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Eight-grade mathematics six-way quadratic equation problem. Kneel for answers and processing speed. 、
I'm in eighth grade (I lied to you about being a pig),

But recently I have a headache when I see math problems (students should understand).

Out of humanitarianism, I will help you with two courses (I hope I can get extra points).

1.x 1=x2=3/2

2:x^2+3x-35=5

x^2+3x-40=0

(x-5)(x+8)=0, so x=5(y=-20) or x=-8 (y=-59) is the root of the original equation.

3. Discriminant△ = b 2-4ac = 9

So (2m+ 1) 2-8m = 9.

4m^2-4m-8=0

(m+ 1)(m-2)=0 m=- 1 or m=2.

When m=- 1 and x=- 1 or 1/2.

When m=2, x=2 or 1/2.

4. It is known that the quadratic equation kx2-4kx+k-5=0 has two equal real roots. Find the value of k and the real root of the equation.

Solution: a=k, b=-4k, c=k-5.

So delta = B2-4ac =16k2-4k (k-5) =16k2-4k2+20k =12k2+20k = 4 (3k2+5k) ........... can omit the steps.

Because the original equation has two equal real roots.

So 4 (3k 2+5k) = 0 means 3k 2+5k=0, so k=0 or k=-3/5.

When k=0, the original equation is not a quadratic equation, so it is discarded.

When k=-3/5, the original equation is ............ (all the brain cells are dead, so you just have to find a way to solve the equation yourself).