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(x + 1)(x - 15) gives roots of -1 and 15, and a = 14;

(x + 5)(x - 3) gives roots of -5 and 3, and a = 2.

Q: Find 'a' when xsquared plus ax minus 15 equals 0 has two integer roots?

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Doesn't have integer roots. Quadratic formula gives roots as 3.71 and -5.38.

1 and the positive and negative square roots of 2

It is a quadratic equation in X, with two real roots.

(x - 1)(x - 11) x = 1,11

Quadratic formula gives roots as 1.53 and 10.47 (to nearest hundredth.)

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Well it does not have any roots at the range of real numbers (cause 25-4*46 is less than 0..) and the imaginary roots shall be + and - : -5+sqrt(184)*i -5-sqrt(184)*i ------------------ ------------------- 2 2

Doesn't have integer roots. Quadratic formula gives roots as 3.71 and -5.38.

1 and the positive and negative square roots of 2

There are no integer roots of this equation. Using the quadratic formula gives roots of 1.34 and 3.04 plus or minus loose change in each case.

It is a quadratic equation in X, with two real roots.

It is a quadratic equation with one unknown variable, x which has no real roots.

(x - 1)(x - 11) x = 1,11

a perfect square

perfect squares

Quadratic formula gives roots as 1.53 and 10.47 (to nearest hundredth.)

Use the quadratic formula, with a = 1, b = -3, c = 2.

The roots are -1/2 of [ 1 plus or minus sqrt(5) ] . When rounded: 0.61803 and -1.61803. Their absolute values are the limits of the Fibonacci series, or the so-called 'Golden Ratio'.