If α and β are the roots of x2−4x−3, then find1(α−4)2+1(β−4)2

You did not factorise x2−4x−3 correctly. Actually, x2−4x−3=(x−2)2−4−3=(x−2)2−7. We find that (x−2)2=7, and so x−2=±7. Therefore, x−4=−2±7, and (x−4)2=11±47. Hence1(α−4)2+1(β−4)2=111+47+111−47=11−47+11+47121−112=229,.An alternative method (which would be slower in this case), would be to make the substitution w=x−4. The resulting quadratic in w will have roots α−4, and β−4.

Expert answers to "If α and β are the roots of..."

High School
Algebra
Algebra I
Quadratic function and equation

Kale Mcclain

Answered question

2022-04-20

α

and

β

are the roots of

x^{2} - 4 x - 3

, then find

\frac{1}{{(α - 4)}^{2}} + \frac{1}{{(β - 4)}^{2}}

Answer & Explanation

diehaterzh8s

Beginner2022-04-21Added 11 answers

You did not factorise

x^{2} - 4 x - 3

correctly. Actually,

x^{2} - 4 x - 3 = {(x - 2)}^{2} - 4 - 3 = {(x - 2)}^{2} - 7

. We find that

{(x - 2)}^{2} = 7

, and so

x - 2 = \pm \sqrt{7}

. Therefore,

x - 4 = - 2 \pm \sqrt{7}

, and

{(x - 4)}^{2} = 11 \pm 4 \sqrt{7}

. Hence

\frac{1}{{(α - 4)}^{2}} + \frac{1}{{(β - 4)}^{2}} = \frac{1}{11 + 4 \sqrt{7}} + \frac{1}{11 - 4 \sqrt{7}} = \frac{11 - 4 \sqrt{7} + 11 + 4 \sqrt{7}}{121 - 112} = \frac{22}{9}, .

An alternative method (which would be slower in this case), would be to make the substitution

w = x - 4

. The resulting quadratic in w will have roots

α - 4

, and

β - 4

Olive Guzman

Beginner2022-04-22Added 16 answers

x^{2} - 4 x - 3 = {(x - 2)}^{2} - 7

This is the mistake in your analysis.

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