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05 Nov 2020, 23:52
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rushabh27
There is only one equation -
\(ax(cx+d)=-b(cx+d)\)
Taking the RHS term to LHS, we get
\(ax(cx+d) + b(cx+d) = 0\)
Note that if you open the brackets, you get a regular quadratic equation with x^2, x and constant terms.
But we should not open the brackets here. Because we solve the quadratic by making factors. Here, the factors are right there in front to our eyes - (cx + d) is common to both terms. Take it out.
(cx + d)(ax + b) = 0
So the roots are -d/c and -b/a
10 Mar 2021, 23:37
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Hi experts
I have seen quadratic equations question using the term 'constants'. What impact does the word 'real numbers' have on this problem? *asking out of curiosity.
I have seen quadratic equations question using the term 'constants'. What impact does the word 'real numbers' have on this problem? *asking out of curiosity.
27 Mar 2021, 07:56
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Vegita
Hey Vegita, a real number is a number that can be represented on a number line. It just essentially means that the number is not imaginary (such as the square root of a negative number).
