Question

What is the solution set of the equation using the quadratic formula?

Answer 1

(1)

we are given

$x^2+6x+10=0$

we can use quadratic formula

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

now, we can compare and find a , b and c

we get

$a=1,b=6,c=10$

now, we can plug these values into quadratic formula

$x=\frac{-6\pm \sqrt{6^2-4(1)(10)}}{2(1)}$

$x=\frac{-6+\sqrt{6^2-4\cdot \:1\cdot \:10}}{2\cdot \:1}$

we can simplify it

$x=-3+i$

$x=\frac{-6-\sqrt{6^2-4\cdot \:1\cdot \:10}}{2\cdot \:1}$

$x=-3-i$

so, we will get solution

{−3+i, −3−i}.........Answer

(2)

we are given equation as

$x^2-8x+14=0$

Since, Jamal solve this equation by completing square

so, firstly we will move constant term on right side

so,  subtract both sides by 14

$x^2-8x+14-14=0-14$

$x^2-8x=-14$

we can write

$-8x=-2\times 4\times x$

so, we will add both sides by 4^2

$x^2-8x+4^2=-14+4^2$

we get

$x^2-8x+16=-14+16$..............Answer