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What are the roots of x^2 - 4x - 7 = 0

1 Answer

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For the given equation, the solutions are
x=2 \pm√(11) .

Explanation:

Step 1:

For an equation of the form
ax^(2) +bx+c=0 the solution is
x=\frac{-b \pm \sqrt{b^(2)-4 a c}}{2 a}.

Here a is the coefficient of
x^(2), b is the coefficient of x and c is the constant term.

Comparing
x^(2) -4x-7=0 with
ax^(2) +bx+c=0, we get that a is 1, b is -4 and c is -7.

To get the solution, we substitute the values of a, b, and c in
x=\frac{-b \pm \sqrt{b^(2)-4 a c}}{2 a}.

Step 2:

Substituting the values, we get


x=\frac{-b \pm \sqrt{b^(2)-4 a c}}{2 a}=\frac{-(-4) \pm \sqrt{(-4)^(2)-4(1)(-7)}}{2(1)}.


\frac{-(-4) \pm \sqrt{(-4)^(2)-4(1)(-7)}}{2(1)}= (4 \pm √(16+28))/(2).


(4 \pm √(16+28))/(2) = (4 \pm √(44))/(2).


(4 \pm √(44))/(2) = (4 \pm 2√(11))/(2) = 2 \pm√(11) .

So
x=2 \pm√(11) .

User Martin Lehmann
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