Roots Of The Equation Are: X² - 5x + 6 = 0?

Roots of the equation are: x² - 5x + 6 = 0?

x = 1,6
x = 2,3
x = -2,-3
None of these

Explanation

x^2 - 5x + 6 = 0

Find 2 numbers: add = 5, multiply = 6

Those numbers = 2 and 3

So (x-2)(x-3) = 0

x = 2, 3

Last verified on 10-06-2026

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If √4x-3=4, find the value of x.

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If AX=0, then it has only the trivial solution when?

∣A∣  =0
∣A∣=0
A=0
X  =0

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Explanation

If the determinant $∣ A ∣ \neq = 0$ , then $A$ is non-singular (invertible), so the homogeneous system $AX=0$ has only the trivial solution $X = 0$ .

Last updated 24-07-2026 by NA

1/x3 +1 =(A/x+1 ) + (Bx + C/ x2-x+c) find the value of A is?

1
1/3
-1/3
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Explanation

Given:

1/x ^{3} + 1 = A/ x + 1 + B x + C/ x ^{2} - x + 1

Since:

x^{3} + 1 = (x + 1) (x^{2} - x + 1)

Multiplying both sides by: $(x + 1) (x^{2} - x + 1)$ :

1 = A (x^{2} - x + 1) + (B x + C) (x + 1)

Expanding:

1 = (A + B) x^{2} + (- A + B + C) x + (A + C)

Comparing coefficients:

- A + B + C = 0

A + C = 1

A+B=0

B = - A

From $A + C = 1$ :

C = 1 - A

Substitute into the second equation:

- A - A + (1 - A) = 0

1 - 3 A = 0

A =1/ 3

Last updated on 23-07-2026 by NA

Evaluate (1/3)+(1/4)=?

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Simplify (60)² = ? × 72

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