = \frac{4 \pm \sqrt{16 + 12}}{2} x = 2 4 ± 16 + 12 x = 4 ± 28 2 x = \frac{4 \pm \sqrt{28}}{2} x = 2 4 ± 28 x = 4 ± 2 7 2 x = \frac{4 \pm 2\sqrt{7}}{2} x = 2 4 ± 2 7 x = 2 ± 7 x = 2 \pm \sqrt{7} x = 2 ± 7 The solution set is { 2 + 7 , 2 − 7 } \{2 + \sqrt{7}, 2 - \sqrt{7}\} { 2 + 7 , 2 − 7 } .
d) x 2 = x + 5 x^2 = x + 5 x 2 = x + 5 . First, rearrange the equation to standard form: x 2 − x − 5 = 0 x^2 - x - 5 = 0 x 2 − x − 5 = 0 Here, a = 1 a = 1 a = 1 , b = − 1 b = -1 b = − 1 , and c = − 5 c = -5 c = − 5 . Substituting these values into the quadratic formula:
x = − ( − 1 ) ± ( − 1 ) 2 − 4 ( 1 ) ( − 5 ) 2 ( 1 ) x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-5)}}{2(1)} x = 2 ( 1 ) − ( − 1 ) ± ( − 1 ) 2 − 4 ( 1 ) ( − 5 ) x = 1 ± 1 + 20 2 x = \frac{1 \pm \sqrt{1 + 20}}{2} x = 2 1 ± 1 + 20 x = 1 ± 21 2 x = \frac{1 \pm \sqrt{21}}{2} x = 2 1 ± 21 The solution set is { 1 + 21 2 , 1 − 21 2 } \{\frac{1 + \sqrt{21}}{2}, \frac{1 - \sqrt{21}}{2}\} { 2 1 + 21 , 2 1 − 21 } . 