Explanation

Since y ∝ x, we have y = kx for some constant k.

When x = 4, y = 16:  

16 = k × 4 → k = 4  

Now for x = 6:  

y = 4 × 6 = 24  

Last verified 13-05-2026

Explanation

4x - 3x = 7 + 5  

x = 12

Last verified on 13-05-2026

Explanation

Roots of x² - 5x + 4 = 0:  

x² - 5x + 4 = (x - 1)(x - 4) → roots are 1 and 4  

Increase each by 1:  

New roots = 1 + 1 = 2, and 4 + 1 = 5  

New equation with roots 2 and 5:  

(x - 2)(x - 5) = x² - 7x + 10

Last verified on 13-05-2026

Explanation

Let the roots be r and r + 2, since they differ by 2.

For x² - px + 35 = 0:

Product of roots = r(r + 2) = 35  

Sum of roots = r + (r + 2) = 2r + 2 = p  

Solve r(r + 2) = 35:  

r² + 2r - 35 = 0  

(r + 7)(r - 5) = 0  

So r = 5 or r = -7  

Case 1: r = 5 → roots are 5 and 7 → p = 5 + 7 = 12  

Case 2: r = -7 → roots are -7 and -5 → p = -7 + (-5) = -12  

So p = 12 or -12

Last verified on 13-05-2026

Explanation

(a + bi)(a - bi) is a standard identity:  

(a + bi)(a - bi) = a² - (bi)² = a² - b²i² = a² - b²(-1) = a² + b²

So a² + b² = 37 directly.

Last verified on 13-05-2026

Explanation

(2x + y) + (x − y) = 11 + 1

3x = 12

x = 4

Last verified on 13-05-2026

Explanation

3x − 4 = 17

3x = 17 + 4 = 21

x = 21 ÷ 3 = 7

Last verified on 13-05-2026

Explanation

x² - 9x + 20 = 0  

Factor: (x - 4)(x - 5) = 0  

So x = 4 or x = 5

Larger value = 5

Last verified on 12-05-2026

Explanation

5x − 15 = 4x + 14

5x − 4x = 14 + 15

x = 29

Last verified on 11-05-2026

Explanation

Using formula:

$b^2-4ac=0$

Here:

• $a = 2$
• $b=-k$
• $c = 5$

So:

(−k)^2−4(2)(5)=0
k^2−40=0
k^2=40

±2√(10)

$$Last\; updated\; question\; NS\; on\; 10\; MAY\; 2026$$