Solve the equation 21x^2 + 68x - 85 = 0.

Step-by-step solution

1. 1

   Set up the AC test. With $a=21$ and $c=-85$ we would need two integers whose product is

   $ac=21\cdot(-85)=-1785$

   and whose sum is $b=68$.

2. 2

   Show the test genuinely fails. The divisor pairs of $1785=3\cdot 5\cdot 7\cdot 17$ give signed sums $\pm 1784,\pm 592,\pm 352,\pm 248,\pm 104,\pm 88,\pm 64,\pm 16$. The near miss is $85$ and $-21$, which sums to $64$, not $68$ — so there is no integer factorisation, and any 'answer' built from that pair would be wrong.

3. 3

   Compute the discriminant.

   $\Delta=68^2-4(21)(-85)=4624+7140=11764$

   Since $108^2=11664$ and $109^2=11881$, $\Delta$ is not a perfect square, confirming irrational roots.

4. 4

   Apply the formula and simplify. $11764=4\cdot 2941$ and $2941=17\cdot 173$ is squarefree, so $\sqrt{11764}=2\sqrt{2941}$:

   $x=\frac{-68\pm 2\sqrt{2941}}{42}=\frac{-34\pm\sqrt{2941}}{21}$

5. 5

   Check numerically. $\sqrt{2941}\approx 54.231$, so $x\approx 0.9634$ or $x\approx-4.2015$. Then $21(0.9634)^2+68(0.9634)-85\approx 19.49+65.51-85\approx 0$ $\checkmark$.