Question

If one of the zeroes of the quadratic equation (k-1) x^2+kx+1=0 is -3 then the value of k is

Answer 1

Answer:

Quadratic equation = (k-1) x² + kx + 1 = 0

If one zero of this equation is -3 means that when x = -3 Value of the equation = 0

Mathematically,

= (k-1) (-3)² + k(-3) + 1 = 0

= 9k - 9 - 3k + 1 = 0

= 6k = 8

∴ k = 8/6 = 4/3

∴ Value of k = $\frac{4}{3}$

Hope this helps you.

Answer 2

Answer:

Step-by-step explanation:

a and b are the roots of x^2+7x+12=0

By middle term factorization,

x^2+3x+4x+12=0

x(x+3)+4(x+3)=0

(x+3)(x+4)=0

x+3=0, x+4=0

x=-3, x=-4

So a=-3 and b=-4

Given that, the roots of an equation is (a+b) ^2 and (a-b) ^2

So, (a+b) ^2=(-3-4)^2=-7^2=49

(a-b)^2=(-3-(-4))^2=(-3+4)^2=1

We know that, quadratic equation is

x^2-(a+b)x+ab=0

x^2-(49+1)x+49(1)=0

x^2-50x+49=0

So, the quadratic equation is x^2-50x+49=0