Solve the equation by using the Quadratic Formula. u 2 − 10 u + 18 = 0 u ^ { 2 } - 10 u + 18 = 0 u 2 − 10 u + 18 = 0 A) u = 10 ± u = 10 \pm u = 10 ± 7 \sqrt { 7 } 7 B) u = 10 ± u = 10 \pm u = 10 ± 43 \sqrt { 43 } 43 C) u = 5 ± u = 5 \pm u = 5 ± 7 \sqrt { 7 } 7 D) u = − 10 ± u = - 10 \pm u = − 10 ± 43 \sqrt { 43 } 43 E) u = − 5 ± u = - 5 \pm u = − 5 ± 7 \sqrt { 7 } 7

Question

Solve the equation by using the Quadratic Formula.   u 2 − 10 u + 18 = 0 u ^ { 2 } - 10 u + 18 = 0 u 2 − 10 u + 18 = 0 A)    u = 10 ± u = 10 \pm u = 10 ±      7 \sqrt { 7 } 7 ​ B)    u = 10 ± u = 10 \pm u = 10 ±     43 \sqrt { 43 } 43 ​ C)    u = 5 ± u = 5 \pm u = 5 ±     7 \sqrt { 7 } 7 ​ D)    u = − 10 ± u = - 10 \pm u = − 10 ±     43 \sqrt { 43 } 43 ​ E)    u = − 5 ± u = - 5 \pm u = − 5 ±      7 \sqrt { 7 } 7 ​

Brittany Taylor · Accepted Answer

Final Answer : C, Explanation :   The quadratic formula is   u = − b ± b 2 − 4 a c 2 a u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} u = 2 a − b ± b 2 − 4 a c ​ ​  , where   a = 1 a = 1 a = 1  ,   b = − 10 b = -10 b = − 10  , and   c = 18 c = 18 c = 18  . Plugging these values in gives   u = 10 ± ( − 10 ) 2 − 4 ( 1 ) ( 18 ) 2 ( 1 ) = 10 ± 100 − 72 2 = 10 ± 28 2 = 5 ± 7 u = \frac{10 \pm \sqrt{(-10)^2 - 4(1)(18)}}{2(1)} = \frac{10 \pm \sqrt{100 - 72}}{2} = \frac{10 \pm \sqrt{28}}{2} = 5 \pm \sqrt{7} u = 2 ( 1 ) 10 ± ( − 10 ) 2 − 4 ( 1 ) ( 18 ) ​ ​ = 2 10 ± 100 − 72 ​ ​ = 2 10 ± 28 ​ ​ = 5 ± 7 ​  .

Solve the equation by using the Quadratic Formula. $u ^ { 2 } - 10 u + 18 = 0$

A) $\pm$ $7\sqrt { 7 }$
B) $\pm$ $43\sqrt { 43 }$
C) $\pm$ $7\sqrt { 7 }$
D) $\pm$ $43\sqrt { 43 }$
E) $\pm$ $7\sqrt { 7 }$

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