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d to the 4th+32=12d squared? help please
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so d to the 4th mean d with exponent 4 ?
yes!
so than just note d^2 = t and will get an quadratic equation in t hope you can solve it
but dont forget it that t_1 and t_2 you need make equal d^2
t_1 and t_2 mean roots of quadratic equation in t ok ?
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\(\bf d^4+32=12d^2\implies d^4-12d^2+32=0\implies (d^{\color{red}{ 2}})^2-12(d^{\color{red}{2}})+32=0\) notice, tis just a quadratic
shoot, got truncated... anyhow \(\bf d^4+32=12d^2\implies d^4-12d^2+32=0\\ \quad \\\implies (d^{\color{red}{ 2}})^2-12(d^{\color{red}{2}})+32=0\)
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