For t > −1, let αₜ ​and βₜ be the roots of the equation ((t + 2)¹/⁶ − 1)x² +((t + 2)¹/⁶ − 1)x + ((t + 2)¹/²¹ − 1) = 0. If lim t → −1⁺​ αₜ ​= a and lim t → −1​⁺ βₜ ​= b, then 72(a + b)² is equal to:

 ❓ Question:

For $t > -1$, let $\alpha_t$ and $\beta_t$ be the roots of the equation:

$$((t+2{)}^{1\mathrm{/}6}-1)x^2+((t+2{)}^{1\mathrm{/}6}-1)x+((t+2{)}^{1\mathrm{/}21}-1)=0.$$

If

$$\underset{t\to -1^{+}}{\lim }{\alpha }_t=a\text{and}\underset{t\to -1^{+}}{\lim }{\beta }_t=b,$$

then find

$$72(a+b{)}^2\mathrm{.}$$

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🖼️ Question Image

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✍️ Short Solution

1. Write the quadratic in standard form:

$$A(t)x^2+B(t)x+C(t)=0$$

where

$$A(t)=(t+2{)}^{1\mathrm{/}6}-1,B(t)=(t+2{)}^{1\mathrm{/}6}-1,C(t)=(t+2{)}^{1\mathrm{/}21}-1.$$

2. Observe the limits as $t \to -1^+$:

$$A(t)\to (1)-1=0,B(t)\to 0,C(t)\to 2^{1\mathrm{/}21}-1$$

but we must use L’Hospital-type analysis since coefficients vanish.

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Step 1: Factor out $A(t)$ from the quadratic

$$x^2+x+\frac{C(t)}{A(t)}=0\Rightarrow x^2+x+\frac{(t+2{)}^{1\mathrm{/}21}-1}{(t+2{)}^{1\mathrm{/}6}-1}=0.$$

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Step 2: Evaluate the limit

Set $h = t+2$, then as $t\to -1^+$, $h\to 1^+$. Then:

$$\frac{(t+2{)}^{1\mathrm{/}21}-1}{(t+2{)}^{1\mathrm{/}6}-1}=\frac{h^{1\mathrm{/}21}-1}{h^{1\mathrm{/}6}-1}\mathrm{.}$$

Use the approximation for $h \to 1$: $h^k - 1 \sim k(h-1)$

$$\frac{h^{1\mathrm{/}21}-1}{h^{1\mathrm{/}6}-1}\sim \frac{\frac{1}{21}(h-1)}{\frac{1}{6}(h-1)}=\frac{1\mathrm{/}21}{1\mathrm{/}6}=\frac{6}{21}=\frac{2}{7}\mathrm{<br>}$$

So the limiting quadratic is:

$$x^2+x+\frac{2}{7}=0.$$

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Step 3: Solve the quadratic

$$x = \frac{-1 \pm \sqrt{1 - 4\cdot \frac{2}{7}}}{2} = \frac{-1 \pm \sqrt{1 - \frac{8}{7}}}{2} = \frac{-1 \pm \sqrt{-1/7}}{2}.$$

Hence the roots are complex:

$$a=\frac{-1+i\mathrm{/}\sqrt{7}}{2},b=\frac{-1-i\mathrm{/}\sqrt{7}}{2}\mathrm{.}$$

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Step 4: Compute $a+b$

$$a+b=\frac{-1+i\mathrm{/}\sqrt{7}}{2}+\frac{-1-i\mathrm{/}\sqrt{7}}{2}=\frac{-2}{2}=-1.$$

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Step 5: Compute $72(a+b)^2$

$$72(a+b{)}^2=72(-1{)}^2=72.$$

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🖼️ Image Solution

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✅ Conclusion & Video Solution

• By factoring out the vanishing coefficient and taking the limit, we reduced the problem to a simple quadratic.

• Roots were complex but sum was real.

• Final answer:

$$\boxed{72}.$$