Question

What are the roots of the quadratic equation x2 − 15x + 56 = 0?

Answer 1

SOLUTION :

TO DETERMINE

The roots of the quadratic equation

$\sf{} {x}^{2} - 15x + 56 = 0$

EVALUATION

The given Quadratic Equation is

$\sf{} {x}^{2} - 15x + 56 = 0$

$\implies \sf{} {x}^{2} - (7 + 8)x + 56 = 0$

$\implies \sf{} {x}^{2} - 7x - 8x + 56 = 0$

$\implies \sf{} x(x - 7) - 8(x - 7) = 0$

$\implies \sf{} (x - 7) (x - 8) = 0$

We know that if product of two real numbers are zero then either one of them must be zero

So

Either x - 7 = 0 or x - 8 = 0

Now

$\sf{}x - 7 = 0 \: \: gives \: \: x = 7$

$\sf{}x - 8 = 0 \: \: gives \: \: x = 8$

Hence the required solution is x = 7, 8

FINAL ANSWER

The roots of the quadratic equation

$\sf{} {x}^{2} - 15x + 56 = 0$ are 7, 8

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ADDITIONAL INFORMATION

A general equation of quadratic equation is

$a {x}^{2} + bx + c = 0$

Now one of the way to solve this equation is by SRIDHAR ACHARYYA formula

For any quadratic equation

$a {x}^{2} + bx + c = 0$

The roots are given by

$\displaystyle \: x = \frac{ - b \pm \: \sqrt{ {b}^{2} - 4ac } }{2a}$

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