How to convert a quadratic solution to an unusual format "Solve the quadratic equation 3x2+4x−5 giving your answer in the form ab±19, where a and b are integers."

In general, the quadratic polynomial ax2+bx+c has roots d−b+b2−4ac2a and d−b−b2−4ac2a. If you multiply the first root by d−b−b2−4ac−b−b2−4ac=1, you get db2−(b2−4ac)−2ab−2ab2−4ac=d−2cb+b2−4ac. Doing the same with the other root, you for the desired expressions.Note: I'm assuming you can actually do the division because the roots are non zero, which is what happens in your problem.

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High School
Algebra
Algebra I
Quadratic function and equation

Adrien Barron

Answered question

2022-01-31

How to convert a quadratic solution to an unusual format
"Solve the quadratic equation

3 x^{2} + 4 x - 5

giving your answer in the form

\frac{a}{b \pm \sqrt{19}}

, where a and b are integers."

Answer & Explanation

Larissa Hogan

Beginner2022-02-01Added 10 answers

In general, the quadratic polynomial

a x^{2} + b x + c

has roots

d \frac{- b + \sqrt{b^{2} - 4 a c}}{2 a} and d \frac{- b - \sqrt{b^{2} - 4 a c}}{2 a}

. If you multiply the first root by

d \frac{- b - \sqrt{b^{2} - 4 a c}}{- b - \sqrt{b^{2} - 4 a c}} = 1

, you get

d \frac{b^{2} - (b^{2} - 4 a c)}{- 2 a b - 2 a \sqrt{b^{2} - 4 a c}} = d \frac{- 2 c}{b + \sqrt{b^{2} - 4 a c}}

. Doing the same with the other root, you for the desired expressions.
Note: I'm assuming you can actually do the division because the roots are non zero, which is what happens in your problem.

Brynn Ortiz

Beginner2022-02-02Added 12 answers

Another approach would be to apply Vietes

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