Answer:
Please check the explanation.
Step-by-step explanation:
Given the quadratic equation
5x²−2x−9 = 0
To factor the quadratic function 5x²−2x−9, we should solve the corresponding quadratic equation 5x²−2x−9 = 0.
Indeed, if x₁ and x₂ are the roots of the quadratic equation ax²+bx+c=0, then
ax²+bx+c = a(x-x₁)(x-x₂)
Now,
solving the quadratic function 5x²−2x−9 = 0
[tex]-5x^2+2x+9=0[/tex]
subtract 9 from both sides
[tex]-5x^2+2x+9-9=0-9[/tex]
Simplify
[tex]-5x^2+2x=-9[/tex]
Divide both sides by -5
[tex]\frac{-5x^2+2x}{-5}=\frac{-9}{-5}[/tex]
[tex]x^2-\frac{2x}{5}=\frac{9}{5}[/tex]
Add (-1/5)² to both sides
[tex]x^2-\frac{2x}{5}+\left(-\frac{1}{5}\right)^2=\frac{9}{5}+\left(-\frac{1}{5}\right)^2[/tex]
[tex]x^2-\frac{2x}{5}+\left(-\frac{1}{5}\right)^2=\frac{46}{25}[/tex]
[tex]\left(x-\frac{1}{5}\right)^2=\frac{46}{25} \\[/tex]
[tex]\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}[/tex]
so solving
[tex]x-\frac{1}{5}=\sqrt{\frac{46}{25}}[/tex]
[tex]x-\frac{1}{5}=\frac{\sqrt{46}}{\sqrt{25}}[/tex]
Add 1/5 to both sides
[tex]x-\frac{1}{5}+\frac{1}{5}=\frac{\sqrt{46}}{5}+\frac{1}{5}[/tex]
[tex]x=\frac{\sqrt{46}+1}{5}[/tex]
similarly solving
[tex]x-\frac{1}{5}=-\sqrt{\frac{46}{25}}[/tex]
[tex]x-\frac{1}{5}=-\frac{\sqrt{46}}{5}[/tex]
Add 1/5 to both sides
[tex]x-\frac{1}{5}+\frac{1}{5}=-\frac{\sqrt{46}}{5}+\frac{1}{5}[/tex]
[tex]x=\frac{-\sqrt{46}+1}{5}[/tex]
Thus, the roots are:
[tex]x_{1} =\frac{\sqrt{46}+1}{5},\:x_{2}=\frac{-\sqrt{46}+1}{5}[/tex]
Conclusion:
Since the roots are irrational, we do not factor further.
Therefore, we leave 5x²−2x−9 as it is.
