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I got the square root of 14 and 11 but the answer book states that these answers are wrong. Can someone help me? I used this formula to find the individual roots

$x = -\frac{p}{2} \pm \sqrt{(\frac{p}{2})^2 - q}$

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  • $\begingroup$ Try this: define a new variable z=x^2+5 and solve z^2-15z+54 using the quadratic formula. Then solve for x in terms of z. $\endgroup$
    – bill s
    May 15, 2013 at 10:24
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    $\begingroup$ OP please mark one of these answers right - it's good SE practice and people will answer you more often! $\endgroup$
    – Soham Chowdhury
    May 17, 2013 at 12:44

3 Answers 3

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Let $a = (x^2 + 5)$. Then $$(a-9)(a-6)=0$$

$$\implies a = 9\; or \;a = 6$$ $$\implies x^2+5 = 9 \; or \;6$$ $$\implies x = \pm\sqrt4\; or \pm\sqrt1$$ $$\implies x = \pm2\; or \pm1$$

This kind of equation is called a biquadratic equation. Cheers!

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Using middle term factor, $$(x^2+5)^2-(9+6)(x^2+5)+6\cdot9=0$$

$$\implies (x^2+5)(x^2+5-9)-6(x^2+5-9)=0$$

$$\implies (x^2-4)(x^2-1)=0$$

$\implies x^2-1=0$ or $x^2-4=0$

Alternatively, using quadratic equation formula for $x^2+5=\frac{15\pm\sqrt{15^2-4\cdot1\cdot 54}}{2\cdot1}=\frac{15\pm 3}2=9$ or $6$

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  • $\begingroup$ I really know this is offtopic, but it's nice to see a Bengali here :) $\endgroup$
    – Soham Chowdhury
    May 15, 2013 at 11:37
  • $\begingroup$ @SohamChowdhury, where are you actually from? $\endgroup$
    – lab bhattacharjee
    May 15, 2013 at 11:40
  • $\begingroup$ I'm currently in 9th grade (in Bosco, if you know it) and I'm sitting in an apartment in Dum Dum :) $\endgroup$
    – Soham Chowdhury
    May 15, 2013 at 11:52
  • $\begingroup$ This answer has one-way implications. Presumably they should be two-way implications, i.e., logical equivalences; otherwise, you could be obtaining extraneous roots or numbers that are not roots at all. $\endgroup$
    – murray
    May 15, 2013 at 14:01
  • $\begingroup$ @murray, not sure how the above method can introduce extraneous roots? $\endgroup$
    – lab bhattacharjee
    May 15, 2013 at 14:38
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Given:

$(x^2 + 5)^2 - 15(x^2 + 5) + 54 = 0$

Let a = $(x^2+5)$ which gives us a simple quadratic trinomial of the form $ax^2+bx+c=0.$

$a^2-15a+54=0$

Factoring this we get:

$(a-9)(a-6)=0$

$a-9=0$

$a-6=0$

$\implies a=9$ or $a=6$

$x^2+5=9$

$x^2+5=6$

$\implies x = \pm2$ or $\pm 1$

You can check that these solutions are correct by simply plugging values back in to see if you get $0$.

$a^2-15a+54=0 \implies$ $(9)^2 - 15(9) + 54 = 0 \implies 81-135+54=0$

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