Question 155165
a)



From {{{x^2+6x-7}}} we can see that {{{a=1}}}, {{{b=6}}}, and {{{c=-7}}}



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



{{{D=(6)^2-4(1)(-7)}}} Plug in {{{a=1}}}, {{{b=6}}}, and {{{c=-7}}}



{{{D=36-4(1)(-7)}}} Square {{{6}}} to get {{{36}}}



{{{D=36--28}}} Multiply {{{4(1)(-7)}}} to get {{{(4)(-7)=-28}}}



{{{D=36+28}}} Rewrite {{{D=36--28}}} as {{{D=36+28}}}



{{{D=64}}} Add {{{36}}} to {{{28}}} to get {{{64}}}



Since the discriminant is greater than zero, this means that there are two real solutions.


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b)



From {{{z^2+z+1}}} we can see that {{{a=1}}}, {{{b=1}}}, and {{{c=1}}}



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



{{{D=(1)^2-4(1)(1)}}} Plug in {{{a=1}}}, {{{b=1}}}, and {{{c=1}}}



{{{D=1-4(1)(1)}}} Square {{{1}}} to get {{{1}}}



{{{D=1-4}}} Multiply {{{4(1)(1)}}} to get {{{(4)(1)=4}}}



{{{D=-3}}} Subtract {{{4}}} from {{{1}}} to get {{{-3}}}



Since the discriminant is less than zero, this means that there are two complex solutions. In other words, there are no real solutions.


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c)


*[Tex \LARGE 3^{\frac{1}{2}}y^2-4y-7\left(3\right)^{\frac{1}{2}}] Start with the given expression



{{{sqrt(3)y^2-4y-7*sqrt(3)}}} Rewrite  *[Tex \LARGE 3^{\frac{1}{2}}] as {{{sqrt(3)}}}



From {{{sqrt(3)y^2-4y-7*sqrt(3)}}}, we can see that {{{a=sqrt(3)}}}, {{{b=-4}}}, and {{{c=7*sqrt(3)}}}



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


<img src="http://i150.photobucket.com/albums/s91/jim_thompson5910/ziggyweek44bsolution1c.jpg" alt="Photobucket - Video and Image Hosting">


{{{D=100}}} Add


Since the discriminant is greater than zero, this means that there are two real solutions.



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d)




From {{{2x^2-10x+25}}} we can see that {{{a=2}}}, {{{b=-10}}}, and {{{c=25}}}



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



{{{D=(-10)^2-4(2)(25)}}} Plug in {{{a=2}}}, {{{b=-10}}}, and {{{c=25}}}



{{{D=100-4(2)(25)}}} Square {{{-10}}} to get {{{100}}}



{{{D=100-200}}} Multiply {{{4(2)(25)}}} to get {{{(8)(25)=200}}}



{{{D=-100}}} Subtract {{{200}}} from {{{100}}} to get {{{-100}}}



Since the discriminant is less than zero, this means that there are two complex solutions. In other words, there are no real solutions.


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e)



From {{{2x^2-6x+5}}} we can see that {{{a=2}}}, {{{b=-6}}}, and {{{c=5}}}



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



{{{D=(-6)^2-4(2)(5)}}} Plug in {{{a=2}}}, {{{b=-6}}}, and {{{c=5}}}



{{{D=36-4(2)(5)}}} Square {{{-6}}} to get {{{36}}}



{{{D=36-40}}} Multiply {{{4(2)(5)}}} to get {{{(8)(5)=40}}}



{{{D=-4}}} Subtract {{{40}}} from {{{36}}} to get {{{-4}}}



Since the discriminant is less than zero, this means that there are two complex solutions. In other words, there are no real solutions.


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f)




From {{{s^2-4s+4}}} we can see that {{{a=1}}}, {{{b=-4}}}, and {{{c=4}}}



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



{{{D=(-4)^2-4(1)(4)}}} Plug in {{{a=1}}}, {{{b=-4}}}, and {{{c=4}}}



{{{D=16-4(1)(4)}}} Square {{{-4}}} to get {{{16}}}



{{{D=16-16}}} Multiply {{{4(1)(4)}}} to get {{{(4)(4)=16}}}



{{{D=0}}} Subtract {{{16}}} from {{{16}}} to get {{{0}}}



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


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g)



From {{{(5/6)x^2-7x-(6/5)}}} we can see that {{{a=5/6}}}, {{{b=-7}}}, and {{{c=-6/5}}}



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



{{{D=(-7)^2-4(5/6)(-6/5)}}} Plug in {{{a=5/6}}}, {{{b=-7}}}, and {{{c=-6/5}}}



{{{D=49-4(5/6)(-6/5)}}} Square {{{-7}}} to get {{{49}}}



{{{D=49--4}}} Multiply {{{4(5/6)(-6/5)}}} to get {{{(10/3)(-6/5)=-4}}}



{{{D=49+4}}} Rewrite {{{D=49--4}}} as {{{D=49+4}}}



{{{D=53}}} Add {{{49}}} to {{{4}}} to get {{{53}}}



Since the discriminant is greater than zero, this means that there are two real solutions.


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h)



From {{{7a^2+8a+2}}} we can see that {{{a=7}}}, {{{b=8}}}, and {{{c=2}}}



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



{{{D=(8)^2-4(7)(2)}}} Plug in {{{a=7}}}, {{{b=8}}}, and {{{c=2}}}



{{{D=64-4(7)(2)}}} Square {{{8}}} to get {{{64}}}



{{{D=64-56}}} Multiply {{{4(7)(2)}}} to get {{{(28)(2)=56}}}



{{{D=8}}} Subtract {{{56}}} from {{{64}}} to get {{{8}}}



Since the discriminant is greater than zero, this means that there are two real solutions.