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I am unsure how to tackle making $t$ the subject of this formula

$D = ut + kt^2$

Because it appears twice and is squared, does this mean I have to factorise quadratically? I have tried but how is this possible when you only have algebra coefficients? Also surely if it was factorised into 2 brackets then you still wouldn't be able to single out the subject onto one side, so I guess you would have to factorise it into one bracket?

Please can someone make sense of this for me, explaining simply? Thankyou

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  • $\begingroup$ Sorry; what is it you're trying to do? $\endgroup$
    – pancini
    Jun 3, 2019 at 19:50
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    $\begingroup$ Complete the squares. $kt^2+ut=k(t^2+\frac ukt)=k\left(t^2+\frac uk t+\frac {u^2}{4k^2}-\frac {u^2}{4k^2}\right)=k(t+\frac {u}{2k})^2-k\times \frac {u^2}{4k^2}$ $\endgroup$
    – lulu
    Jun 3, 2019 at 19:52
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    $\begingroup$ You could also use the quadratic formula after subtracting D on both sides. $\endgroup$
    – randomgirl
    Jun 3, 2019 at 19:53
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    $\begingroup$ Yes. As has been remarked, this is just the same as using the quadratic formula. In fact, this is how the quadratic formula is demonstrated. $\endgroup$
    – lulu
    Jun 3, 2019 at 20:02
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    $\begingroup$ Well, at a minimum your expression should allow both signs for the square root. But let's check your expression using the quadratic formula. Rearrange the original to get $kt^2+ut-D=0$. The quadratic formula then tells us that $t=\frac {-u\pm \sqrt {u^2+4kD}}{2k}$. Can you check that this is the same as your expression (other than the signs of the square root)? $\endgroup$
    – lulu
    Jun 3, 2019 at 20:16

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