Question

find the value of x if p+7=0,q-12=0,and x²+px+q=0​

Answer 1

$\large \green {\underline{answer}} \\ \\ \large \implies{x \: = 3 \: \: and \: \: x = 4} \\ \\ \large \implies \orange {\underline{find}} \\ \\ \large \implies{values \: \: of \: \: x} \\ \\ \large \implies \green{ \underline{step - \: by \: - \: step \: \: explanation}} \\ \\ \large \implies \pink{ \underline{given}} \\ \\ \large \implies{p + 7 = 0} \\ \\ \large \implies{q - 12 = 0} \\ \\ \large \implies{equation \: = {x}^{2} + px + q = 0 } \\ \\ \large \implies{put \: \: the \: \: value \: \: of \: \: p \: \: and \: \: q \: \: in \: \: above \: \: equations} \\ \\ \large \implies{ {x}^{2} - 7x + 12 = 0 } \\ \\ \large \implies{ {x}^{2} - 4x - 3x + 12 = 0 } \\ \\ \large \implies{x(x - 4) - 3(x - 4) = 0} \\ \\ \large \implies{(x - 3)(x - 4) = 0} \\ \\ \large \implies \therefore \green{ \boxed{x = 3 \: \: \: \: and \: \: \: x = 4 }}$

Answer 2

QUESTION:

find the value of x if p+7=0,q-12=0,and x²+px+q=0

GIVEN :

$p + 7 = 0$

$q - 12 = 0$

${x}^{2} + px + q = 0$

So,

$p + 7 = 0$

$</u><u>\</u><u>h</u><u>u</u><u>g</u><u>e</u><u>\</u><u>o</u><u>r</u><u>a</u><u>n</u><u>g</u><u>e</u><u> </u><u>{</u><u>p = - 7</u><u>}</u><u>$

$q - 12 = 0$

$</u><u>\</u><u>h</u><u>u</u><u>g</u><u>e</u><u>\</u><u>b</u><u>l</u><u>u</u><u>e</u><u> </u><u>{</u><u>q = 12</u><u>}</u><u>$

now we put the value of p and q in the x^2 +px +q =0 to find the value of x.

Hence,

${x}^{2} + ( - 7)x + 12 = 0$

$\huge\red { {x}^{2} - 7x + 12 = 0}$

It is a quadratic equation we have to factorised it.

splitting the middle term.

In such a way that it's sum equal to -7 and product equal to 12.

so;

${x}^{2} - 4x - 3x + 12 = 0 \\ x(x - 4) - 3(x - 4) = 0$

$\huge\purple {(x - 3)(x - 4) = 0}$

so,

x - 3 = 0

$\huge\blue {x = 3}$

or

x-4 = 0

$\huge\red {x = 4}$