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2 years ago

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Joshua’s computer loan statement said he would pay $1,244.34 in simple interest for a three-year loan at 12.4%. How much did

Joshua borrow to buy the computer?

1 answer:

alexandr402 [8]2 years ago

7 0

$~~~~~~ \textit{Simple Interest Earned Amount} \\\\ A=P(1+rt)\qquad \begin{cases} A=\textit{accumulated amount}\dotfill & \$1244.34\\ P=\textit{original amount deposited}\\ r=rate\to 12.4\%\to \frac{12.4}{100}\dotfill &0.124\\ t=years\dotfill &3 \end{cases} \\\\\\ 1244.34=P[1+(0.124)(3)]\implies 1244.34=P(1.372) \\\\\\ \cfrac{1244.34}{1.372}=P\implies 906.95\approx P$

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Answer:

$(x-43.45)(x+7.45)=0$

Step-by-step explanation:

We have the quadratic equation $d^{2}=36d+324$ i.e. $d^{2}-36d-324=0$

As, the roots of the quadratic equation $ax^{2}+bx+c=0$ are given by $x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$ .

So, from the given equation, we have,

a = 1, b = -36 , c = -324.

Substituting the values in $x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$ , we get,

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i.e. $x=\frac{36\pm \sqrt{1296+1296}}{2}$

i.e. $x=\frac{36\pm \sqrt{2592}}{2}$

i.e. $x=\frac{36\pm 50.9}{2}$

i.e. $x=\frac{36+50.9}{2}$ and $x=\frac{36-50.9}{2}$

i.e. $x=\frac{86.9}{2}$ and $x=\frac{-14.9}{2}$

i.e. x = 43.45 and x = -7.45

Thus, the roots of the equation are 43.45 and -7.45.

Hence, the factored form of the given expression will be $(x-43.45)(x+7.45)=0$

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