A triangle has angles of 124°, 28° and 28°.
1 answer:
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- subtract the 12 from both sides so that it becomes the last constant term in the quadratic equation which should now equal 0.
- take the 4x
- half the coefficient of 4 (2)
- square it (4)
- add it to the equation (+4)
- subtract it from the equation (-4)
- factorise the square (x+2)^2 expands to (x^2 + 4x + 2) as {a+b}^2={a^2 + ab + ba + b^2}
- now the equation is in turning point form.
- to find x, add 16 and square root 16 and (x+2)
- subtract 2 from positive or negative 4 (as -4^2 and 4^2 both equal 16).
- This should give you two values for x, -6 and 2.
I really hope that this helped :)