+0

# 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

0
607
2

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

#2
+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]

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

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

Melody  Feb 11, 2015
#1
+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
+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]

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

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

Melody  Feb 11, 2015