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A quadratic equation ax²+bx+c=0 has -10 and 7 as solutions. Find the values of b and c if the value of a is 1. (Hint: Use the zero-factor property in reverse.)
Question 10, 1.4.95 >

Sagot :

Semsee45idn

Answer:

b = 3

c = -70

Step-by-step explanation:

[tex]\boxed{\begin{minipage}{8.5 cm}\underline{Zero Product Property}\\\\If $a \cdot b = 0$ then either $a = 0$ or $b = 0$ (or both).\\\end{minipage}}[/tex]

If a quadratic equation [tex]ax^2+bx+c=0[/tex] has -10 and 7 as its solutions, then applying the zero product property in reverse gives its factors:

[tex]x=-10 \implies (x+10)=0[/tex]

[tex]x=7 \implies (x-7)=0[/tex]

Therefore,

[tex]\implies a(x+10)(x-7)=0[/tex]

As a = 1, then:

[tex]\implies (x+10)(x-7)=0[/tex]

Expand the brackets:

[tex]\begin{aligned}\implies (x+10)(x-7)&=0\\x(x-7)+10(x-7)&=0\\x^2-7x+10x-70&=0\\x^2+3x-70&=0\end{aligned}[/tex]

Compare with [tex]ax^2+bx+c=0[/tex] :

  • a = 1
  • b = 3
  • c = -70
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