How to form a quadratic equation when only one root is given?

2x^2 - 3x + 4 = 0 x^2 - 3/2x + 2 = 0 x^2 - 3/2x = - 2 x^2 - 3/2x + 9/16 = - 2 + 9/16 (x - 3/4)^2 = -23/16 x - 3/4 = ±?(-23/16) x = 3/4 ± i?(23)/4 a = 3/4 + i?(23)/4 B = 3/4 - i?(23)/4 a + 2 = 3/4 + i?(23)/4 + 2 = 11/4 + i?(23)/4 answer B + 2 = 3/4 - i?(23)/4 + 2 = 11/4 - i?(23)/4 answer.

When you have a leading coefficient of 1, then the coefficient of x is the negative of the sum of the roots and the constant term is the product of the roots. That is, first do this. Divide everything by 2: x^2 - (3/2)x + 2 = 0 Then the sum of the roots is +3/2 (the negative of the coefficient -3/2).

A + be = 3/2. The product of the roots is 2. Ab = 2.

The new quadratic will have -(a+2) - (b+2) as the coefficient of x, because it's the negative of the sum of the new roots. That's -a - 2 - be - 2 = -a - be - 4 = -(a + b) - 4. You know what a + be is, so you can calculate this.

The new quadratic will have a constant term of (a + 2)(b + 2), the product of the new roots. (a + 2)(b + 2) = ab + 2a + 2b + 4 = ab + 2(a + b) + 4 and you know what ab and (a + b) are, so you can calculate this too. Of course you could also just use the quadratic formula on the original equation and find out exactly what a and be are.

Then form the new equation (x - (a+2)) (x - (b+2)) = 0. That's because (x - r1)(x - r2) is a quadratic with roots r1 and r2.

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