+0  
 
0
350
2
avatar

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+91517 
+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
Sort: 

2+0 Answers

 #1
avatar+18845 
+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+91517 
+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

7 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details