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Solve equation $\sqrt{4t + 1} = 3-3t$

→ I squared both sides and got ► $4t + 1 = 9 - 18t- 3t²$

→ I then moved the 3t² to the left side and combined like pairs and got ► $3t² + 12t - 8 = 0$

I'm stuck at that point. Can someone tell me what I am doing wrong?

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  • $\begingroup$ $4t+18t=22t$, is that troubling you? $\endgroup$
    – Inceptio
    May 17, 2013 at 8:27
  • $\begingroup$ When you square the right, you should get $9-18t+9t^2$ Then when you combine the two sides you should have $22t$, not $12t$ $\endgroup$
    – Ross Millikan
    May 17, 2013 at 18:23

4 Answers 4

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For starters, squaring $\sqrt{4t+1}=3-3t$ gives $4t+1=9-18t+9t^2$. A little rearranging and we have $9t^2-22t+8=0$.

With a bit of insight we can see that this is $(9x-4)(x-2)$.

However, I'm pretty crap at factoring so lets use the quadratic formula:

$$ax^2+bx+c=0$$ implies

$$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$

So we take $$x=\frac{-(-22)\pm\sqrt{22^2-4(9)(8)}}{2(9)}$$ now step-by-step we have

$$x=\frac{22\pm\sqrt{484-288}}{18}=\frac{22\pm\sqrt{196}}{18}=\frac{22\pm14}{18} =\frac{11\pm7}{9}$$ So for "$+$" we have $x=\frac{18}{9}=2$ and for "$-$" we have $x=\frac 4 9$.

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    $\begingroup$ I think its also quickly worth mentioning that this only shows that if $\sqrt{4t+1}=3-3t$, then $9t^2-22t+8=0$ - i.e. that a solution of the quadratic is not necessarily a solution to the original equation, so you still need to check. $\endgroup$
    – Andrew D
    May 17, 2013 at 8:47
  • $\begingroup$ Thanks so much! I usually use this formula: x^2 + px + q thus x = p/2 ± √ (p/2)^2 - q Will have to memorize this formula. I guess it's quicker way to solving problems such as these. $\endgroup$
    – Kalina Kama
    May 17, 2013 at 9:38
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You made a mistake in squaring the right hand side: $$ (3-3t)^2=9-18t+9t^2 $$ However you have also to impose the condition $3-3t\ge0$, because the left hand side is non negative by definition. So you get

$$ 9t^2-22t+8=0,\qquad t\le1 $$

The quadratic equation has roots $2$ and $4/9$, as it's easy to see. Hence the unique solution to your equation is $t=4/9$.

If you "back substitute", you can understand what happens: for $t=2$ the left hand side is $\sqrt{4\cdot2+1}=3$, while the right hand side is $3-3\cdot2=-3$.

For $t=4/9$, the left hand side is $\sqrt{16/9+1}=5/3$, and the right hand side is $3-3\cdot(4/9)=3-4/3=5/3$.

The problem is that upon squaring you also add the solutions to $\sqrt{4t+1}=-(3-3t)$ that is another equation.

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Given: $\sqrt{4t+1}=3-3t$

$$\bigg(\sqrt{4t+1}\bigg)^2=\bigg(3-3t\bigg)^2$$

$$4t+1=\bigg(3-3t\bigg)\bigg(3-3t\bigg)$$

$$4t+1=9-9t-9t+9t^2$$

$$4t+1=9-18t+9t^2$$

$$0=9-1-18t-4t+9t^2$$

$$0=9t^2-22t+8$$

Which is a quadratic in the form:$$0 =ax^2+bx+c$$ Use the quadratic equation: $$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$

$$x = \frac{ -(-22)\pm \sqrt{(-22)^2-4(9)(8)}}{2(9)}$$

$$x = \frac{22\pm\sqrt{484-288}}{18}$$

$$x = \frac{22\pm\sqrt{196}}{18}$$

$$x = \frac{22\pm{14}}{18}$$

$$x = 2 $$ $$or$$ $$x =\frac{4}{9} $$

Understand that it still be checked that these are indeed solutions to your initial equation.

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  • $\begingroup$ Of course $2$ is not a solution for the original equation, because $\sqrt{9}\ne3-3\cdot2$. $\endgroup$
    – egreg
    May 17, 2013 at 17:22
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Hint:

If $ax^2+bx+c=0$, $\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.

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