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Question: Let $\alpha$ and $\beta$ are two real roots of $x^2 + 10x - 7 = 0$. Then...

Let α\alpha and β\beta are two real roots of x2+10x−7=0x^2 + 10x - 7 = 0. Then

A

α20+β20−7(α18+β18)α19+β19=−10\frac{\alpha^{20} + \beta^{20} - 7(\alpha^{18} + \beta^{18})}{\alpha^{19} + \beta^{19}} = -10

B

αβ18−7αβ16−10α17βα16+β16=70\frac{\alpha\beta^{18} - 7\alpha\beta^{16} - 10\alpha^{17}\beta}{\alpha^{16} + \beta^{16}} = 70

C

(α−7α)(β−7β)=10\sqrt{(\alpha - \frac{7}{\alpha})(\beta - \frac{7}{\beta})} = 10

D

α3+9α2−17α+14β3+11β2+3β−8=7\frac{\alpha^3 + 9\alpha^2 - 17\alpha + 14}{\beta^3 + 11\beta^2 + 3\beta - 8} = 7

Answer

A, B, C

Explanation

Solution

The given equation is x2+10x−7=0x^2 + 10x - 7 = 0. Let α\alpha and β\beta be the real roots of this equation. By Vieta's formulas, we have: α+β=−10\alpha + \beta = -10 αβ=−7\alpha \beta = -7 Since α\alpha and β\beta are roots of the equation, they satisfy the equation: α2+10α−7=0  ⟹  α2=−10α+7\alpha^2 + 10\alpha - 7 = 0 \implies \alpha^2 = -10\alpha + 7 β2+10β−7=0  ⟹  β2=−10β+7\beta^2 + 10\beta - 7 = 0 \implies \beta^2 = -10\beta + 7

Let's evaluate each option:

A) α20+β20−7(α18+β18)α19+β19\frac{\alpha^{20} + \beta^{20} - 7(\alpha^{18} + \beta^{18})}{\alpha^{19} + \beta^{19}} Numerator = α20−7α18+β20−7β18=α18(α2−7)+β18(β2−7)\alpha^{20} - 7\alpha^{18} + \beta^{20} - 7\beta^{18} = \alpha^{18}(\alpha^2 - 7) + \beta^{18}(\beta^2 - 7). From α2+10α−7=0\alpha^2 + 10\alpha - 7 = 0, we have α2−7=−10α\alpha^2 - 7 = -10\alpha. From β2+10β−7=0\beta^2 + 10\beta - 7 = 0, we have β2−7=−10β\beta^2 - 7 = -10\beta. Numerator = α18(−10α)+β18(−10β)=−10α19−10β19=−10(α19+β19)\alpha^{18}(-10\alpha) + \beta^{18}(-10\beta) = -10\alpha^{19} - 10\beta^{19} = -10(\alpha^{19} + \beta^{19}). The expression is −10(α19+β19)α19+β19=−10\frac{-10(\alpha^{19} + \beta^{19})}{\alpha^{19} + \beta^{19}} = -10. The roots are x=−10±100+282=−5±32=−5±42x = \frac{-10 \pm \sqrt{100+28}}{2} = -5 \pm \sqrt{32} = -5 \pm 4\sqrt{2}. Let α=−5+42=42−5\alpha = -5 + 4\sqrt{2} = 4\sqrt{2} - 5 and β=−5−42=−(5+42)\beta = -5 - 4\sqrt{2} = -(5 + 4\sqrt{2}). Since 42=324\sqrt{2} = \sqrt{32} and 5=255 = \sqrt{25}, 42>54\sqrt{2} > 5, so α>0\alpha > 0. Also, 5+42>05 + 4\sqrt{2} > 0, so β<0\beta < 0. α19>0\alpha^{19} > 0 and β19<0\beta^{19} < 0. ∣β∣=5+42|\beta| = 5 + 4\sqrt{2} and ∣α∣=42−5|\alpha| = 4\sqrt{2} - 5. Since 5+42>42−55 + 4\sqrt{2} > 4\sqrt{2} - 5, ∣β∣>∣α∣|\beta| > |\alpha|. So β19\beta^{19} is a negative number with a larger magnitude than the positive number α19\alpha^{19}. Hence, α19+β19≠0\alpha^{19} + \beta^{19} \neq 0. So option A is correct.

