Quadratic Equation

Definition

The quadratic formula gives us the solutions, or roots, to a quadratic equation of the form ax2+bx+cax^2 + bx + c:

x=b±b24ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Technique

What is the sum of the possible solutions of x22x3=0 x^2 - 2x - 3 = 0 ?

Using the quadratic formula:

x=(2)±(2)24(1)(3)2(1)x=1±2x=3 or x=1 \begin{aligned} x &= \frac{-(-2) \pm \sqrt{(2)^2 - 4(1)(-3)}}{2(1)} \\ x &= 1 \pm 2 \\ x &=3 \text{ or } x=-1 \end{aligned}

The the answer is 31=2 3-1=2 . _\square

The quadratic formula is helpful even when the solutions are complex numbers:

 

The roots of x24x+5 x^2 - 4x + 5 are two complex numbers, z1 z_1 and z2 z_2 . What is the sum of z1 z_1 and z2 z_2 ?

x=(4)±(4)24(1)(5)2(1)x=4±42x=2±i \begin{aligned} x &= \frac{-(-4) \pm \sqrt{(-4)^2-4(1)(5)}}{2(1)} \\ x &= \frac{4 \pm \sqrt{-4}}{2} \\ x &= 2 \pm i \end{aligned}

Thus, the two roots are z1=2i z_1 = 2-i and z2=2+i z_2 = 2 + i and their sum is (2i)+(2+i)=4 (2-i) + (2 + i ) = 4 . _\square

Whether the roots are real or complex depends on the quadratic formula's discriminant, b24ac b^2 - 4ac, the expression inside the square root. The roots are real when the discriminant is positive and complex when the discriminant is negative.

Application and Extensions

For what value of cc will 2x2+7x+c=0 2x^2 + 7x + c = 0 have only a single real root?

The quadratic formula's ±\pm tells us that there will always be two roots unless the discriminant is equal to 0. So,

b24ac=0(7)24(2)c=0c=498 \begin{aligned} b^2 - 4ac &= 0\\ (7)^2 - 4(2)c &= 0\\ c &= \tfrac{49}{8} \end{aligned} _\square

 

xx is a negative number such that x2+9x22=0 x^2+9x -22 = 0 . What is the sum of all possible values of yy which satisfy the equation x=y213y+24 x = y^2 - 13y + 24 ?

Since x2+9x22=0 x^2+9x -22 = 0 , we know that

x=9±814(22)2=9±1692=9±132=2 or 11 \begin{aligned} x &= \frac{-9 \pm \sqrt{81-4(-22)}}{2} \\ &= \frac{-9 \pm \sqrt{169}}{2} \\ &= \frac{-9 \pm 13}{2} \\ &= 2 \text{ or } -11 \end{aligned}

Since x x is negative, x=11 x=-11 . So now we need to solve 11=y213y+24 -11 = y^2 - 13y + 24 .

This produces the following quadratic equation:

y213y+35=0(y5)(y7)=0 \begin{aligned} y^2 - 13y + 35 &=0 \\ (y-5)(y-7)&=0 \end{aligned}

Thus y=5 or 7 y=5 \text{ or } 7, and the answer is 12. _\square

Note by Arron Kau
5 years, 5 months ago

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