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if a and b are the roots of the quadratic equation x^2+4x-k=0 and a>b, find the value of a-b

Guest Feb 11, 2015

Best Answer 

 #2
avatar+94202 
+5

if a and b are the roots of the quadratic equation x^2+4x-k=0 and a>b, find the value of a-b

[I have replace a with $$\alpha$$  and b with   $$\beta$$   because I wanted a and be to stand for the quadratic coefficients]

 

For the quadratic  

 

$$\boxed{ax^2+bx+c=0 \;\:If\; the\; roots\; are\; \alpha \;and\; \beta\; \\\;then\;
\alpha+\beta=\frac{-b}{a}\;\;and\;\;\alpha+\beta=\frac{c}{a}}$$

 

 

$$\\\alpha+\beta=-4\qquad \alpha\beta=-k\\\\
(\alpha+\beta)^2-4\alpha\beta=\alpha^2+\beta^2+2\alpha\beta-4\alpha\beta=
\alpha^2+\beta^2-2\alpha\beta=(\alpha-\beta)^2\\\\
so\\\\
(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta\\\\
(\alpha-\beta)^2=(-4)^2-4*-k\\\\
(\alpha-\beta)^2=16+4k\\\\
\alpha-\beta=\pm\sqrt{16+4k}\\\\
But\; \alpha\; is \;bigger\; than\; \beta\; so\\\\
\alpha-\beta=\sqrt{16+4k}\\\\
\alpha-\beta=2\sqrt{4+k}\\\\$$

Melody  Feb 11, 2015
 #1
avatar+20711 
+5

If a and b are the roots of the quadratic equation x^2+4x-k=0 and a>b, find the value of a-b

$$\\\small{\text{
$
x_{1,2}
=\dfrac{-4\pm\sqrt{16-4*(-k)}}{2}
=\dfrac{-4\pm\sqrt{16+4k}}{2}
=\dfrac{-4\pm\sqrt{4(4+k)}}{2}
$
}}\\\\
\small{\text{
$
x_{1,2}=\dfrac{-4\pm 2\sqrt{4+k}}{2}
=-2\pm \sqrt{4+k}
$
}}\\\\
\small{\text{
$
a=-2+ \sqrt{4+k} \quad | \quad a > b
$
}}\\
\small{\text{
$
b=-2- \sqrt{4+k}
$
}}\\
\small{\text{
$
a-b=-2+ \sqrt{4+k} -( -2- \sqrt{4+k} )=-2+ \sqrt{4+k} + 2 + \sqrt{4+k} )=2\sqrt{4+k}
$
}}\\
\small{\text{
$
a-b =2\sqrt{4+k}
$
}}$$

heureka  Feb 11, 2015
 #2
avatar+94202 
+5
Best Answer

if a and b are the roots of the quadratic equation x^2+4x-k=0 and a>b, find the value of a-b

[I have replace a with $$\alpha$$  and b with   $$\beta$$   because I wanted a and be to stand for the quadratic coefficients]

 

For the quadratic  

 

$$\boxed{ax^2+bx+c=0 \;\:If\; the\; roots\; are\; \alpha \;and\; \beta\; \\\;then\;
\alpha+\beta=\frac{-b}{a}\;\;and\;\;\alpha+\beta=\frac{c}{a}}$$

 

 

$$\\\alpha+\beta=-4\qquad \alpha\beta=-k\\\\
(\alpha+\beta)^2-4\alpha\beta=\alpha^2+\beta^2+2\alpha\beta-4\alpha\beta=
\alpha^2+\beta^2-2\alpha\beta=(\alpha-\beta)^2\\\\
so\\\\
(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta\\\\
(\alpha-\beta)^2=(-4)^2-4*-k\\\\
(\alpha-\beta)^2=16+4k\\\\
\alpha-\beta=\pm\sqrt{16+4k}\\\\
But\; \alpha\; is \;bigger\; than\; \beta\; so\\\\
\alpha-\beta=\sqrt{16+4k}\\\\
\alpha-\beta=2\sqrt{4+k}\\\\$$

Melody  Feb 11, 2015

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