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Given the quadratic equation 2x² − 4x + 7 = 0.

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Answers

5 parts
i

−67-\frac{6}{7}

ii

47\frac{4}{7}

iii
-3
iv
-13
v

−1249-\frac{12}{49}

Step-by-step solution

15 steps

Here are the evaluations for question 2a:

Given the quadratic equation 2x2−4x+7=02x^2 - 4x + 7 = 0.
For a quadratic equation ax2+bx+c=0ax^2 + bx + c = 0, the sum of the roots is α+β=−ba\alpha + \beta = -\frac{b}{a} and the product of the roots is αβ=ca\alpha\beta = \frac{c}{a}.
In this case, a=2a=2, b=−4b=-4, c=7c=7.

Sum of roots:

α+β=−−42=42=2\alpha + \beta = -\frac{-4}{2} = \frac{4}{2} = 2

Product of roots:

αβ=72\alpha\beta = \frac{7}{2}

Part i· 4 steps

Evaluate αβ+βα\frac{\alpha}{\beta} + \frac{\beta}{\alpha}

  1. Combine the fractions.

    αβ+βα=α2+β2αβ\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta}

  2. Express α2+β2\alpha^2 + \beta^2 in terms of α+β\alpha + \beta and αβ\alpha\beta.

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

  3. Substitute the expressions and values.

    (α+β)2−2αβαβ=(2)2−2(72)72\frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta} = \frac{(2)^2 - 2\left(\frac{7}{2}\right)}{\frac{7}{2}}

  4. Simplify the expression.

    4−772=−372=−3×27=−67\frac{4 - 7}{\frac{7}{2}} = \frac{-3}{\frac{7}{2}} = -3 \times \frac{2}{7} = \boxed{-\frac{6}{7}}

Part ii

47\frac{4}{7}

Evaluate 1α+1β\frac{1}{\alpha} + \frac{1}{\beta}

  1. Combine the fractions.

    1α+1β=β+ααβ\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\beta + \alpha}{\alpha\beta}

  2. Substitute the values.

    272=2×27=47\frac{2}{\frac{7}{2}} = 2 \times \frac{2}{7} = \boxed{\frac{4}{7}}

Part iii-3

Evaluate α2+β2\alpha^2 + \beta^2

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

  2. Substitute the values.

    (2)2−2(72)=4−7=−3(2)^2 - 2\left(\frac{7}{2}\right) = 4 - 7 = \boxed{-3}

Part iv-13

Evaluate α3+β3\alpha^3 + \beta^3

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

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

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

  3. Substitute the values.

    (2)((2)2−3(72))=2(4−212)(2)\left((2)^2 - 3\left(\frac{7}{2}\right)\right) = 2\left(4 - \frac{21}{2}\right)

  4. Simplify the expression.

    2(82−212)=2(−132)=−132\left(\frac{8}{2} - \frac{21}{2}\right) = 2\left(-\frac{13}{2}\right) = \boxed{-13}

Part v

−1249-\frac{12}{49}

Evaluate 1α2+1β2\frac{1}{\alpha^2} + \frac{1}{\beta^2}

  1. Combine the fractions.

    1α2+1β2=β2+α2(αβ)2\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\beta^2 + \alpha^2}{(\alpha\beta)^2}

  2. Substitute the value of α2+β2=−3\alpha^2 + \beta^2 = -3 (from part iii) and αβ=72\alpha\beta = \frac{7}{2}.

    −3(72)2=−3494\frac{-3}{\left(\frac{7}{2}\right)^2} = \frac{-3}{\frac{49}{4}}

  3. Simplify the expression.

    −3×449=−1249-3 \times \frac{4}{49} = \boxed{-\frac{12}{49}}
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