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The equation (k+3)x^2 + 6x + k =5 has two distinct real solutions for x. Prove that k^2-2k-24<0

We argue using the discriminant of a quadratic polynomial. For a quadratic ax^2 + bx + c=0, the discriminant D is D = b^2 - 4ac. When this quadratic has two distinct real roots, we have that D > 0. So ...

Answered by Huw D. Maths tutor
13959 Views

I don't understand why the function "f(x)=x^2 for all real values of x" has no inverse. Isn't sqrt(x) the inverse?

I don't understand why the function "f(x)=x^2 for all real values of x" has no inverse. Isn't sqrt(x) the inverse?
I like to think of a function a bit like a machine. It takes in a number, ...

Answered by Benjamin M. Maths tutor
4035 Views

x^2-9x+20=0

x2-9x+20=0x2-5x-4x+20=0x(x-5)-4(x-5)=0(x-4)(x-5)=0x-4=0x=4x-5=0x=5

Answered by Georgia L. Maths tutor
2937 Views

Solve these simultaneously to find values for a and b: 6a + b = 16 and 5a - 2b = 19

In order to tackle questions like this with two letters of unknown value, first what we try to do is eliminate one of the variables completely from an equation. If we call 6a + b = 16 eqn 1 and 5a - 2b = ...

Answered by Malvika P. Maths tutor
4313 Views

Show that (x + 1)(x + 2)(x + 3) can be written in the form ax3 + bx2 + cx + d where a, b, c and d are positive integers.

(x+1)(x+2) = ( x^2 + 3x + 2) - multiplying out the first 2 terms(x^2 + 3x + 2)(x + 3) = x^3 + 3x^2 + 2x + 3x^2 + 9x + 6 - multiplying the product of the first two terms by the last termx^3 + 6x^2 + 11x + ...

Answered by Rachel K. Maths tutor
6053 Views

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