Find all real solutions of the quadratic equation

Calculator Use

This online calculator is a quadratic equation solver that will solve a second-order polynomial equation such as ax2 + bx + c = 0 for x, where a ≠ 0, using the quadratic formula.

The calculator solution will show work using the quadratic formula to solve the entered equation for real and complex roots. Calculator determines whether the discriminant \( (b^2 - 4ac) \) is less than, greater than or equal to 0.

 When \( b^2 - 4ac = 0 \) there is one real root.

 When \( b^2 - 4ac > 0 \) there are two real roots.

 When \( b^2 - 4ac < 0 \) there are two complex roots.

Quadratic Formula:

The quadratic formula

\( x = \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{ 2a } \)

is used to solve quadratic equations where a ≠ 0 (polynomials with an order of 2)

\( ax^2 + bx + c = 0 \)

Examples using the quadratic formula

Example 1: Find the Solution for \( x^2 + -8x + 5 = 0 \), where a = 1, b = -8 and c = 5, using the Quadratic Formula.

\( x = \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{ 2a } \)

\( x = \dfrac{ -(-8) \pm \sqrt{(-8)^2 - 4(1)(5)}}{ 2(1) } \)

\( x = \dfrac{ 8 \pm \sqrt{64 - 20}}{ 2 } \)

\( x = \dfrac{ 8 \pm \sqrt{44}}{ 2 } \)

The discriminant \( b^2 - 4ac > 0 \) so, there are two real roots.

Simplify the Radical:

\( x = \dfrac{ 8 \pm 2\sqrt{11}\, }{ 2 } \)

\( x = \dfrac{ 8 }{ 2 } \pm \dfrac{2\sqrt{11}\, }{ 2 } \)

Simplify fractions and/or signs:

\( x = 4 \pm \sqrt{11}\, \)

which becomes

\( x = 7.31662 \)

\( x = 0.683375 \)

Example 2: Find the Solution for \( 5x^2 + 20x + 32 = 0 \), where a = 5, b = 20 and c = 32, using the Quadratic Formula.

\( x = \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{ 2a } \)

\( x = \dfrac{ -20 \pm \sqrt{20^2 - 4(5)(32)}}{ 2(5) } \)

\( x = \dfrac{ -20 \pm \sqrt{400 - 640}}{ 10 } \)

\( x = \dfrac{ -20 \pm \sqrt{-240}}{ 10 } \)

The discriminant \( b^2 - 4ac < 0 \) so, there are two complex roots.

Simplify the Radical:

\( x = \dfrac{ -20 \pm 4\sqrt{15}\, i}{ 10 } \)

\( x = \dfrac{ -20 }{ 10 } \pm \dfrac{4\sqrt{15}\, i}{ 10 } \)

Simplify fractions and/or signs:

\( x = -2 \pm \dfrac{ 2\sqrt{15}\, i}{ 5 } \)

which becomes

\( x = -2 + 1.54919 \, i \)

\( x = -2 - 1.54919 \, i \)

calculator updated to include full solution for real and complex roots

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I am suppose to find all the solutions to this problem, I think some theorem states that there can only be as many solutions to the problem as the highest degree. I know that calculus reinforces this so I know that

$2x^2 + 4x + 1 = 0$

Can have at most two solutions. In calculus this is proven by the derivative being zero at only somewhere. I can't remember and it isn't important yet.

Anyways I have no idea what to do with this problem. I don't think I can factor it conventionally because of the 2 coefficient so what is the method at this point? I tried guessing and it didn't work at all for -2 - 3.

asked Dec 2, 2012 at 23:49

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$$ax^2 + bx + c = 0, \text{ roots?}$$
This is where the quadratic formula comes in handy (you should memorize this!):

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

The discriminant d of the quadratic formula is the term $d= (b^2 - 4ac)$. When $b^2\geq 4ac$, you have two real roots.

At the very least, try to memorize how to compute the discriminant. That will allow you to easily determine whether or not a quadratic equation has real roots.

answered Dec 2, 2012 at 23:56

amWhyamWhy

205k158 gold badges266 silver badges491 bronze badges

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If you don't want to use the formula, you can do the following.(Actually this is how the formula is derived though. I used to use this method when I didn't memorize the formula.)

$2x^2+4x+1=0$

$x^2+2x+1/2=0$

$(x+1)^2-1/2=0$

$(x+1)^2=1/2$

$x+1=+1/\sqrt 2$ or $-1/\sqrt 2$

answered Dec 3, 2012 at 0:08

TenguTengu

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Use the Quadratic Formula. One cannot expect all quadratic polynomials to factor "nicely." The Quadratic Formula needs to become a completely standard tool in your arsenal.

answered Dec 2, 2012 at 23:54

André NicolasAndré Nicolas

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The quadratic formula works for all quadratic equations. For any equation [in the form] $ax^2 + bx + c = 0$, this is true:$$x = {-b \pm \sqrt{b^2 - 4ac} \over 2a}$$

answered Dec 3, 2012 at 14:36

P.K.P.K.

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You should definitely commit the formula to memory. Here's a cool video to the tune of row-row-row-your-boat, but there are plenty more out there. http://www.youtube.com/watch?v=HRcj9slciqM. Once you get on youtube, search around as there are lots of great videos in there that show you how to use the formula.

answered Dec 4, 2012 at 6:54

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What is a real solution of a quadratic?

How do you know how many real solutions?