**2.6 Quadratic Equations, SPM Practice (Paper 2)**

**Question 4:**

Given 3

*t*and (*t*– 7) are the roots of the quadratic equation 4*x*^{2}*– 4**x*+*m*= 0 where*m*is a constant.(a) Find the values of

*t*and*m*.(b) Hence, form the quadratic equation with roots 4

*t*and 2*t*+ 6.

*Solutions:***(a)**

Given 3

*t*and (*t*– 7) are the roots of the quadratic equation 4*x*^{2}*– 4**x*+*m*= 0*a*= 4,

*b*= – 4,

*c*=

*m*

Sum of roots =
$-\frac{b}{a}$

3

*t*+ (*t*– 7) = $-\frac{-4}{4}$3

*t*+*t*– 7 = 14

*t*= 8

*t***= 2**

Product of roots =
$\frac{c}{a}$

3

*t*(*t*– 7) = $\frac{m}{4}$4 [3(2) (2 – 7)] =

*m*← (substitute*t*= 2)4 [3(2) (2 – 7)] =

*m*4 (–30) =

*m*

*m***= –120**

**(b)**

*t***= 2**

4

*t*= 4(2) = 82

*t*+ 6 = 2(2) + 6 = 10Sum of roots = 8 + 10 = 18

Product of roots = 8(10) = 80

Using the formula,

*x*^{2}– (sum of roots)*x*+ product of roots = 0Thus, the quadratic equation is,

*x*

^{2}

**– 18**

*x*+ 80 = 0