The solutions for ‘x’ in the given equation are – 3 and - 7
Step-by-step explanation:
Given equation:
[tex]x^{2} + 10 x + 21 = 0[/tex]
To find the ‘x’ value, try to factor, because in this case it works, it's fast. By using factor method, we get
(x + 3) (x + 7) = 0 (adding both value we get 10 and multiply as 21 as in equation and check with signs also while factoring)
x = - 3, -7
Verify above values by multiply both terms,
(x + 3) (x + 7) = 0
[tex]x^{2} + 7 x + 3 x + 21 = 0[/tex]
[tex]x^{2} + 10 x + 21 = 0[/tex] (so values obtained from factor method are correct)
Or, can use quadratic formula, for [tex]a x^{2} + b x + c=0[/tex], the solutions are given by:
[tex]x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}[/tex]
In the given equation, a = 1, b = 10, c = 21, apply these in above formula
[tex]x=\frac{-10 \pm \sqrt{10^{2}-(4 \times 1 \times 21)}}{2(1)}=\frac{-10 \pm \sqrt{100-84}}{2}[/tex]
[tex]x=\frac{-10 \pm \sqrt{16}}{2}=\frac{-10 \pm 4}{2}[/tex]
So,
When [tex]x=\frac{-10+4}{2}=\frac{-6}{2}=-3[/tex]
When [tex]x=\frac{-10-4}{2}=\frac{-14}{2}=-7[/tex]
Hence, the values for ‘x’ are - 3 and - 7
