Math  /  Algebra

Questionx2675x+8=0x^{2}-675 x+8=0
If AA and β\beta are the roots of the quadratic equation 4x2+x12=04 x^{2}+x-12=0, find the quadratic equation whose root are a3βa^{3} \beta and ap3a p^{3}

Studdy Solution

STEP 1

Assumptions
1. The quadratic equation 4x2+x12=04x^2 + x - 12 = 0 has roots aa and β\beta.
2. We need to find the quadratic equation whose roots are a3βa^3 \beta and aβ3a \beta^3.

STEP 2

First, we use Vieta's formulas to find the sum and product of the roots aa and β\beta of the quadratic equation 4x2+x12=04x^2 + x - 12 = 0.
According to Vieta's formulas: a+β=baandaβ=caa + \beta = -\frac{b}{a} \quad \text{and} \quad a\beta = \frac{c}{a}
For the equation 4x2+x12=04x^2 + x - 12 = 0, we have a=4a = 4, b=1b = 1, and c=12c = -12.

STEP 3

Calculate the sum of the roots aa and β\beta: a+β=14a + \beta = -\frac{1}{4}

STEP 4

Calculate the product of the roots aa and β\beta: aβ=124=3a\beta = \frac{-12}{4} = -3

STEP 5

Now, we need to find the new roots a3βa^3 \beta and aβ3a \beta^3.

STEP 6

Calculate the sum of the new roots a3βa^3 \beta and aβ3a \beta^3: a3β+aβ3a^3 \beta + a \beta^3

STEP 7

Factor out aβa\beta from the expression: a3β+aβ3=aβ(a2+β2)a^3 \beta + a \beta^3 = a \beta (a^2 + \beta^2)

STEP 8

To find a2+β2a^2 + \beta^2, use the identity: a2+β2=(a+β)22aβa^2 + \beta^2 = (a + \beta)^2 - 2a\beta

STEP 9

Substitute the values of a+βa + \beta and aβa\beta: a2+β2=(14)22(3)a^2 + \beta^2 = \left(-\frac{1}{4}\right)^2 - 2(-3)

STEP 10

Simplify the expression: a2+β2=116+6=116+9616=9716a^2 + \beta^2 = \frac{1}{16} + 6 = \frac{1}{16} + \frac{96}{16} = \frac{97}{16}

STEP 11

Substitute back into the expression for the sum of the new roots: a3β+aβ3=aβ(9716)a^3 \beta + a \beta^3 = a \beta \left(\frac{97}{16}\right)

STEP 12

Since aβ=3a\beta = -3, we have: a3β+aβ3=3(9716)=29116a^3 \beta + a \beta^3 = -3 \left(\frac{97}{16}\right) = -\frac{291}{16}

STEP 13

Next, calculate the product of the new roots a3βa^3 \beta and aβ3a \beta^3: (a3β)(aβ3)=a4β4(a^3 \beta)(a \beta^3) = a^4 \beta^4

STEP 14

Since aβ=3a\beta = -3, we have: a4β4=(3)4=81a^4 \beta^4 = (-3)^4 = 81

STEP 15

Now, we have the sum and product of the new roots: Sum of the roots=29116\text{Sum of the roots} = -\frac{291}{16} Product of the roots=81\text{Product of the roots} = 81

STEP 16

Using these values, we can form the new quadratic equation. For a quadratic equation x2+px+q=0x^2 + px + q = 0, the sum of the roots is p-p and the product of the roots is qq.

STEP 17

Thus, the new quadratic equation is: x2(29116)x+81=0x^2 - \left(-\frac{291}{16}\right)x + 81 = 0

STEP 18

Simplify the equation: x2+29116x+81=0x^2 + \frac{291}{16}x + 81 = 0

STEP 19

To clear the fraction, multiply the entire equation by 16: 16x2+291x+1296=016x^2 + 291x + 1296 = 0
The quadratic equation whose roots are a3βa^3 \beta and aβ3a \beta^3 is: 16x2+291x+1296=016x^2 + 291x + 1296 = 0

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