Find the largest value of $x$ such that $3x^2 + 17x + 15 = 5.$

Find the largest value of $x$ such that $3x^2 + 17x + 15 = 5.$[tex]Find the largest value of $x$ such that $3x^2 + 17x + 15 = 5.$[/tex]

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  1. Hello, please consider the following.

    [tex]\begin{aligned}5(9x^2+9x+10) &= x(9x-40)\\&=9x^2-40x\end{aligned}\\\\ 45x^2+45x+50=9x^2-40x\\\\ (45-9)x^2+(45+40)x+50=0\\\\ 36x^2+85x+50=0[/tex]

    We can estimate the discriminant, and then, the solutions and we take the largest one.

    [tex]\Delta=b^2-4ac=85^2-4*36*50=25=5^2\\\\x_1=\dfrac{-85-5}{2*36}=\dfrac{-18*5}{18*4}=\dfrac{-5}{4}\\\\x_2=\dfrac{-85+5}{2*36}=\dfrac{-80}{72}=\dfrac{-8*10}{8*9}=\boxed{\dfrac{-10}{9}}[/tex]

    Thank you

  2. The largest x that satisfies

    [tex]3x^2+17x+15=5[/tex]

    is the largest solution to

    [tex]3x^2+17x+10=0[/tex]

    We have

    [tex]3x^2+17x+10=(3x+2)(x+5)=0[/tex]

    [tex]\implies x=-\dfrac23\text{ or }x=-5[/tex]

    and so the largest value of x we want is -2/3.

  3. The expression [tex]\frac{x+1}{8x^{2}-65x+8}[/tex] is undefined when its denominator is zero. The denominator factors as
      (x-8)(8x -1)
    so it will be zero for x = 1/8 and x = 8.

    The largest value of x for which the expression is undefined is 8.

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