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Mathematics Question on Quadratic Equations

Let α,β\alpha, \beta be the roots of the equation
x2+22x−1=0.x^2 + 2\sqrt{2}x - 1 = 0.
The quadratic equation, whose roots are α4+β4\alpha^4 + \beta^4 and 110(α6+β6)\frac{1}{10} \left( \alpha^6 + \beta^6 \right), is:

A

x2−190x+9466=0x^2 - 190x + 9466 = 0

B

x2−195x+9466=0x^2 - 195x + 9466 = 0

C

x2−195x+9506=0x^2 - 195x + 9506 = 0

D

x2−180x+9506=0x^2 - 180x + 9506 = 0

Answer

x2−195x+9506=0x^2 - 195x + 9506 = 0

Explanation

Solution

Step 1: Roots of the quadratic equation Given:

x2+22x−1=0.x^2 + 2\sqrt{2}x - 1 = 0.

The sum of the roots:

α+β=−coefficient of xcoefficient of x2=−22.\alpha + \beta = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2} = -2\sqrt{2}.

The product of the roots:

αβ=constant termcoefficient of x2=−1.\alpha \beta = \frac{\text{constant term}}{\text{coefficient of } x^2} = -1.

Step 2: Compute α4+β4\alpha^4 + \beta^4 Using the identity:

α4+β4=(α2+β2)2−2(αβ)2.\alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha \beta)^2.

First, calculate α2+β2\alpha^2 + \beta^2:

α2+β2=(α+β)2−2αβ.\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta.

Substitute α+β=−22\alpha + \beta = -2\sqrt{2} and αβ=−1\alpha \beta = -1:

α2+β2=(−22)2−2(−1).\alpha^2 + \beta^2 = (-2\sqrt{2})^2 - 2(-1).

α2+β2=8+2=10.\alpha^2 + \beta^2 = 8 + 2 = 10.

Now substitute into α4+β4\alpha^4 + \beta^4:

α4+β4=(10)2−2(−1)2.\alpha^4 + \beta^4 = (10)^2 - 2(-1)^2.

α4+β4=100−2=98.\alpha^4 + \beta^4 = 100 - 2 = 98.

Step 3: Compute α6+β6\alpha^6 + \beta^6 Using the identity:

α6+β6=(α3+β3)2−2(α3β3).\alpha^6 + \beta^6 = (\alpha^3 + \beta^3)^2 - 2(\alpha^3 \beta^3).

First, calculate α3+β3\alpha^3 + \beta^3 using:

α3+β3=(α+β)((α2+β2)−αβ).\alpha^3 + \beta^3 = (\alpha + \beta)((\alpha^2 + \beta^2) - \alpha \beta).

Substitute α+β=−22\alpha + \beta = -2\sqrt{2}, α2+β2=10\alpha^2 + \beta^2 = 10, and αβ=−1\alpha \beta = -1:

α3+β3=(−22)(10−(−1)).\alpha^3 + \beta^3 = (-2\sqrt{2})(10 - (-1)).

α3+β3=(−22)(11)=−222.\alpha^3 + \beta^3 = (-2\sqrt{2})(11) = -22\sqrt{2}.

Now calculate α3β3\alpha^3 \beta^3:

α3β3=(αβ)3=(−1)3=−1.\alpha^3 \beta^3 = (\alpha \beta)^3 = (-1)^3 = -1.

Substitute into α6+β6\alpha^6 + \beta^6:

α6+β6=(−222)2−2(−1).\alpha^6 + \beta^6 = (-22\sqrt{2})^2 - 2(-1).

α6+β6=(484⋅2)+2=968+2=970.\alpha^6 + \beta^6 = (484 \cdot 2) + 2 = 968 + 2 = 970.

Thus:

110(α6+β6)=97010=97.\frac{1}{10}(\alpha^6 + \beta^6) = \frac{970}{10} = 97.

Step 4: Quadratic equation The roots of the quadratic equation are α4+β4=98\alpha^4 + \beta^4 = 98 and:

110(α6+β6)=97.\frac{1}{10}(\alpha^6 + \beta^6) = 97.

The quadratic equation is:

x2−(sum of roots)x+(product of roots)=0.x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0.

Sum of roots:

98+97=195.98 + 97 = 195.

Product of roots:

98⋅97=9506.98 \cdot 97 = 9506.

Thus, the quadratic equation is:

x2−195x+9506=0.x^2 - 195x + 9506 = 0.

Final Answer: Option (3).