B) αβ18−7αβ16−10α17βα16+β16\frac{\alpha\beta^{18} - 7\alpha\beta^{16} - 10\alpha^{17}\beta}{\alpha^{16} + \beta^{16}} Numerator = αβ16(β2−7)−10α17β\alpha\beta^{16}(\beta^2 - 7) - 10\alpha^{17}\beta. Using β2−7=−10β\beta^2 - 7 = -10\beta, the numerator becomes αβ16(−10β)−10α17β=−10αβ17−10α17β\alpha\beta^{16}(-10\beta) - 10\alpha^{17}\beta = -10\alpha\beta^{17} - 10\alpha^{17}\beta. Factor out −10αβ-10\alpha\beta: −10αβ(β16+α16)-10\alpha\beta(\beta^{16} + \alpha^{16}). Using αβ=−7\alpha\beta = -7, the numerator becomes −10(−7)(α16+β16)=70(α16+β16)-10(-7)(\alpha^{16} + \beta^{16}) = 70(\alpha^{16} + \beta^{16}). The expression is 70(α16+β16)α16+β16=70\frac{70(\alpha^{16} + \beta^{16})}{\alpha^{16} + \beta^{16}} = 70. Since α=42−5>0\alpha = 4\sqrt{2} - 5 > 0 and β=−(5+42)<0\beta = -(5 + 4\sqrt{2}) < 0, α16>0\alpha^{16} > 0 and β16=(−(5+42))16=(5+42)16>0\beta^{16} = (-(5 + 4\sqrt{2}))^{16} = (5 + 4\sqrt{2})^{16} > 0. So α16+β16>0\alpha^{16} + \beta^{16} > 0. So option B is correct.

C) (α−7α)(β−7β)\sqrt{(\alpha - \frac{7}{\alpha})(\beta - \frac{7}{\beta})} From α2+10α−7=0\alpha^2 + 10\alpha - 7 = 0, dividing by α\alpha (since α≠0\alpha \neq 0), we get α+10−7α=0\alpha + 10 - \frac{7}{\alpha} = 0, so α−7α=−10\alpha - \frac{7}{\alpha} = -10. From β2+10β−7=0\beta^2 + 10\beta - 7 = 0, dividing by β\beta (since β≠0\beta \neq 0), we get β+10−7β=0\beta + 10 - \frac{7}{\beta} = 0, so β−7β=−10\beta - \frac{7}{\beta} = -10. The expression is (−10)(−10)=100=10\sqrt{(-10)(-10)} = \sqrt{100} = 10. So option C is correct.

D) α3+9α2−17α+14β3+11β2+3β−8\frac{\alpha^3 + 9\alpha^2 - 17\alpha + 14}{\beta^3 + 11\beta^2 + 3\beta - 8} Let P(x)=x3+9x2−17x+14P(x) = x^3 + 9x^2 - 17x + 14. We want to evaluate P(α)P(\alpha). Since α2+10α−7=0\alpha^2 + 10\alpha - 7 = 0, we can divide P(x)P(x) by x2+10x−7x^2 + 10x - 7. Using polynomial division: x3+9x2−17x+14=(x)(x2+10x−7)−x2−10x+14x^3 + 9x^2 - 17x + 14 = (x)(x^2 + 10x - 7) - x^2 - 10x + 14 =x(x2+10x−7)−(x2+10x−7)+7= x(x^2 + 10x - 7) - (x^2 + 10x - 7) + 7 =(x2+10x−7)(x−1)+7= (x^2 + 10x - 7)(x - 1) + 7. So, α3+9α2−17α+14=(α2+10α−7)(α−1)+7=(0)(α−1)+7=7\alpha^3 + 9\alpha^2 - 17\alpha + 14 = (\alpha^2 + 10\alpha - 7)(\alpha - 1) + 7 = (0)(\alpha - 1) + 7 = 7. Numerator = 7.

Let Q(x)=x3+11x2+3x−8Q(x) = x^3 + 11x^2 + 3x - 8. We want to evaluate Q(β)Q(\beta). Since β2+10β−7=0\beta^2 + 10\beta - 7 = 0, we can divide Q(x)Q(x) by x2+10x−7x^2 + 10x - 7. Using polynomial division: x3+11x2+3x−8=(x)(x2+10x−7)+x2+10x−8x^3 + 11x^2 + 3x - 8 = (x)(x^2 + 10x - 7) + x^2 + 10x - 8 =x(x2+10x−7)+(x2+10x−7)−1= x(x^2 + 10x - 7) + (x^2 + 10x - 7) - 1 =(x2+10x−7)(x+1)−1= (x^2 + 10x - 7)(x + 1) - 1. So, β3+11β2+3β−8=(β2+10β−7)(β+1)−1=(0)(β+1)−1=−1\beta^3 + 11\beta^2 + 3\beta - 8 = (\beta^2 + 10\beta - 7)(\beta + 1) - 1 = (0)(\beta + 1) - 1 = -1. Denominator = -1. The expression is 7−1=−7\frac{7}{-1} = -7. So option D is incorrect.

All options A, B, and C are correct.