Question

3. If roots of one quadratic equation are
(4√2 + 3) and (4√2 - 3) then find the
quadratic equation.​

Answer 1

Answer:

x^2+8root2x-23=0

Step-by-step explanation:

consider alpha=4root2+3 and beta=4root2 -3

use the formula x^2+(alpha+beta)x-alpha ×beta =0

Answer 2

Given:-

• Roots of Quadratic equation are   (4√2 + 3) and (4√2 - 3)

To find:-

The required Quadratic equation       

Solution :-

The required Quadratic equation whose roots are $\alpha, \beta$ are

$x^2-(\alpha+\beta)x+\alpha\beta$

Here,

$\alpha = 4\sqrt{2} +3$

$\beta = 4\sqrt{2} -3$

$\maltese \: \alpha+\beta = (4\sqrt{2}+3)+(4\sqrt{2}-3)$

$\alpha +\beta = 4\sqrt{2} +3+4\sqrt{2} -3$

$\alpha +\beta = 4\sqrt{2} +\not3+4\sqrt{2} \not{-3}$

$\alpha +\beta = 4\sqrt{2} +4\sqrt{2}$

$\red{\boxed{\alpha+\beta = 8\sqrt{2}}}$

$\maltese\: \alpha\beta =(4\sqrt{2} +3)(4\sqrt{2}-3)$

$\alpha\beta = (4\sqrt{2})^2 - (3)^2$

$\alpha\beta = 32-9$

$\alpha\beta = 23$

$\red{\boxed{\alpha\beta=23}}$

Substituting the value in  formula

$x^2-(\alpha+\beta)x+\alpha\beta$

$x^2- (8\sqrt{2} )x+23$

$\red{x^2-8\sqrt{2} x+23}$

So, the required Quadratic equation is $\red{\boxed{x^2-8\sqrt{2} x+23}}$

Verification :-

Hence we got the quadratic equation their roots must be(4√2 + 3) and (4√2 - 3)   $x^2-8\sqrt{2} x+23$

Finding the roots by the quadratic formula

$a = 1\\b= -8\sqrt{2} \\c=23$                            

Quadratic equation :-$\dfrac{-b\pm\sqrt{b^2-4ac} }{2a}$

$x = \dfrac{-(-8\sqrt{2})\pm\sqrt{(-8\sqrt{2} )^2-4(1)(23)} }{2(1)}$

$x =\dfrac{8\sqrt{2}\pm\sqrt{128-92} }{2}$

$x =\dfrac{8\sqrt{2} \pm\sqrt{36} }{2}$

$x =\dfrac{8\sqrt{2} \pm6}{2}$

$x =\dfrac{8\sqrt{2}+6 }{2} , \dfrac{8\sqrt{2}-6 }{2}$

$x = \dfrac{8\sqrt{2} }{2} +\dfrac{6}{2} , \dfrac{8\sqrt{2} }{2} -\dfrac{6}{2}$

$x = 4\sqrt{2}+3,4\sqrt{2} -3$

Hence verified !!