Question 151472
{{{16-24x+9x^2=0}}} Start with the given equation.



{{{9x^2-24x+16=0}}} Rearrange the terms.



From {{{9x^2-24x+16}}} we can see that {{{a=9}}}, {{{b=-24}}}, and {{{c=16}}}



{{{D=b^2-4ac}}} Start with the discriminant formula.



{{{D=(-24)^2-4(9)(16)}}} Plug in {{{a=9}}}, {{{b=-24}}}, and {{{c=16}}}



{{{D=576-4(9)(16)}}} Square {{{-24}}} to get {{{576}}}



{{{D=576-576}}} Multiply {{{4(9)(16)}}} to get {{{(36)(16)=576}}}



{{{D=0}}} Subtract {{{576}}} from {{{576}}} to get {{{0}}}



Since the discriminant is equal to zero, this means that there is one real solution.



So you are correct. However, I'm not sure where you got a=-9 b=24 c=-16 from.



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To find the one solution, you have 2 options


Option # 1 Quadratic Formula (preferred method)


{{{9x^2-24x+16=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=9}}}, {{{b=-24}}}, and {{{c=16}}}



Let's use the quadratic formula to solve for x



{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{x = (-(-24) +- sqrt( (-24)^2-4(9)(16) ))/(2(9))}}} Plug in  {{{a=9}}}, {{{b=-24}}}, and {{{c=16}}}



{{{x = (24 +- sqrt( (-24)^2-4(9)(16) ))/(2(9))}}} Negate {{{-24}}} to get {{{24}}}. 



{{{x = (24 +- sqrt( 576-4(9)(16) ))/(2(9))}}} Square {{{-24}}} to get {{{576}}}. 



{{{x = (24 +- sqrt( 576-576 ))/(2(9))}}} Multiply {{{4(9)(16)}}} to get {{{576}}}



{{{x = (24 +- sqrt( 0 ))/(2(9))}}} Subtract {{{576}}} from {{{576}}} to get {{{0}}}



{{{x = (24 +- sqrt( 0 ))/(18)}}} Multiply {{{2}}} and {{{9}}} to get {{{18}}}. 



{{{x = (24 +- 0)/(18)}}} Take the square root of {{{0}}} to get {{{0}}}. 



{{{x = (24 + 0)/(18)}}} or {{{x = (24 - 0)/(18)}}} Break up the expression. 



{{{x = (24)/(18)}}} or {{{x =  (24)/(18)}}} Combine like terms. 



{{{x = 4/3}}} or {{{x = 4/3}}} Simplify. 



So our answer is {{{x = 4/3}}} (with a multiplicity of 2)

  
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Option # 2 Factoring:


{{{9x^2-24x+16=0}}} Start with the given equation


{{{(3x-4)(3x-4)=0}}} Factor the left side (note: if you need help with factoring, check out this <a href=http://www.algebra.com/algebra/homework/playground/change-this-name4450.solver>solver</a>)




Now set each factor equal to zero:

{{{3x-4=0}}} or  {{{3x-4=0}}} 


{{{x=4/3}}} or  {{{x=4/3}}}    Now solve for x in each case



Since we have a repeating answer, our only answer is {{{x=4/3}}}