Which are the solutions of the quadratic equation x2 7x 4

x^{2}-7x=4

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

x^{2}-7x-4=4-4

Subtract 4 from both sides of the equation.

x^{2}-7x-4=0

Subtracting 4 from itself leaves 0.

x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\left(-4\right)}}{2}

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -7 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

x=\frac{-\left(-7\right)±\sqrt{49-4\left(-4\right)}}{2}

Square -7.

x=\frac{-\left(-7\right)±\sqrt{49+16}}{2}

Multiply -4 times -4.

x=\frac{-\left(-7\right)±\sqrt{65}}{2}

Add 49 to 16.

x=\frac{7±\sqrt{65}}{2}

The opposite of -7 is 7.

x=\frac{\sqrt{65}+7}{2}

Now solve the equation x=\frac{7±\sqrt{65}}{2} when ± is plus. Add 7 to \sqrt{65}.

x=\frac{7-\sqrt{65}}{2}

Now solve the equation x=\frac{7±\sqrt{65}}{2} when ± is minus. Subtract \sqrt{65} from 7.

x=\frac{\sqrt{65}+7}{2} x=\frac{7-\sqrt{65}}{2}

The equation is now solved.

x^{2}-7x=4

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

x^{2}-7x+\left(-\frac{7}{2}\right)^{2}=4+\left(-\frac{7}{2}\right)^{2}

Divide -7, the coefficient of the x term, by 2 to get -\frac{7}{2}. Then add the square of -\frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

x^{2}-7x+\frac{49}{4}=4+\frac{49}{4}

Square -\frac{7}{2} by squaring both the numerator and the denominator of the fraction.

x^{2}-7x+\frac{49}{4}=\frac{65}{4}

Add 4 to \frac{49}{4}.

\left(x-\frac{7}{2}\right)^{2}=\frac{65}{4}

Factor x^{2}-7x+\frac{49}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

\sqrt{\left(x-\frac{7}{2}\right)^{2}}=\sqrt{\frac{65}{4}}

Take the square root of both sides of the equation.

x-\frac{7}{2}=\frac{\sqrt{65}}{2} x-\frac{7}{2}=-\frac{\sqrt{65}}{2}

Simplify.

x=\frac{\sqrt{65}+7}{2} x=\frac{7-\sqrt{65}}{2}

Add \frac{7}{2} to both sides of the equation.

x^{2}+7x-4=0

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

x=\frac{-7±\sqrt{7^{2}-4\left(-4\right)}}{2}

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 7 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

x=\frac{-7±\sqrt{49-4\left(-4\right)}}{2}

Square 7.

x=\frac{-7±\sqrt{49+16}}{2}

Multiply -4 times -4.

x=\frac{-7±\sqrt{65}}{2}

Add 49 to 16.

x=\frac{\sqrt{65}-7}{2}

Now solve the equation x=\frac{-7±\sqrt{65}}{2} when ± is plus. Add -7 to \sqrt{65}.

x=\frac{-\sqrt{65}-7}{2}

Now solve the equation x=\frac{-7±\sqrt{65}}{2} when ± is minus. Subtract \sqrt{65} from -7.

x=\frac{\sqrt{65}-7}{2} x=\frac{-\sqrt{65}-7}{2}

The equation is now solved.

x^{2}+7x-4=0

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

x^{2}+7x-4-\left(-4\right)=-\left(-4\right)

Add 4 to both sides of the equation.

x^{2}+7x=-\left(-4\right)

Subtracting -4 from itself leaves 0.

x^{2}+7x=4

Subtract -4 from 0.

x^{2}+7x+\left(\frac{7}{2}\right)^{2}=4+\left(\frac{7}{2}\right)^{2}

Divide 7, the coefficient of the x term, by 2 to get \frac{7}{2}. Then add the square of \frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

x^{2}+7x+\frac{49}{4}=4+\frac{49}{4}

Square \frac{7}{2} by squaring both the numerator and the denominator of the fraction.

x^{2}+7x+\frac{49}{4}=\frac{65}{4}

Add 4 to \frac{49}{4}.

\left(x+\frac{7}{2}\right)^{2}=\frac{65}{4}

Factor x^{2}+7x+\frac{49}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

\sqrt{\left(x+\frac{7}{2}\right)^{2}}=\sqrt{\frac{65}{4}}

Take the square root of both sides of the equation.

x+\frac{7}{2}=\frac{\sqrt{65}}{2} x+\frac{7}{2}=-\frac{\sqrt{65}}{2}

Simplify.

x=\frac{\sqrt{65}-7}{2} x=\frac{-\sqrt{65}-7}{2}

Subtract \frac{7}{2} from both sides of the equation.

x ^ 2 +7x -4 = 0

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.

r + s = -7 rs = -4

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

r = -\frac{7}{2} - u s = -\frac{7}{2} + u

Two numbers r and s sum up to -7 exactly when the average of the two numbers is \frac{1}{2}*-7 = -\frac{7}{2}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width:100%;max-width:700px' /></div>

(-\frac{7}{2} - u) (-\frac{7}{2} + u) = -4

To solve for unknown quantity u, substitute these in the product equation rs = -4

\frac{49}{4} - u^2 = -4

Simplify by expanding (a -b) (a + b) = a^2 – b^2

-u^2 = -4-\frac{49}{4} = -\frac{65}{4}

Simplify the expression by subtracting \frac{49}{4} on both sides

u^2 = \frac{65}{4} u = \pm\sqrt{\frac{65}{4}} = \pm \frac{\sqrt{65}}{2}

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

r =-\frac{7}{2} - \frac{\sqrt{65}}{2} = -7.531 s = -\frac{7}{2} + \frac{\sqrt{65}}{2} = 0.531

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

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