Question

fins the zeroes of the quadratic equation : 4√ 5 x^ 2 - 24x - 9√ 5

10th grade Cbse chapter 2 : Polynomials

pls give the step by step explanation so that i can understand​

Answer 1

Answer:

Step-by-step explanation:

Hello,

First of all, we know that the solution of the following equation

   $ax^2+bx+c=0$

are

   $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} \ \text{ when } \Delta=b^2-4ac \geq 0$

I would suggest that you try to apply this formula first and check the solution only after you try.

Let's apply it in this case, we have:

$a=4\sqrt{5} \\ \\ b=-24 \\ \\ c= -9\sqrt{5}$

$\Delta=b^2-4ac=24^2+4*9*5=1296 \\ \\ \sqrt{\Delta}=36 \ \text{ and the solutions are } \\ \\ \\ x_1=\dfrac{24-36}{8\sqrt{5}}=\dfrac{-12}{8\sqrt{5}}=\boxed{\dfrac{-3}{2\sqrt{5}}} \\ \\x_2=\dfrac{24+36}{8\sqrt{5}}=\dfrac{60}{8\sqrt{5}}=\boxed{\dfrac{15}{2\sqrt{5}}}$

Hope this helps.

Do not hesitate if you need further explanation.

Thank you