1. **State the problem:** Given roots $a$ and $b$ of an equation with conditions $a^2+b^2=21$ and $a+b=-3$, find the equation whose roots are $\frac{a}{b-2}$ and $\frac{b}{a-2}$. 2. **Recall formulas and rules:** For roots $a$ and $b$ of a quadratic equation, the sum and product are related to coefficients by: $$a+b = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2}, \quad ab = \frac{\text{constant term}}{\text{coefficient of } x^2}$$ Also, note the identity: $$a^2 + b^2 = (a+b)^2 - 2ab$$ 3. **Find $ab$ using given data:** $$21 = a^2 + b^2 = (a+b)^2 - 2ab = (-3)^2 - 2ab = 9 - 2ab$$ Rearranged: $$2ab = 9 - 21 = -12 \implies ab = -6$$ 4. **Define new roots:** Let $$x_1 = \frac{a}{b-2}, \quad x_2 = \frac{b}{a-2}$$ We want the quadratic equation with roots $x_1$ and $x_2$. 5. **Find sum of new roots:** $$x_1 + x_2 = \frac{a}{b-2} + \frac{b}{a-2} = \frac{a(a-2) + b(b-2)}{(b-2)(a-2)}$$ Calculate numerator: $$a(a-2) + b(b-2) = a^2 - 2a + b^2 - 2b = (a^2 + b^2) - 2(a+b) = 21 - 2(-3) = 21 + 6 = 27$$ Calculate denominator: $$(b-2)(a-2) = ab - 2a - 2b + 4 = (-6) - 2a - 2b + 4 = (-6) - 2(a+b) + 4 = (-6) - 2(-3) + 4 = -6 + 6 + 4 = 4$$ So, $$x_1 + x_2 = \frac{27}{4}$$ 6. **Find product of new roots:** $$x_1 x_2 = \frac{a}{b-2} \times \frac{b}{a-2} = \frac{ab}{(b-2)(a-2)} = \frac{-6}{4} = -\frac{3}{2}$$ 7. **Form the quadratic equation:** If roots are $x_1$ and $x_2$, the quadratic equation is: $$x^2 - (x_1 + x_2)x + x_1 x_2 = 0$$ Substitute values: $$x^2 - \frac{27}{4}x - \frac{3}{2} = 0$$ Multiply through by 4 to clear denominators: $$4x^2 - 27x - 6 = 0$$ **Final answer:** $$\boxed{4x^2 - 27x - 6 = 0}$$