If a trinomial is a perfect square, then the discriminant will equal 0. The discriminant is equal to B^2-4AC. The variables come from the standard form of a quadratic which is Ax^2+Bx+C In this problem, A=81, B=-72, and C=16 so the discriminant is: (-72)^2-4(81)(16)=5,184-5,184=0 so this is a perfect square trinomial. To factor, notice that 81=9^2 and 16=4^2, so 81x^2=(9x)^2. We can then factor the trinomial into (9x+4)(9x-4)
false, in order to be one 8 would have to be a square number which it is not. So the constant term in the trinomial would need to be 4, 9, 16 etc.
I'm going to go out on a limb and assume that y2 8y c actually means y^2 + 8y + c c = 16 makes a perfect square: (y + 4)^2 = (y+4)*(y+4) = y^2 + 8y + 16
16 is a perfect squarebecause 16 = 4 x 4
The largest perfect square factor of 48 is 16.
True
Yes; the factored form would be (9c+4)(9c+4) or just (9c+4)2 Since the two factors are the same, the beginning trinomial 81c2+72c+16 is a perfect square trinomial
yes, (7x - 4)2
If a trinomial is a perfect square, then the discriminant will equal 0. The discriminant is equal to B^2-4AC. The variables come from the standard form of a quadratic which is Ax^2+Bx+C In this problem, A=81, B=-72, and C=16 so the discriminant is: (-72)^2-4(81)(16)=5,184-5,184=0 so this is a perfect square trinomial. To factor, notice that 81=9^2 and 16=4^2, so 81x^2=(9x)^2. We can then factor the trinomial into (9x+4)(9x-4)
false, in order to be one 8 would have to be a square number which it is not. So the constant term in the trinomial would need to be 4, 9, 16 etc.
True
(b/2)^2= 64
No. But it's only (8x) bigger than one perfect square, (24x) bigger than another one, and (48) smaller than another one.
16 is a perfect square (of 4) and the square root of 256.
16 is a perfect square.
It is; (2x+4)^2.
the perfect square of 4 is 16